( ) ( ) 1. (a) The amplitude is half the range of the displacement, or x m = 1.0 mm.

Transcription

1 1. (a) The aplitude is half the range of the displaceent, or x = 1.0. (b) The axiu speed v is related to the aplitude x by v = ωx, where ω is the angular frequency. Since ω = πf, where f is the frequency, (c) The axiu acceleration is v a x f x 3 ( )( ) = π fx =π 10 Hz = 0.75 /s. ( ) ( ) ( ) ( ) 3 = ω = π = π 10 Hz = /s.

2 . (a) The acceleration aplitude is related to the axiu force by Newton s second law: F ax = a. The textbook notes (in the discussion iediately after Eq. 15-7) that the acceleration aplitude is a = ω x, where ω is the angular frequency (ω = πf since there are π radians in one cycle). The frequency is the reciprocal of the period: f = 1/T = 1/0.0 = 5.0 Hz, so the angular frequency is ω = 10π (understood to be valid to two significant figures). Therefore, (b) Using Eq. 15-1, we obtain F ax = x = 0.1 kg 10π rad / s = 10 N. ω b gb gb g k ( )( ) ω = k = 0.1kg 10 π rad/s = N/.

3 3. (a) The angular frequency ω is given by ω = πf = π/t, where f is the frequency and T is the period. The relationship f = 1/T was used to obtain the last for. Thus ω = π/( s) = rad/s. (b) The axiu speed v and axiu displaceent x are related by v = ωx, so x v = = 5 ω /s rad / s 3 =

4 4. The textbook notes (in the discussion iediately after Eq. 15-7) that the acceleration aplitude is a = ω x, where ω is the angular frequency (ω = πf since there are π radians in one cycle). Therefore, in this circustance, we obtain c b ghb g. a = π 6.60 Hz 0.00 = 37.8 / s

5 5. (a) The otion repeats every s so the period ust be T = s. (b) The frequency is the reciprocal of the period: f = 1/T = 1/(0.500 s) =.00 Hz. (c) The angular frequency ω is ω = πf = π(.00 Hz) = 1.6 rad/s. (d) The angular frequency is related to the spring constant k and the ass by ω = k. We solve for k: k = ω = (0.500 kg)(1.6 rad/s) = 79.0 N/. (e) Let x be the aplitude. The axiu speed is v = ωx = (1.6 rad/s)(0.350 ) = 4.40 /s. (f) The axiu force is exerted when the displaceent is a axiu and its agnitude is given by F = kx = (79.0 N/)(0.350 ) = 7.6 N.

6 6. (a) The proble describes the tie taken to execute one cycle of the otion. The period is T = 0.75 s. (b) Frequency is siply the reciprocal of the period: f = 1/T 1.3 Hz, where the SI unit abbreviation Hz stands for Hertz, which eans a cycle-per-second. (c) Since π radians are equivalent to a cycle, the angular frequency ω (in radians-persecond) is related to frequency f by ω = πf so that ω 8.4 rad/s.

7 7. (a) During siple haronic otion, the speed is (oentarily) zero when the object is at a turning point (that is, when x = +x or x = x ). Consider that it starts at x = +x and we are told that t = 0.5 second elapses until the object reaches x = x. To execute a full cycle of the otion (which takes a period T to coplete), the object which started at x = +x ust return to x = +x (which, by syetry, will occur 0.5 second after it was at x = x ). Thus, T = t = 0.50 s. (b) Frequency is siply the reciprocal of the period: f = 1/T =.0 Hz. (c) The 36 c distance between x = +x and x = x is x. Thus, x = 36/ = 18 c.

8 8. (a) Since the proble gives the frequency f = 3.00 Hz, we have ω = πf = 6π rad/s (understood to be valid to three significant figures). Each spring is considered to support one fourth of the ass car so that Eq leads to k 1 ( 1450kg )( 6 rad/s ) ω = k = π = N/. /4 4 car (b) If the new ass being supported by the four springs is total = [ (73)] kg = 1815 kg, then Eq leads to ω new k = fnew = = /4 π (1815 / 4) kg total N/.68Hz.

