Chapter 11 Simple Harmonic Motion

Transcription

1 Chapter 11 Siple Haronic Motion "We are to adit no ore causes of natural things than such as are both true and sufficient to explain their appearances." Isaac Newton 11.1 Introduction to Periodic Motion Periodic otion is any otion that repeats itself in equal intervals of tie. The uniforly rotating earth represents a periodic otion that repeats itself every 4 hours. The otion of the earth around the sun is periodic, repeating itself every 1 onths. A vibrating spring and a pendulu also exhibit periodic otion. The period of the otion is defined as the tie for the otion to repeat itself. A special type of periodic otion is siple haronic otion and we now proceed to investigate it. 11. Siple Haronic Motion An exaple of siple haronic otion is the vibration of a ass, attached to a spring of negligible ass, as the ass slides on a frictionless surface, as shown in figure We say that the ass, in the unstretched position, figure 11.1(a), is in its equilibriu position. If an applied force FA acts on the ass, the ass will be displaced to the right of its equilibriu position a distance x, figure 11.1(b). The distance that the spring stretches, obtained fro Hooke s law, is FA = kx The displaceent x is defined as the distance the body oves fro its equilibriu position. Because FA is a force that pulls the ass to the right, it is also the force that pulls the spring to the right. By Newton s third law there is an equal but opposite elastic force exerted by the spring on the ass pulling the ass toward the left. Since this force tends to restore the ass to its original position, we call it the restoring force FR. Because the restoring force is opposite to the applied force, it is given by FR = FA = kx (11.1) When the applied force FA is reoved, the elastic restoring force FR is then the only force acting on the ass, figure 11.1(c), and it tries to restore to its equilibriu position. We can then find the acceleration of the ass fro Newton s second law as Thus, a = FR = kx a = k x (11.) Equation 11. is the defining equation for siple haronic otion. Siple haronic otion is otion in which the acceleration of a body is directly proportional to its displaceent fro the equilibriu position but in the opposite direction. A vibrating syste that executes siple haronic otion is soeties called a haronic oscillator. Because the acceleration is directly proportional to the displaceent x Figure 11.1 The vibrating spring. in siple haronic otion, the acceleration of the syste is not constant but varies with x. At large displaceents, the acceleration is large, at sall displaceents the acceleration is sall. Describing the vibratory otion of the ass requires soe new techniques because the kineatic equations derived in chapter 3 were based on the assuption that the acceleration of the syste was a constant. As we can see fro equation 11., this assuption is no longer valid. We need to derive a new set of kineatic equations to describe siple haronic otion, and we will do so in section However, let us first look at the otion fro a physical point of view. The ass in Chapter 11 Siple Haronic Motion 11-1

2 figure 11.(a) is pulled a distance x = A to the right, and is then released. The axiu restoring force on acts to the left at this position because FRax = kxax = ka The axiu displaceent A is called the aplitude of the otion. At this position the ass experiences its axiu acceleration to the left. Fro equation 11. we obtain a = k A The ass continues to ove toward the left while the acceleration continuously decreases. At the equilibriu position, figure 11.(b), x = 0 and hence, fro equation 11., the acceleration is also zero. However, because the ass has inertia it oves past the equilibriu position to negative values of x, thereby copressing the spring. The restoring force FR now points to the right, since for negative values of x, Figure 11. Detailed otion of the vibrating spring. FR = k( x) = kx The force acting toward the right causes the ass to slow down, eventually coing to rest at x = A. At this point, figure 11.(c), there is a axiu restoring force pointing toward the right and hence a axiu acceleration FRax = k( A)ax = ka aax = k ( A) = k A also to the right. The ass oves to the right while the force FR and the acceleration a decreases with x until x is again equal to zero, figure 11.(d). Then FR and a are also zero. Because of the inertia of the ass, it oves past the equilibriu position to positive values of x. The restoring force again acts toward the left, slowing down the ass. When the displaceent x is equal to A, figure 11.(e), the ass oentarily coes to rest and then the otion repeats itself. One coplete otion fro x = A and back to x = A is called a cycle or an oscillation. The period T is the tie for one coplete oscillation, and the frequency f is the nuber of coplete oscillations or cycles ade in unit tie. The period and the frequency are reciprocal to each other, that is, f = 1 (11.3) T 11- Vibratory Motion, Wave Motion and Fluids

3 The unit for a period is the second, while the unit for frequency, called a hertz, is one cycle per second. The hertz is abbreviated, Hz. Also note that a cycle is a nuber not a diensional quantity and can be dropped fro the coputations whenever doing so is useful Analysis of Siple Haronic Motion -- The Reference Circle As pointed out in section 11., the acceleration of the ass in the vibrating spring syste is not a constant, but rather varies with the displaceent x. Hence, the kineatic equations of chapter 3 can not be used to describe the otion. (We derived those equations on the assuption that the acceleration was constant.) Thus, a new set of equations ust be derived to describe siple haronic otion. Siple haronic otion is related to the unifor circular otion studied in chapter 6. An analysis of unifor circular otion gives us a set of equations to describe siple haronic otion. As an exaple, consider a point Q oving in unifor circular otion with an angular velocity ω, as shown in figure 11.3(a). At a particular instant of tie t, the angle θ that Q has turned through is The projection of point Q onto the x-axis gives the point P. As Q rotates in the circle, P oscillates back and forth along the x-axis, figure 11.3(b). That is, when Q is at position 1, P is at 1. As Q oves to position on the circle, P oves to the left along the x-axis to position.as Q oves to position 3, P oves on the x-axis to position 3, which is of course the value of x = 0. As Q oves to position 4 on the circle, P oves along the negative x-axis to position 4.When Q arrives at position 5, P is also there. As Q oves to position 6 on the circle, P oves to position 6 on the x- axis. Then finally, as Q oves through positions 7, 8, and 1, P oves through 7, 8, and 1, respectively. The oscillatory otion of point P on the x-axis corresponds to the siple haronic otion of a body oving under the influence of an elastic restoring force, as shown in figure 11.. The position of P on the x-axis and hence the position of the ass is described in ters of the point Q and the angle θ found in figure 11.3(a) as x = A cos θ (11.5) Here A is the aplitude of the vibratory otion and using the value of θ fro equation 11.4 we have θ = ωt (11.4) Figure 11.3 Siple haronic otion and the reference circle. x = A cos ωt (11.6) Equation 11.6 is the first kineatic equation for siple haronic otion; it gives the displaceent of the vibrating body at any instant of tie t. The angular velocity ω of point Q in the reference circle is related to the frequency of the siple haronic otion. Because the angular velocity was defined as ω = θ (11.7) t then, for a coplete rotation of point Q, θ rotates through an angle of π rad. But this occurs in exactly the tie for P to execute one coplete vibration. We call this tie for one coplete vibration the period T. Hence, we can Chapter 11 Siple Haronic Motion 11-3