9 9. (a) Making sure our calculator is in radians ode, we find F HG b g π x =6.0cos 3π.0 + = (b) Differentiating with respect to tie and evaluating at t =.0 s, we find (c) Differentiating again, we obtain b g b g dx F π v = = = 49. dt H G I π sin π 3 K J / s b gb g b g dv F π a = = = dt H G I π cos π 3 K J / s (d) In the second paragraph after Eq. 15-3, the textbook defines the phase of the otion. In this case (with t =.0 s) the phase is 3π(.0) + π/3 0 rad. (e) Coparing with Eq. 15-3, we see that ω = 3π rad/s. Therefore, f = ω/π = 1.5 Hz. (f) The period is the reciprocal of the frequency: T = 1/f 0.67 s. I K J

10 10. We note (fro the graph) that x = 6.00 c. Also the value at t = 0 is x o =.00 c. Then Eq leads to f = cos 1 (.00/6.00) = rad or 4.37 rad. The other root (+4.37 rad) can be rejected on the grounds that it would lead to a positive slope at t = 0.

11 11. When displaced fro equilibriu, the net force exerted by the springs is kx acting in a direction so as to return the block to its equilibriu position (x = 0). Since the acceleration a = d x/dt, Newton s second law yields d x = kx. dt Substituting x = x cos(ωt + φ) and siplifying, we find ω = k where ω is in radians per unit tie. Since there are π radians in a cycle, and frequency f easures cycles per second, we obtain f ω 1 k 1 (7580 N/) = = = = 39.6 Hz. π π π 0.45 kg

12 1. We note (fro the graph) that v = ωx = 5.00 c/s. Also the value at t = 0 is v o = 4.00 c/s. Then Eq leads to φ = sin 1 ( 4.00/5.00) = 0.97 rad or rad. The other root (+4.07 rad) can be rejected on the grounds that it would lead to a positive slope at t = 0.

13 13. The agnitude of the axiu acceleration is given by a = ω x, where ω is the angular frequency and x is the aplitude. (a) The angular frequency for which the axiu acceleration is g is given by ω = g/ x, and the corresponding frequency is given by f ω 1 g /s = = = = 498 Hz. 6 π π x π (b) For frequencies greater than 498 Hz, the acceleration exceeds g for soe part of the otion.

14 14. Fro highest level to lowest level is twice the aplitude x of the otion. The period is related to the angular frequency by Eq Thus, x = 1 d and ω = rad/h. The phase constant φ in Eq is zero since we start our clock when x o = x (at the highest point). We solve for t when x is one-fourth of the total distance fro highest to lowest level, or (which is the sae) half the distance fro highest level to iddle level (where we locate the origin of coordinates). Thus, we seek t when the ocean surface is at 1 1 x = x = d 4. x = x F cos( ωt+ φ) 1 d = d t H G 1 I K J cos b = cos( t ) which has t =.08 h as the sallest positive root. The calculator is in radians ode during this calculation. g

15 15. The axiu force that can be exerted by the surface ust be less than µ s F N or else the block will not follow the surface in its otion. Here, µ s is the coefficient of static friction and F N is the noral force exerted by the surface on the block. Since the block does not accelerate vertically, we know that F N = g, where is the ass of the block. If the block follows the table and oves in siple haronic otion, the agnitude of the axiu force exerted on it is given by F = a = ω x = (πf) x, where a is the agnitude of the axiu acceleration, ω is the angular frequency, and f is the frequency. The relationship ω = πf was used to obtain the last for. We substitute F = (πf) x and F N = g into F < µ s F N to obtain (πf) x < µ s g. The largest aplitude for which the block does not slip is x µ = s g = πf b g b0.50gc9.8 /s h = π.0 Hz b A larger aplitude requires a larger force at the end points of the otion. The surface cannot supply the larger force and the block slips. g

16 16. The stateent that the spring does not affect the collision justifies the use of elastic collision forulas in section We are told the period of SHM so that we can find the ass of block : kt T = = = k 4π π kg. At this point, the rebound speed of block 1 can be found fro Eq : v 1 f = (8.00 /s) = 4.00 /s This becoes the initial speed v 0 of the projectile otion of block 1. A variety of choices for the positive axis directions are possible, and we choose left as the +x direction and down as the +y direction, in this instance. With the launch angle being zero, Eq. 4-1 and Eq. 4- (with g replaced with +g) lead to x x 0 = v 0 t = v 0 h g = (4.00) (4.90) 9.8 Since x x 0 = d, we arrive at d = 4.00.

17 17. (a) Eq leads to a a = x = 13 ω ω = x which yields ω = rad/s. Therefore, f = ω/π = 5.58 Hz. (b) Eq provides a relation between ω (found in the previous part) and the ass: k 400 N/ ω = = = 0.35kg. (35.07 rad/s) (c) By energy conservation, 1 kx (the energy of the syste at a turning point) is equal to the su of kinetic and potential energies at the tie t described in the proble. 1 kx = v kx x = +. k v x Consequently, x = (0.35/ 400)(13.6) + (0.100) =

18 18. We note that the ratio of Eq and Eq is v/x = ωtan(ωt + φ) where ω = 1.0 rad/s in this proble. Evaluating this at t = 0 and using the values fro the graphs shown in the proble, we find φ = tan 1 ( v o /x o ω) = tan 1 (+4.00/( 1.0)) =1.03 rad (or 5.5 rad). One can check that the other root (4.17 rad) is unacceptable since it would give the wrong signs for the individual values of v o and x o.

19 19. Eq gives the angular velocity: ω = k 100 N/ 7.07rad/s. =.00 kg = Energy ethods (discussed in 15-4) provide one ethod of solution. Here, we use trigonoetric techniques based on Eq and Eq (a) Dividing Eq by Eq. 15-3, we obtain so that the phase (ωt + φ) is found fro v = ωtanbωt + φ x F I HG K J v ωt + φ = tan = tan ωx 1 1 g F I HG b gb gk J With the calculator in radians ode, this gives the phase equal to 1.31 rad. Plugging this back into Eq leads to 0.19 = x cos( 1.31) x = (b) Since ωt + φ = 1.31 rad at t = 1.00 s, we can use the above value of ω to solve for the phase constant φ. We obtain φ = 8.38 rad (though this, as well as the previous result, can have π or 4π (and so on) added to it without changing the physics of the situation). With this value of φ, we find x o = x cos φ = (c) And we obtain v o = x ω sinφ = 3.06 /s.

20 0. Both parts of this proble deal with the critical case when the axiu acceleration becoes equal to that of free fall. The textbook notes (in the discussion iediately after Eq. 15-7) that the acceleration aplitude is a = ω x, where ω is the angular frequency; this is the expression we set equal to g = 9.8 /s. (a) Using Eq and T = 1.0 s, we have F πi HG T K J gt x = g x = = π (b) Since ω = πf, and x = is given, we find = g =. Hz. π x = ( π f ) x g f 1

21 1. (a) Let x 1 F HG A πt = cos T I K J be the coordinate as a function of tie for particle 1 and x F HG A πt π = cos + T 6 be the coordinate as a function of tie for particle. Here T is the period. Note that since the range of the otion is A, the aplitudes are both A/. The arguents of the cosine functions are in radians. Particle 1 is at one end of its path (x 1 = A/) when t = 0. Particle is at A/ when πt/t + π/6 = 0 or t = T/1. That is, particle 1 lags particle by onetwelfth a period. We want the coordinates of the particles 0.50 s later; that is, at t = 0.50 s, I K J A π 0.50 s x1 = cos = 0.5A 1.5 s and x A π 0.50 s π = cos + = 0.43 A. 1.5 s 6 Their separation at that tie is x 1 x = 0.5A A = 0.18A. (b) The velocities of the particles are given by and dx1 πa πt v1 = = sinf dt T H T I K v dx πa πt π = = sinf +. dt T H T 6 We evaluate these expressions for t = 0.50 s and find they are both negative-valued, indicating that the particles are oving in the sae direction. I K

22 1. They pass each other at tie t, at x 1 = x = x where x = x cos( ωt + φ ) and x = x cos( ωt + φ ) Fro this, we conclude that cos( ωt + φ1) = cos( ωt + φ ) =, and therefore that the phases (the arguents of the cosines) are either both equal to π/3 or one is π/3 while the other is π/3. Also at this instant, we have v 1 = v 0 where v = x ω sin( ωt + φ ) and v = x ω sin( ωt + φ ). 1 1 This leads to sin(ωt + φ 1 ) = sin(ωt + φ ). This leads us to conclude that the phases have opposite sign. Thus, one phase is π/3 and the other phase is π /3; the wt ter cancels if we take the phase difference, which is seen to be π /3 ( π /3) = π /3.