4 also write the angular velocity, equation 11.7, as ω = θ = π (11.8) t T Since the frequency f is the reciprocal of the period T (equation 11.3) we can write equation 11.8 as ω = πf (11.9) Thus, the angular velocity of the unifor circular otion in the reference circle is related to the frequency of the vibrating syste. Because of this relation between the angular velocity and the frequency of the syste, we usually call the angular velocity ω the angular frequency of the vibrating syste. We can substitute equation 11.9 into equation 11.6 to give another for for the first kineatic equation of siple haronic otion, naely x = A cos(πft) (11.10) We can find the velocity of the ass attached to the end of the spring in figure 11. with the help of the reference circle in figure 11.3(c). The point Q oves with the tangential velocity VT. The x-coponent of this velocity is the velocity of the point P and hence the velocity of the ass. Fro figure 11.3(c) we can see that v = VT sin θ (11.11) The inus sign indicates that the velocity of P is toward the left at this position. The linear velocity VT of the point Q is related to the angular velocity ω by equation 9. of chapter 9, that is v = rω For the reference circle, v = VT and r is the aplitude A. Hence, the tangential velocity VT is given by Using equations 11.11, 11.1, and 11.4, the velocity of point P becoes VT = ωa (11.1) v = ωa sin ωt (11.13) Equation is the second of the kineatic equations for siple haronic otion and it gives the speed of the vibrating ass at any tie t. A third kineatic equation for siple haronic otion giving the speed of the vibrating body as a function of displaceent can be found fro equation by using the trigonoetric identity or Fro figure 11.3(a) or equation 11.5, we have Hence, sin θ + cos θ = 1 sin θ =± 1 cos θ cos θ = x A x sin θ=± 1 (11.14) A Substituting equation back into equation 11.13, we get or x v =± ω A 1 A v A x =± ω (11.15) 11-4 Vibratory Motion, Wave Motion and Fluids

5 Equation is the third of the kineatic equations for siple haronic otion and it gives the velocity of the oving body at any displaceent x. The ± sign in equation indicates the direction of the vibrating body. If the body is oving to the right, then the positive sign (+) is used. If the body is oving to the left, then the negative sign ( ) is used. Finally, we can find the acceleration of the vibrating body using the reference circle in figure 11.3(d). The point Q in unifor circular otion experiences a centripetal acceleration ac pointing toward the center of the circle in figure 11.3(d). The x-coponent of the centripetal acceleration is the acceleration of the vibrating body at the point P. That is, a = ac cos θ (11.16) The inus sign again indicates that the acceleration is toward the left. But recall fro chapter 6 that the agnitude of the centripetal acceleration is ac = v (6.1) r where v represents the tangential speed of the rotating object, which in the present case is VT, and r is the radius of the circle, which in the present case is the radius of the reference circle A. Thus, ac = VT A But we saw in equation 11.1 that VT = ωa, therefore ac = ω A The acceleration of the ass, equation 11.16, thus becoes a = ω A cos ωt (11.17) Equation is the fourth of the kineatic equations for siple haronic otion. It gives the acceleration of the vibrating body at any tie t. This equation has no counterpart in chapter 3, because there the acceleration was always a constant. Also, since F = a by Newton s second law, the force acting on the ass, becoes Thus, the force acting on the ass is a variable force. Equations 11.6 and can be cobined into the siple equation F = ω A cos ωt (11.18) If equation is copared with equation 11., a = ω x (11.19) a = k x we see that the acceleration of the ass at P, equation 11.19, is directly proportional to the displaceent x and in the opposite direction. But this is the definition of siple haronic otion as stated in equation 11.. Hence, the projection of a point at Q, in unifor circular otion, onto the x-axis does indeed represent siple haronic otion. Thus, the kineatic equations developed to describe the otion of the point P, also describe the otion of a ass attached to a vibrating spring. An iportant relation between the characteristics of the spring and the vibratory otion can be easily deduced fro equations 11. and That is, because both equations represent the acceleration of the vibrating body they can be equated to each other, giving ω = k or k ω= (11.0) Chapter 11 Siple Haronic Motion 11-5