23 3. (a) The object oscillates about its equilibriu point, where the downward force of gravity is balanced by the upward force of the spring. If is the elongation of the spring at equilibriu, then k = g, where k is the spring constant and is the ass of the object. Thus k= g and a f a f. f = ω π= 1π k = 1 π g Now the equilibriu point is halfway between the points where the object is oentarily at rest. One of these points is where the spring is unstretched and the other is the lowest point, 10 c below. Thus = 50. c = and f = /s. Hz. π = (b) Use conservation of energy. We take the zero of gravitational potential energy to be at the initial position of the object, where the spring is unstretched. Then both the initial potential and kinetic energies are zero. We take the y axis to be positive in the downward direction and let y = The potential energy when the object is at this point is U = ky gy. The energy equation becoes 0 = ky gy + v. We solve for the speed. k g 9.8/s v= gy y = gy y = 9.8/s = = 0.56/s ( )( ) ( ) (c) Let be the original ass and be the additional ass. The new angular frequency is ω = k/( + ). This should be half the original angular frequency, or 1 k. We 1 solve k/( + ) = k/ for. Square both sides of the equation, then take the reciprocal to obtain + = 4. This gives = /3 = (300 g)/3 = 100 g = kg. (d) The equilibriu position is deterined by the balancing of the gravitational and spring forces: ky = ( + )g. Thus y = ( + )g/k. We will need to find the value of the spring constant k. Use k = ω = (π f ). Then ( + ) ( π f ) ( kg kg)( 9.80 /s ) g y = = kg.4 Hz This is easured fro the initial position. ( )( π )

24 4. Let the spring constants be k 1 and k. When displaced fro equilibriu, the agnitude of the net force exerted by the springs is k 1 x + k x acting in a direction so as to return the block to its equilibriu position (x = 0). Since the acceleration a = d x/d, Newton s second law yields d x = kx. 1 kx dt Substituting x = x cos(ωt + φ) and siplifying, we find k + k = ω 1 where ω is in radians per unit tie. Since there are π radians in a cycle, and frequency f easures cycles per second, we obtain f ω = = 1 π π k + k 1. The single springs each acting alone would produce siple haronic otions of frequency 1 k1 1 k f1= = 30 Hz, f = = 45 Hz, π π respectively. Coparing these expressions, it is clear that f = f + f = = 1 (30 Hz) +(45 Hz) 54 Hz.

25 5. To be on the verge of slipping eans that the force exerted on the saller block (at the point of axiu acceleration) is f ax = µ s g. The textbook notes (in the discussion iediately after Eq. 15-7) that the acceleration aplitude is a =ω x, where ω = k/( + M) is the angular frequency (fro Eq. 15-1). Therefore, using Newton s second law, we have which leads to x = 0.. a k = µ g = + M x µ g s s

26 6. We wish to find the effective spring constant for the cobination of springs shown in Fig We do this by finding the agnitude F of the force exerted on the ass when the total elongation of the springs is x. Then k eff = F/ x. Suppose the left-hand spring is elongated by x and the right-hand spring is elongated by x r. The left-hand spring exerts a force of agnitude k x on the right-hand spring and the right-hand spring exerts a force of agnitude k x r on the left-hand spring. By Newton s third law these ust be equal, so x = x r. The two elongations ust be the sae and the total elongation is twice the elongation of either spring: x = x. The left-hand spring exerts a force on the block and its agnitude is F = k x. Thus keff = k x / xr = k/. The block behaves as if it were subject to the force of a single spring, with spring constant k/. To find the frequency of its otion replace k eff in f = 1/ π k / a f eff with k/ to obtain f = 1 k. π With = 0.45 kg and k = 6430 N/, the frequency is f = 18. Hz.

27 7. (a) We interpret the proble as asking for the equilibriu position; that is, the block is gently lowered until forces balance (as opposed to being suddenly released and allowed to oscillate). If the aount the spring is stretched is x, then we exaine force-coponents along the incline surface and find 14.0sin 40.0 kx = g sin θ x = = at equilibriu. The calculator is in degrees ode in the above calculation. The distance fro the top of the incline is therefore ( ) = (b) Just as with a vertical spring, the effect of gravity (or one of its coponents) is siply to shift the equilibriu position; it does not change the characteristics (such as the period) of siple haronic otion. Thus, Eq applies, and we obtain T = π = s. 10

28 8. (a) The energy at the turning point is all potential energy: E kx where E = 1.00 J and x = Thus, = E k = 00 N/. x (b) The energy as the block passes through the equilibriu position (with speed v = 1.0 /s) is purely kinetic: (c) Eq (divided by π) yields = 1 = E E v =1.39 kg. v = 1 f = 1 π k = 191. Hz.

29 9. When the block is at the end of its path and is oentarily stopped, its displaceent is equal to the aplitude and all the energy is potential in nature. If the spring potential energy is taken to be zero when the block is at its equilibriu position, then c ha f = 1 = 1 E kx N/ 0.04 = J.

30 30. The total echanical energy is equal to the (axiu) kinetic energy as it passes through the equilibriu position (x = 0): 1 v = 1 (.0 kg)(0.85 /s) = 0.7 J. Looking at the graph in the proble, we see that U(x=10)=0.5 J. Since the potential function has 3 the for U( x) = bx, the constant is b = J/c. Thus, U(x) = 0.7 J when x = 1 c. (a) Thus, the ass does turn back before reaching x = 15 c. (b) It turns back at x = 1 c.

31 = The total energy is given by E kx, where k is the spring constant and x is the aplitude. We use the answer fro part (b) to do part (a), so it is best to look at the solution for part (b) first. (a) The fraction of the energy that is kinetic is where the result fro part (b) has been used. K E U U 1 3 = =1 =1 = = 0.75 E E E 4 4 (b) When x = x the potential energy is U = kx = kx 8. The ratio is U kx 1 = 0.5. E kx = 4 = (c) Since E= 1 kx and U = 1 kx, U/E = x x. We solve x x = 1/ for x. We should get x = x /.

32 3. We infer fro the graph (since echanical energy is conserved) that the total energy in the syste is 6.0 J; we also note that the aplitude is apparently x = 1 c = 0.1. Therefore we can set the axiu potential energy equal to 6.0 J and solve for the spring constant k: 1 k x = 6.0 J k = N/.

33 33. (a) Eq (divided by π) yields f = 1 π k = N/ 5 π 500kg =. Hz.. (b) With x o = 0.500, we have U 0 1 = kx = 15 J. 0 (c) With v o = 10.0 /s, the initial kinetic energy is K 0 1 = v = 50 J. 0 (d) Since the total energy E = K o + U o = 375 J is conserved, then consideration of the energy at the turning point leads to = 1 E E kx x = = k

34 34. We note that the ratio of Eq and Eq is v/x = ωtan(ωt + φ) where ω is given by Eq Since the kinetic energy is 1 v and the potential energy is 1 kx (which ay be conveniently written as 1 ω x ) then the ratio of kinetic to potential energy is siply (v/x) /ω = tan (ωt + φ), which at t = 0 is tan φ. Since φ = π/6 in this proble, then the ratio of kinetic to potential energy at t = 0 is tan (π/6) = 1/3.

35 35. The textbook notes (in the discussion iediately after Eq. 15-7) that the acceleration aplitude is a = ω x, where ω is the angular frequency and x = is the aplitude. Thus, a = 8000 /s leads to ω = 000 rad/s. Using Newton s second law with = kg, we have F π c a fh a f I H K F= a= acos ωt+ φ = 80 N cos 000t 3 where t is understood to be in seconds. (a) Eq gives T = π/ω = s. (b) The relation v = ωx can be used to solve for v, or we can pursue the alternate (though related) approach of energy conservation. Here we choose the latter. By Eq. 15-1, the spring constant is k = ω = N/. Then, energy conservation leads to (c) The total energy is 1 kx 1 1 k kx = v v = x = 4.0 /s. = v = J. 1 (d) At the axiu displaceent, the force acting on the particle is F = kx= 4 3 ( N/)(.0 10 )=80 N. (e) At half of the axiu displaceent, F = kx= x = 1.0, and the force is 4 3 ( N/)( )=40 N.