6 The value of ω in the kineatic equations is expressed in ters of the force constant k of the spring and the ass attached to the spring. The physics of siple haronic otion is thus connected to the angular frequency ω of the vibration. In suary, the kineatic equations for siple haronic otion are where, fro equations 11.9 and 11.0, we have x = A cos ωt (11.6) v = ωa sin ωt (11.13) v =± ω A x (11.15) a = ω A cos ωt (11.17) F = ω A cos ωt (11.18) ω = πf = k A plot of the displaceent, velocity, and acceleration of the vibrating body as a function of tie are shown in figure We can see that the atheatical description follows the physical description in figure 11.. When x = A, the axiu displaceent, the velocity v is zero, while the acceleration is at its axiu value of ω A. The inus sign indicates that the acceleration is toward the left. The force is at its axiu value of ω A, where the inus sign shows that the restoring force is pulling the ass back toward its equilibriu position. At the equilibriu position x = 0, a = 0, and F = 0, but v has its axiu velocity of ωa toward the left. As x goes to negative values, the force and the acceleration becoe positive, slowing down the otion to the left, and hence v starts to decrease. At x = A the velocity is zero and the force and acceleration take on their axiu values toward the right, tending to restore the ass to its equilibriu position. As x becoes less negative, the velocity to the right increases, until it picks up its axiu value of ωa at x = 0, the equilibriu position, where F and a are both zero. Because of this large velocity, the ass passes the equilibriu position in its otion toward the right. However, as soon as x becoes positive, the force and the acceleration becoe negative thereby slowing down the ass until its velocity becoes zero at the axiu displaceent A. One entire cycle has been copleted, and the otion starts over again. (We should ephasize here that in this Figure 11.4 Displaceent, velocity, and acceleration in siple haronic otion. vibratory otion there are two places where the velocity is instantaneously zero, x = A and x = A, even though the instantaneous acceleration is nonzero there.) Soeties the vibratory otion is so rapid that the actual displaceent, velocity, and acceleration at every instant of tie are not as iportant as the gross otion, which can be described in ters of the frequency or period of the otion. We can find the frequency of the vibrating ass in ters of the spring constant k and the vibrating ass by setting equation 11.9 equal to equation Thus, 11-6 Vibratory Motion, Wave Motion and Fluids

7 Solving for the frequency f, we obtain k ω = πf = 1 k f = π (11.1) Equation 11.1 gives the frequency of the vibration. Because the period of the vibrating otion is the reciprocal of the frequency, we get for the period T = π (11.) k Equation 11. gives the period of the siple haronic otion in ters of the ass in otion and the spring constant k. Notice that for a particular value of and k, the period of the otion reains a constant throughout the otion. Exaple 11.1 An exaple of siple haronic otion. A ass of kg is placed on a vertical spring and the spring stretches by 10.0 c. It is then pulled down an additional 5.00 c and then released. Find (a) the spring constant k, (b) the angular frequency ω, (c) the frequency f, (d) the period T, (e) the axiu velocity of the vibrating ass, (f) the axiu acceleration of the ass, (g) the axiu restoring force, (h) the velocity of the ass at x =.00 c, and (i) the equation of the displaceent, velocity, and acceleration at any tie t. Solution Although the original analysis dealt with a ass on a horizontal frictionless surface, the results also apply to a ass attached to a spring that is allowed to vibrate in the vertical direction. The constant force of gravity on the kg ass displaces the equilibriu position to x = 10.0 c. When the additional force is applied to displace the ass another 5.00 c, the ass oscillates about the equilibriu position, located at the 10.0-c ark. Thus, the force of gravity only displaces the equilibriu position, but does not otherwise influence the result of the dynaic otion. a. The spring constant, found fro Hooke s law, is k = FA = g x x = (0.300 kg)(9.80 /s ) = 9.4 N/ b. The angular frequency ω, found fro equation 11.0, is k ω= 9.4 N/ = kg = 9.90 rad/s c. The frequency of the otion, found fro equation 11.9, is f = ω π = 9.90 rad/s π rad Chapter 11 Siple Haronic Motion 11-7

8 = 1.58 cycles = 1.58 Hz s d. We could find the period fro equation 11. but since we already know the frequency f, it is easier to copute T fro equation Thus, T = 1 = 1 = s f 1.58 cycles/s e. The axiu velocity, found fro equation 11.13, is vax = ωa = (9.90 rad/s)( ) = /s f. The axiu acceleration, found fro equation 11.17, is aax = ω A = (9.90 rad/s) ( ) = 4.90 /s g. The axiu restoring force, found fro Hooke s law, is Fax = kxax = ka = (9.4 N/)( ) = 1.47 N h. The velocity of the ass at x =.00 c, found fro equation 11.15, is v =± ω A x ( 9.90 rad/s ) ( ) ( ) v =± = ± /s where v is positive when oving to the right and negative when oving to the left. i. The equation of the displaceent at any instant of tie, found fro equation 11.6, is x = A cos ωt = ( ) cos(9.90 rad/s)t The equation of the velocity at any instant of tie, found fro equation 11.13, is v = ωa sin ωt = (9.90 rad/s)( )sin(9.90 rad/s)t = (0.495 /s)sin(9.90 rad/s)t The equation of the acceleration at any tie, found fro equation 11.17, is a = ω A cos ωt = (9.90 rad/s) ( )cos(9.90 rad/s)t = (4.90 /s )cos(9.90 rad/s)t To go to this Interactive Exaple click on this sentence Vibratory Motion, Wave Motion and Fluids

9 11.4 The Potential Energy of a Spring In chapter 7 we defined the gravitational potential energy of a body as the energy that a body possesses by virtue of its position in a gravitational field. A body can also have elastic potential energy. For exaple, a copressed spring has potential energy because it has the ability to do work as it expands to its equilibriu configuration. Siilarly, a stretched spring ust also contain potential energy because it has the ability to do work as it returns to its equilibriu position. Because work ust be done on a body to put the body into the configuration where it has the elastic potential energy, this work is used as the easure of the potential energy. Thus, the potential energy of a spring is equal to the work that you, the external agent, ust do to copress (or stretch) the spring to its present configuration. We defined the potential energy as PE = W = Fx (11.3) However, we can not use equation 11.3 in its present for to deterine the potential energy of a spring. Recall that the work defined in this way, in chapter 7, was for a constant force. We have seen in this chapter that the force necessary to copress or stretch a spring is not a constant but is rather a variable force depending on the value of x, (F = kx). We can still solve the proble, however, by using the average value of the force between the value at the equilibriu position and the value at the position x. That is, because the restoring force is directly proportional to the displaceent, the average force exerted in oving the ass fro x = 0 to the value x in figure 11.5(a) is Favg = F0 + F Figure 11.5 The potential energy of a spring. Thus, we find the potential energy in this configuration by using the average force, that is, Hence, PE = W = Favgx F0 + F = W = x 0 + kx = x PE = 1 kx (11.4) Because of the x in equation 11.4, the potential energy of a spring is always positive, whether x is positive or negative. The zero of potential energy is defined at the equilibriu position, x = 0. Note that equation 11.4 could also be derived by plotting the force F acting on the spring versus the displaceent x of the spring, as shown in figure 11.5(b). Because the work is equal to the product of the force F and the displaceent x, the work is also equal to the area under the curve in figure 11.5(b). The area of that triangle is ½ (x)(f) = ½ (x)(kx) = ½kx. (For the ore general proble where the force is not a linear function of the displaceent x, if the force is plotted versus the displaceent x, the work done, and hence the potential energy, will still be equal to the area under the curve.) Chapter 11 Siple Haronic Motion 11-9

10 Exaple 11. The potential energy of a spring. A spring, with a spring constant of 9.4 N/, is stretched 5.00 c. How uch potential energy does the spring possess? Solution The potential energy of the spring, found fro equation 11.4, is PE = 1 kx = 1 (9.4 N/)( ) = J To go to this Interactive Exaple click on this sentence Conservation of Energy and the Vibrating Spring The vibrating spring syste of figure 11. can also be described in ters of the law of conservation of energy. When the spring is stretched to its axiu displaceent A, work is done on the spring, and hence the spring contains potential energy. The ass attached to the spring also has that potential energy. The total energy of the syste is equal to the potential energy at the axiu displaceent because at that point, v = 0, and therefore the kinetic energy is equal to zero, that is, Etot = PE = 1 ka (11.5) When the spring is released, the ass oves to a saller displaceent x, and is oving at a speed v. At this arbitrary position x, the ass will have both potential energy and kinetic energy. The law of conservation of energy then yields Etot = PE + KE Etot = 1 kx + 1 v (11.6) But the total energy iparted to the ass is given by equation Hence, the law of conservation of energy gives Etot = Etot 1 ka = 1 kx + 1 v (11.7) We can also use equation 11.7 to find the velocity of the oving body at any displaceent x. Thus, 1 v = 1 ka 1 kx v = k (A x ) k v =± ( A x ) (11.8) We should note that this is the sae equation for the velocity as derived earlier (equation 11.15). It is inforative to replace the values of x and v fro their respective equations 11.6 and into equation Thus, Etot = 1 k(a cos ωt) + 1 ( ωa sin ωt) Vibratory Motion, Wave Motion and Fluids

11 or but since Etot = 1 ka cos ωt + 1 ω A sin ωt ω = k Etot = 1 ka cos ωt + 1 k A sin ωt = 1 ka cos ωt + 1 ka sin ωt (11.9) These ters are plotted in figure Figure 11.6 Conservation of energy and siple haronic otion. The total energy of the vibrating syste is a constant and this is shown as the horizontal line, Etot. At t = 0 the total energy of the syste is potential energy (v is zero, hence the kinetic energy is zero). As the tie increases the potential energy decreases and the kinetic energy increases, as shown. However, the total energy reains the sae. Fro equation 11.4 and figure 11.6, we see that at x = 0 the potential energy is zero and hence all the energy is kinetic. This occurs when t = T/4. The axiu velocity of the ass occurs here and is easily found by equating the axiu kinetic energy to the total energy, that is, When the oscillating ass reaches x = A, the kinetic energy becoes zero since 1 vax = 1 ka k vax = A = ω A (11.30) 1 ka = 1 ka + 1 v 1 v = 1 ka 1 ka = 0 = KE = 0 As the oscillation continues there is a constant interchange of energy between potential energy and kinetic energy but the total energy of the syste reains a constant. Exaple 11.3 Conservation of energy applied to a spring. A horizontal spring has a spring constant of 9.4 N/. A ass of 300 g is attached to the spring and displaced 5.00 c. The ass is then released. Find (a) the total energy of the syste, (b) the axiu velocity of the syste, and (c) the potential energy and kinetic energy for x =.00 c. Solution a. The total energy of the syste is Etot = 1 ka = 1 (9.4 N/)( ) = J b. The axiu velocity occurs when x = 0 and the potential energy is zero. Therefore, using equation 11.30, Chapter 11 Siple Haronic Motion 11-11

12 c. The potential energy at.00 c is The kinetic energy at.00 c is v v ax = ax 1 k A 9.4 N/ = kg = /s ( ) PE = 1 kx = 1 (9.4 N/)( ) = J KE = 1 v = 1 k (A x ) = 1 (9.4 N/)[( ) ( ) ] = J Note that the su of the potential energy and the kinetic energy is equal to the sae value for the total energy found in part a. To go to this Interactive Exaple click on this sentence The Siple Pendulu Another exaple of periodic otion is a pendulu. A siple pendulu is a bob that is attached to a string and allowed to oscillate, as shown in figure 11.7(a). The bob oscillates because there is a restoring force, given by Figure 11.7 The siple pendulu. Restoring force = g sin θ (11.31) This restoring force is just the coponent of the weight of the bob that is perpendicular to the string, as shown in figure 11.7(b). If Newton s second law, F = a, is applied to the otion of the pendulu bob, we get g sin θ = a The tangential acceleration of the pendulu bob is thus a = g sin θ (11.3) Note that although this pendulu otion is periodic, it is not, in general, siple haronic otion because the acceleration is not directly proportional to the displaceent of the pendulu bob fro its equilibriu position Vibratory Motion, Wave Motion and Fluids