36 36. We note that the spring constant is k = 4π 1 /T = N/. It is iportant to deterine where in its siple haronic otion (which phase of its otion) block is when the ipact occurs. Since ω = π/t and the given value of t (when the collision takes place) is one-fourth of T, then ωt = π/ and the location then of block is x = x cos(ωt + φ) where φ = π/ which gives x = x cos(π/ + π/) = x. This eans block is at a turning point in its otion (and thus has zero speed right before the ipact occurs); this eans, too, that the spring is stretched an aount of 1 c = 0.01 at this oent. To calculate its after-collision speed (which will be the sae as that of block 1 right after the ipact, since they stick together in the process) we use oentu conservation and obtain (4.0 kg)(6.0 /s)/(6.0 kg) = 4.0 /s. Thus, at the end of the ipact itself (while block 1 is still at the sae position as before the ipact) the syste (consisting now of a total ass M = 6.0 kg) has kinetic energy 1 (6.0 kg)(4.0 /s) = 48 J and potential energy 1 ( N/)(0.010 ) 10 J, eaning the total echanical energy in the syste at this stage is approxiately 58 J. When the syste reaches its new turning point (at the new aplitude X ) then this aount ust equal its (axiu) potential energy there: 1 ( ) X. Therefore, we find X = (58)/(1.97 x 10 5 ) = 0.04.

37 37. The proble consists of two distinct parts: the copletely inelastic collision (which is assued to occur instantaneously, the bullet ebedding itself in the block before the block oves through significant distance) followed by siple haronic otion (of ass + M attached to a spring of spring constant k). (a) Moentu conservation readily yields v = v/( + M). With = 9.5 g, M = 5.4 kg and v = 630 /s, we obtain v ' = 1.1 /s. (b) Since v occurs at the equilibriu position, then v = v for the siple haronic otion. The relation v = ωx can be used to solve for x, or we can pursue the alternate (though related) approach of energy conservation. Here we choose the latter: which siplifies to x v 1 + M v = kx + M = kx ( )( ') ( ) ( + M ) 3 v ( kg)(630 /s) = = = 3 k + (6000 N/)( kg + 5.4kg) ( M)

38 38. Fro Eq (in absolute value) we find the torsion constant: τ 00. κ = = = θ 085. in SI units. With I = R /5 (the rotational inertia for a solid sphere fro Chapter 11), Eq leads to bgb R T = π = π = 1 s. κ g

39 39. (a) We take the angular displaceent of the wheel to be θ = θ cos(πt/t), where θ is the aplitude and T is the period. We differentiate with respect to tie to find the angular velocity: Ω = (π/t)θ sin(πt/t). The sybol Ω is used for the angular velocity of the wheel so it is not confused with the angular frequency. The axiu angular velocity is Ω = π = π πrad T s θ a fa f (b) When θ = π/, then θ/θ = 1/, cos(πt/t) = 1/, and = 39.5 rad / s. ( πtt) ( πtt) ( ) sin = 1 cos = 1 1 = 3 where the trigonoetric identity cos θ + sin θ = 1 is used. Thus, I KJ = π 3 Ω F πti π H K = F I F θ sin H = K π rad H G rad / s T T s During another portion of the cycle its angular speed is +34. rad/s when its angular displaceent is π/ rad. (c) The angular acceleration is a f d θ π cos ( t/ T) π α = = θ π = θ. dt T T When θ = π/4, π π α = s 4 = 14 rad/s, or α = 14 rad/s.

40 40. (a) Referring to Saple Proble 15-5, we see that the distance between P and C is 1 1 h = 3 L L = 6 L. The parallel axis theore (see Eq ) leads to And Eq gives F H = 1 + = = 1 I L h L L I K I L / 9 T = π = π = π gh gl / 6 L 3g which yields T = 1.64 s for L = (b) We note that this T is identical to that coputed in Saple Proble As far as the characteristics of the periodic otion are concerned, the center of oscillation provides a pivot which is equivalent to that chosen in the Saple Proble (pivot at the edge of the stick).

41 41. We require Lo T = π = π g I gh siilar to the approach taken in part (b) of Saple Proble 15-5, but treating in our case a ore general possibility for I. Canceling π, squaring both sides, and canceling g leads directly to the result; L o = I/h.

42 4. (a) Coparing the given expression to Eq (after changing notation x θ ), we see that ω = 4.43 rad/s. Since ω = g/l then we can solve for the length: L = (b) Since v = ωx = ωlθ = (4.43 rad/s)(0.499 )( rad) and = kg, then we can find the axiu kinetic energy: 1 v = J.