13 However, if the angle θ of the siple pendulu is sall, then the sine of θ can be replaced by the angle θ itself, expressed in radians. (The discrepancy in using θ rather than the sin θ is less than 0.% for angles less than 10 degrees.) That is, for sall angles sin θ θ The acceleration of the bob is then a = gθ (11.33) Fro figure 11.7 and the definition of an angle in radians (θ = arc length/radius), we have θ = s l where s is the actual path length followed by the bob. Thus a = g s (11.34) l The path length s is curved, but if the angle θ is sall, the arc length s is approxiately equal to the chord x, figure 11.7(c). Hence, a = g x (11.35) l which is an equation having the sae for as that of the equation for siple haronic otion. Therefore, if the angle of oscillation θ is sall, the pendulu will execute siple haronic otion. For siple haronic otion of a spring, the acceleration was found to be a = k x (11.) We can now use the equations developed for the vibrating spring to describe the otion of the pendulu. We find an equivalent spring constant of the pendulu by setting equation 11. equal to equation That is or k = g l kp = g (11.36) l Equation states that the otion of a pendulu can be described by the equations developed for the vibrating spring by using the equivalent spring constant of the pendulu kp. Thus, the period of otion of the pendulu, found fro equation 11., is Tp = π k = π T p p g / l l = π (11.37) g The period of otion of the pendulu is independent of the ass of the bob but is directly proportional to the square root of the length of the string. If the angle θ is equal to 15 0 on either side of the central position, then the true period differs fro that given by equation by less than 0.5%. The pendulu can be used as a very siple device to easure the acceleration of gravity at a particular location. We easure the length l of the pendulu and then set the pendulu into otion. We easure the period by a clock and obtain the acceleration of gravity fro equation as Chapter 11 Siple Haronic Motion 11-13

14 g = 4π l (11.38) Tp Exaple 11.4 The period of a pendulu. Find the period of a siple pendulu 1.50 long. Solution The period, found fro equation 11.37, is T p = π l g 1.50 = π 9.80 /s =.46 s To go to this Interactive Exaple click on this sentence. Exaple 11.5 The length of a pendulu. Find the length of a siple pendulu whose period is 1.00 s. Solution The length of the pendulu, found fro equation 11.37, is l = T p g 4π = (1.00 s) (9.80 /s ) 4π = 0.48 To go to this Interactive Exaple click on this sentence. Exaple 11.6 The pendulu and the acceleration due to gravity. A pendulu 1.50 long is observed to have a period of.47 s at a certain location. Find the acceleration of gravity there. Solution The acceleration of gravity, found fro equation 11.38, is g = 4π l Tp = 4π (1.50 ) (.47 s) = 9.71 /s To go to this Interactive Exaple click on this sentence Vibratory Motion, Wave Motion and Fluids

15 We can also use a pendulu to easure an acceleration. If a pendulu is placed on board a rocket ship in interstellar space and the rocket ship is accelerated at 9.80 /s, the pendulu oscillates with the sae period as it would at rest on the surface of the earth. An enclosed person or thing on the rocket ship could not distinguish between the acceleration of the rocket ship at 9.80 /s and the acceleration due to gravity of 9.80 /s on the earth. (This is an exaple of Einstein s principle of equivalence in general relativity.) An oscillating pendulu of easured length l can be placed in an elevator and the period T easured. We can use equation to easure the resultant acceleration experienced by the pendulu in the elevator Springs in Parallel and in Series Soeties ore than one spring is used in a vibrating syste. The otion of the syste will depend on the way the springs are connected. As an exaple, suppose there are three assless springs with spring constants k1, k, and k3. These springs can be connected in parallel, as shown in figure 11.8(a), or in series, as in figure 11.8(b). The period of otion of either configuration can be found by using an equivalent spring constant ke. Figure 11.8 Springs in parallel and in series. Springs in Parallel If the total force pulling the ass a distance x to the right is Ftot, this force will distribute itself aong the three springs such that there will be a force F1 on spring 1, a force F on spring, and a force F3 on spring 3. If the displaceent of each spring is equal to x, then the springs are said to be in parallel. Then we can write the total force as Ftot = F1 + F + F3 (11.39) However, since we assued that each spring was displaced the sae distance x, Hooke s law for each spring is Substituting equation into equation gives F1 = k1x F = kx F3 = k3x (11.40) Ftot = k1x + kx + k3x = (k1 + k + k3)x We now define an equivalent spring constant ke for springs connected in parallel as ke = k1 + k + k3 (11.41) Hooke s law for the cobination of springs is given by Ftot = kex (11.4) The springs in parallel will execute a siple haronic otion whose period, found fro equation 11., is Chapter 11 Siple Haronic Motion 11-15

16 T = π = π k k + k + k E 1 3 (11.43) Springs in Series If the sae springs are connected in series, as in figure 11.8(b), the total force Ftot displaces the ass a distance x to the right. But in this configuration, each spring stretches a different aount. Thus, the total displaceent x is the su of the displaceents of each spring, that is, The displaceent of each spring, found fro Hooke s law, is x = x1 + x + x3 (11.44) x1 = F1 k1 x = F k x3 = F3 (11.45) k3 Substituting these values of the displaceent into equation 11.44, yields x = F1 + F + F3 (11.46) k1 k k3 But because the springs are in series the total applied force is transitted equally fro spring to spring. Hence, Ftot = F1 = F = F3 (11.47) Substituting equation into equation 11.46, gives and x = Ftot + Ftot + Ftot (11.46) k1 k k x = + + F k1 k k3 tot (11.48) We now define the equivalent spring constant for springs connected in series as 1 = (11.49) ke k1 k k3 Thus, the total displaceent, equation 11.48, becoes x = Ftot (11.50) and Hooke s law becoes ke Ftot = kex (11.51) where ke is given by equation Hence, the cobination of springs in series executes siple haronic otion and the period of that otion, given by equation 11., is T = π = π + + ke k1 k k3 (11.5) Vibratory Motion, Wave Motion and Fluids

17 Exaple 11.7 Springs in parallel. Three springs with force constants k1 = 10.0 N/, k = 1.5 N/, and k3 = 15.0 N/ are connected in parallel to a ass of kg. The ass is then pulled to the right and released. Find the period of the otion. Solution The period of the otion, found fro equation 11.43, is T = π T = π k + k + k kg 10.0 N/ N/ N/ = 0.76 s To go to this Interactive Exaple click on this sentence. Exaple 11.8 Springs in series. The sae three springs as in exaple 11.7 are now connected in series. Find the period of the otion. Solution The period, found fro equation 11.5, is T = π + + k 1 k k = π ( kg) N/ 1.5 N/ 15.0 N/ =.1 s To go to this Interactive Exaple click on this sentence. The Language of Physics Periodic otion Motion that repeats itself in equal intervals of tie (p. ). Displaceent The distance a vibrating body oves fro its equilibriu position (p. ). Siple haronic otion Periodic otion in which the acceleration of a body is directly proportional to its displaceent fro the equilibriu position but in the opposite direction. Because the acceleration is directly proportional to the displaceent, the acceleration of the body is not constant. The kineatic equations developed in chapter 3 are no longer valid to describe this type of otion (p. ). Aplitude The axiu displaceent of the vibrating body (p. ). Cycle One coplete oscillation or vibratory otion (p. ). Chapter 11 Siple Haronic Motion 11-17

18 Period The tie for the vibrating body to coplete one cycle (p. ). Frequency The nuber of coplete cycles or oscillations in unit tie. The frequency is the reciprocal of the period (p. ). Reference circle A body executing unifor circular otion does so in a circle. The projection of the position of the rotating body onto the x- or y-axis is equivalent to siple haronic otion along that axis. Thus, vibratory otion is related to otion in a circle, the reference circle (p. ). Angular velocity The angular velocity of the unifor circular otion in the reference circle is related to the frequency of the vibrating syste. Hence, the angular velocity is called the angular frequency of the vibrating syste (p. ). Potential energy of a spring The energy that a body possesses by virtue of its configuration. A copressed spring has potential energy because it has the ability to do work as it expands to its equilibriu configuration. A stretched spring can also do work as it returns to its equilibriu configuration (p. ). Siple pendulu A bob that is attached to a string and allowed to oscillate to and fro under the action of gravity. If the angle of the pendulu is sall the pendulu will oscillate in siple haronic otion (p. ). Suary of Iportant Equations Restoring force in a spring FR = kx (11.1) Defining relation for siple haronic otion a = k x (11.) Frequency f = 1 (11.3) T Displaceent in siple haronic otion x = A cos ωt (11.6) Angular frequency ω = πf (11.9) Velocity as a function of tie in siple haronic otion v = ωa sin ωt (11.13) Velocity as a function of displaceent v =± ω A x (11.15) Acceleration as a function of tie a = ω A cos ωt (11.17) Angular frequency of a spring k ω= (11.0) Frequency in siple haronic 1 k otion f = π (11.1) Period in siple haronic otion T = π (11.) k Potential energy of a spring PE = 1 kx (11.4) Conservation of energy for a vibrating spring 1 ka = 1 kx + 1 v (11.7) Period of otion of a siple l pendulu Tp = π (11.37) g Equivalent spring constant for springs in parallel ke = k1 + k + k3 (11.41) Period of otion for springs in parallel T = π (11.43) k + k + k 1 3 Equivalent spring constant for springs in series 1 = (11.49) ke k1 k k3 Period of otion for springs in series T = π + + (11.5) k 1 k k 3 1. Can the periodic otion of the earth be considered to be an exaple of siple haronic otion?. Can the kineatic equations derived in chapter 3 be used to describe siple haronic otion? Questions for Chapter In the siple haronic otion of a ass attached to a spring, the velocity of the ass is equal to zero when the acceleration has its axiu value. How is this possible? Can you think of other exaples in which a body has zero velocity with a nonzero acceleration? 4. What is the characteristic of the restoring force that akes siple haronic otion possible? 5. Discuss the significance of the reference circle in the analysis of siple haronic otion Vibratory Motion, Wave Motion and Fluids

19 6. How can a ass that is undergoing a one-diensional translational siple haronic otion have anything to do with an angular velocity or an angular frequency, which is a characteristic of two or ore diensions? 7. How is the angular frequency related to the physical characteristics of the spring and the vibrating ass in siple haronic otion? *8. In the entire derivation of the equations for siple haronic otion we have assued that the springs are assless and friction can be neglected. Discuss these assuptions. Describe qualitatively what you would expect to happen to the otion if the springs are not sall enough to be considered assless. *9. Describe how a geological survey for iron ight be undertaken on the oon using a siple pendulu. *10. How could a siple pendulu be used to ake an acceleroeter? *11. Discuss the assuption that the displaceent of each spring is the sae when the springs are in parallel. Under what conditions is this assuption valid and when would it be invalid? Probles for Chapter Siple Haronic Motion and 11.3 Analysis of Siple Haronic Motion 1. A ass of 0.00 kg is attached to a spring of spring constant 30.0 N/. If the ass executes siple haronic otion, what will be its frequency?. A 30.0-g ass is attached to a vertical spring and it stretches 10.0 c. It is then stretched an additional 5.00 c and released. Find its period of otion and its frequency. 3. A 0.00-kg ass on a spring executes siple haronic otion at a frequency f. What ass is necessary for the syste to vibrate at a frequency of f? 4. A siple haronic oscillator has a frequency of.00 Hz and an aplitude of 10.0 c. What is its axiu acceleration? What is its acceleration at t = 0.5 s? 5. A ball attached to a string travels in unifor circular otion in a horizontal circle of 50.0 c radius in 1.00 s. Sunlight shining on the ball throws its shadow on a wall. Find the velocity of the shadow at (a) the end of its path and (b) the center of its path. 6. A 50.0-g ass is attached to a spring of force constant 10.0 N/. The spring is stretched 0.0 c and then released. Find the displaceent, velocity, and acceleration of the ass at 0.00 s. 7. A 5.0-g ass is attached to a vertical spring and it stretches 15.0 c. It is then stretched an additional 10.0 c and then released. What is the axiu velocity of the ass? What is its axiu acceleration? 8. The displaceent of a body in siple haronic otion is given by x = (0.15 )cos[(5.00 rad/s)t]. Find (a) the aplitude of the otion, (b) the angular frequency, (c) the frequency, (d) the period, and (e) the displaceent at 3.00 s. 9. A 500-g ass is hung fro a coiled spring and it stretches 10.0 c. It is then stretched an additional 15.0 c and released. Find (a) the frequency of vibration, (b) the period, and (c) the velocity and acceleration at a displaceent of 10.0 c. 10. A ass of 0.00 kg is placed on a vertical spring and the spring stretches by 15.0 c. It is then pulled down an additional 10.0 c and then released. Find (a) the spring constant, (b) the angular frequency, (c) the frequency, (d) the period, (e) the axiu velocity of the ass, (f) the axiu acceleration of the ass, (g) the axiu restoring force, and (h) the equation of the displaceent, velocity, and acceleration at any tie t Conservation of Energy and the Vibrating Spring 11. A siple haronic oscillator has a spring constant of 5.00 N/. If the aplitude of the otion is 15.0 c, find the total energy of the oscillator. 1. A body is executing siple haronic otion. At what displaceent is the potential energy equal to the kinetic energy? 13. A 0.0-g ass is attached to a horizontal spring on a sooth table. The spring constant is 3.00 N/. The spring is then stretched 15.0 c and then released. What is the total energy of the otion? What is the potential and kinetic energy when x = 5.00 c? 14. A body is executing siple haronic otion. At what displaceent is the speed v equal to one-half the axiu speed? 11.6 The Siple Pendulu 15. Find the period and frequency of a siple pendulu 0.75 long. 16. If a pendulu has a length L and a period T, what will be the period when (a) L is doubled and (b) L is halved? 17. Find the frequency of a child s swing whose ropes have a length of What is the period of a pendulu on the oon where g = (1/6)ge? 19. What is the period of a pendulu long on a spaceship (a) accelerating at 4.90 /s and (b) oving at constant velocity? 0. What is the period of a pendulu in free-fall? Chapter 11 Siple Haronic Motion 11-19

20 1. A pendulu has a period of s at the equator at sea level. The pendulu is carried to another place on the earth and the period is now found to be s. Find the acceleration due to gravity at this location Springs in Parallel and in Series *. Springs with spring constants of 5.00 N/ and 10.0 N/ are connected in parallel to a kg ass. Find (a) the equivalent spring constant and (b) the period of the otion. *3. Springs with spring constants 5.00 N/ and 10.0 N/ are connected in series to a 5.00-kg ass. Find (a) the equivalent spring constant and (b) the period of the otion. Additional Probles 4. A 500-g ass is attached to a vertical spring of spring constant 30.0 N/. How far should the spring be stretched in order to give the ass an upward acceleration of 3.00 /s? 5. A ball is caused to ove in a horizontal circle of 40.0-c radius in unifor circular otion at a speed of 5.0 c/s. Its projection on the wall oves in siple haronic otion. Find the velocity and acceleration of the shadow of the ball at (a) the end of its otion, (b) the center of its otion, and (c) halfway between the center and the end of the otion. *6. The otion of the piston in the engine of an autoobile is approxiately siple haronic. If the stroke of the piston (twice the aplitude) is equal to 0.3 c and the engine turns at 1800 rp, find (a) the acceleration at x = A and (b) the speed of the piston at the idpoint of the stroke. *7. A 535-g ass is dropped fro a height of 5.0 c above an uncopressed spring of k = 0.0 N/. By how uch will the spring be copressed? 8. A siple pendulu is used to operate an electrical device. When the pendulu bob sweeps through the idpoint of its swing, it causes an electrical spark to be given off. Find the length of the pendulu that will give a spark rate of 30.0 sparks per inute. *9. The general solution for the period of a siple pendulu, without aking the assuption of sall angles of swing, is given by T = π 1 ( ) sin θ 1 l g ( ) ( ) sin θ Find the period of a pendulu for θ = , , and and copare with the period obtained with the sall angle approxiation. Deterine the percentage error in each case by using the sall angle approxiation. 30. A pendulu clock on the earth has a period of 1.00 s. Will this clock run slow or fast, and by how uch if taken to (a) Mars, (b) Moon, and (c) Venus? *31. A pendulu bob, 355 g, is raised to a height of 1.5 c before it is released. At the botto of its path it akes a perfectly elastic collision with a 500-g ass that is connected to a horizontal spring of spring constant 15.8 N/, that is at rest on a sooth surface. How far will the spring be copressed? Diagra for proble 31. *3. A 500-g block is in siple haronic otion as shown in the diagra. A ass is added to the top of the block when the block is at its axiu extension. How uch ass should be added to change the frequency by 5%? Diagra for proble 3. *33. A pendulu clock keeps correct tie at a location at sea level where the acceleration of gravity is equal to 9.80 /s. The clock is then taken up to the top of a ountain and the clock loses 3.00 s per day. How high is the ountain? *34. Three people, who together weigh 1880 N, get into a car and the car is observed to ove 5.08 c closer to the ground. What is the spring constant of the car springs? *35. In the accopanying diagra, the ass is pulled down a distance of 9.50 c fro its equilibriu position and is then released. The ass then executes siple haronic otion. Find (a) the total potential energy of the ass with respect to the ground when the ass is located at positions 1,, and 3; (b) the total energy of the ass at positions 1,, and 3; and (c) the speed of the ass at position. Assue = 55.6 g, k = 5.0 N/, h0 = 50.0 c. Diagra for proble 35. *36. A 0.0-g ball rests on top of a vertical spring gun whose spring constant is 0 N/. The spring is copressed 10.0 c and the gun is 11-0 Vibratory Motion, Wave Motion and Fluids

21 then fired. Find how high the ball rises in its vertical trajectory. *37. A toy spring gun is used to fire a ball as a projectile. Find the value of the spring constant, such that when the spring is copressed 10.0 c, and the gun is fired at an angle of 6.5 0, the range of the projectile will be The ass of the ball is 5. g. *38. In the siple pendulu shown in the diagra, find the tension in the string when the height of the pendulu is (a) h, (b) h/, and (c) h = 0. The ass = 500 g, the initial height h = 15.0 c, and the length of the pendulu l = Diagra for proble 38. *39. A ass is attached to a horizontal spring. The ass is given an initial aplitude of 10.0 c on a rough surface and is then released to oscillate in siple haronic otion. If 10.0% of the energy is lost per cycle due to the friction of the ass oving over the rough surface, find the axiu displaceent of the ass after 1,, 4, 6, and 8 coplete oscillations. *40. Find the axiu aplitude of vibration after periods for a 85.0-g ass executing siple haronic otion on a rough horizontal surface of µk = The spring constant is 4.0 N/ and the initial aplitude is 0.0 c. 41. A 40-g ass slides down a circular chute without friction and collides with a horizontal spring, as shown in the diagra. If the original position of the ass is 5.0 c above the table top and the spring has a spring constant of 18 N/, find the axiu distance that the spring will be copressed. Diagra for proble 41. *4. A 35-g block slides down a frictionless inclined plane, 1.5 long, that akes an angle of with the horizontal. At the botto of the plane the block slides along a rough horizontal surface 1.50 long. The coefficient of kinetic friction between the block and the rough horizontal surface is The block then collides with a horizontal spring of k = 0.0 N/. Find the axiu copression of the spring. Diagra for proble 4. *43. A 335-g disk that is free to rotate about its axis is connected to a spring that is stretched 35.0 c. The spring constant is 10.0 N/. When the disk is released it rolls without slipping as it oves toward the equilibriu position. Find the speed of the disk when the displaceent of the spring is equal to 17.5 c. *44. A 5.0-g ball oving at a velocity of 00 c/s to the right akes an inelastic collision with a 00-g block that is initially at rest on a frictionless surface. There is a hole in the large block for the sall ball to fit into. If k = 10 N/, find the axiu copression of the spring. Diagra for proble 43. Diagra for proble 44. *45. Show that the period of siple haronic otion for the ass shown is equivalent to the period for two springs in parallel. Diagra for proble 45. *46. A nail is placed in the wall at a distance of l/ fro the top, as shown in the diagra. A pendulu of length 85.0 c is released fro position 1. (a) Find the tie it takes for the pendulu bob to reach position. When the bob of the pendulu reaches position, the string hits the nail. (b) Find the tie it takes for the pendulu bob to reach position 3. Diagra for proble 46. Chapter 11 Siple Haronic Motion 11-1

22 *47. A spring is attached to the top of an Atwood s achine as shown. The spring is stretched to A = 10 c before being released. Find the velocity of when x = A/. Assue 1 = 8.0 g, = 43.0 g, and k = 10.0 N/. Diagra for proble 47. *48. A 80-g block is connected to a spring on a rough inclined plane that akes an angle of with the horizontal. The block is pulled down the plane a distance A = 0.0 c, and is then released. The spring constant is 40.0 N/ and the coefficient of kinetic friction is Find the speed of the block when the displaceent x = A/. Diagra for proble The rotational analog of siple haronic otion, is angular siple haronic otion, wherein a body rotates periodically clockwise and then counterclockwise. Hooke s law for rotational otion is given by τ = C θ where τ is the torque acting on the body, θ is the angular displaceent, and C is a constant, like the spring constant. Use Newton s second law for rotational otion to show α = C θ I Use the analogy between the linear result, a = ω x, to show that the frequency of vibration of an object executing angular siple haronic otion is given by 1 f = π C I Interactive Tutorials 50. Siple Pendulu. Calculate the period T of a siple pendulu located on a planet having a gravitational acceleration of g = 9.80 /s, if its length l = 1.00 is increased fro 1 to 10 in steps of Plot the results as the period T versus the length l. 51. Siple Haronic Motion. The displaceent x of a body undergoing siple haronic otion is given by the forula x = A cos ωt, where A is the aplitude of the vibration, ω is the angular frequency in rad/s, and t is the tie in seconds. Plot the displaceent x as a function of t for an aplitude A = and an angular frequency ω = 5.00 rad/s as t increases fro 0 to s in 0.10 s increents. 5. The Vibrating Spring. A ass = kg is attached to a spring on a sooth horizontal table. An applied force FA = 4.00 N is used to stretch the spring a distance x0 = (a) Find the spring constant k of the spring. The ass is returned to its equilibriu position and then stretched to a value A = 0.15 and then released. The ass then executes siple haronic otion. Find (b) the angular frequency ω, (c) the frequency f, (d) the period T, (e) the axiu velocity vax of the vibrating ass, (f) the axiu acceleration aax of the vibrating ass, (g) the axiu restoring force FRax, and (h) the velocity of the ass at the displaceent x = (i) Plot the displaceent x, velocity v, acceleration a, and the restoring force FR at any tie t. 53. Conservation of Energy and the Vibrating Horizontal Spring. A ass = kg is attached to a horizontal spring. The ass is then pulled a distance x = A = 0.00 fro its equilibriu position and when released the ass executes siple haronic otion. Find (a) the total energy Etot of the ass when it is at its axiu displaceent A fro its equilibriu position. When the ass is at the displaceent x = 0.10 find, (b) its potential energy PE, (c) its kinetic energy KE, and (d) its speed v. (e) Plot the total energy, potential energy, and kinetic energy of the ass as a function of the displaceent x. The spring constant k = 35.5 N/. 54. Conservation of Energy and the Vibrating Vertical Spring. A ass = kg is attached to a vertical spring. The ass is at a height h0 = 1.50 fro the floor. The ass is then pulled down a distance A = 0.0 fro its equilibriu position and when released executes siple haronic otion. Find (a) the total energy of the ass when it is at its axiu displaceent A below its equilibriu position, (b) the gravitational potential energy when it is at the displaceent x = 0.10, (c) the elastic potential energy when it is at the sae displaceent x, (d) the kinetic energy at the displaceent x, and (e) the speed of the ass when it is at the displaceent x. The spring constant k = 35.5 N/. To go to these Interactive Tutorials click on this sentence. To go to another chapter, return to the table of contents by clicking on this sentence. 11- Vibratory Motion, Wave Motion and Fluids