43 1 43. (a) A unifor disk pivoted at its center has a rotational inertia of Mr, where M is its ass and r is its radius. The disk of this proble rotates about a point that is displaced fro its center by r+ L, where L is the length of the rod, so, according to the parallel-axis 1 1 theore, its rotational inertia is Mr + M( L+ r). The rod is pivoted at one end and has a rotational inertia of L /3, where is its ass. The total rotational inertia of the disk and rod is 1 1 I = Mr + M( L+ r) + L = (0.500kg)(0.100) + (0.500kg)( ) + (0.70kg)(0.500) 3 = 0.05kg. (b) We put the origin at the pivot. The center of ass of the disk is d = L+ r= = away and the center of ass of the rod is r = L / = ( )/ = away, on the sae line. The distance fro the pivot point to the center of ass of the disk-rod syste is a fa f a fa f M d + r kg kg 0.50 d = = M kg kg (c) The period of oscillation is = T I 0.05 kg = π = π = 1.50 s. + gd (0.500 kg kg)(9.80 /s )(0.447 ) ( M )

44 44. We use Eq and the parallel-axis theore I = I c + h where h = d. For a solid disk of ass, the rotational inertia about its center of ass is I c = R /. Therefore, R / d R d (.35 c) +(1.75 c) T π s. gd π + = = = gd π = (980 c/s )(1.75 c)

45 45. We use Eq and the parallel-axis theore I = I c + h where h = d, the unknown. For a eter stick of ass, the rotational inertia about its center of ass is I c = L /1 where L = 1.0. Thus, for T =.5 s, we obtain T = π L / 1 + d gd L d = π +. 1gd g Squaring both sides and solving for d leads to the quadratic forula: a f a f 4 gt/ π ± d T/ π L /3 d =. Choosing the plus sign leads to an ipossible value for d (d = 1.5 > L). If we choose the inus sign, we obtain a physically eaningful result: d =

46 46. Fro Eq. 15-8, we find the length of the pendulu when the period is T = 8.85 s: L = gt. 4 π The new length is L = L d where d = The new period is which yields T = 8.77 s. L' L d T d T ' = π = π = π g g g 4π g

47 47. To use Eq we need to locate the center of ass and we need to copute the rotational inertia about A. The center of ass of the stick shown horizontal in the figure is at A, and the center of ass of the other stick is 0.50 below A. The two sticks are of 1 equal ass so the center of ass of the syste is h = (0.50 ) = 0.5 below A, as shown in the figure. Now, the rotational inertia of the syste is the su of the rotational inertia I 1 of the stick shown horizontal in the figure and the rotational inertia I of the stick shown vertical. Thus, we have I = I + = = 5 1 I ML ML ML where L = 1.00 and M is the ass of a eter stick (which cancels in the next step). Now, with = M (the total ass), Eq yields 5 1 ML T = π = π Mgh 5L 6g where h = L/4 was used. Thus, T = 1.83 s.

48 48. (a) The rotational inertia of a unifor rod with pivot point at its end is I = L /1 + L = 1/3ML. Therefore, Eq leads to so that L = (b) By energy conservation 1 3 ML 3gT T = π Mg L 8π E botto of swing ( ) K = E = U end of swing where U = Mg ( 1 cos θ ) with being the distance fro the axis of rotation to the center of ass. If we use the sall angle approxiation ( cosθ 1 1 θ with θ in radians (Appendix E)), we obtain U L 1 = θ b gb g F H G I K J F H G where θ = 0.17 rad. Thus, K = U = J. If we calculate (1 cosθ) straightforwardly (without using the sall angle approxiation) then we obtain within 0.3% of the sae answer. I K J

49 49. If the torque exerted by the spring on the rod is proportional to the angle of rotation of the rod and if the torque tends to pull the rod toward its equilibriu orientation, then the rod will oscillate in siple haronic otion. If τ = Cθ, where τ is the torque, θ is the angle of rotation, and C is a constant of proportionality, then the angular frequency of oscillation is ω = C/ I and the period is T = π/ ω = π I / C, where I is the rotational inertia of the rod. The plan is to find the torque as a function of θ and identify the constant C in ters of given quantities. This iediately gives the period in ters of given quantities. Let 0 be the distance fro the pivot point to the wall. This is also the equilibriu length of the spring. Suppose the rod turns through the angle θ, with the left end oving away fro the wall. This end is now (L/) sin θ further fro the wall and has oved a distance (L/)(1 cos θ) to the right. The length of the spring is now ( L/ ) ( 1 cos ) θ + [ + ( L/ )sin θ]. If the angle θ is sall we ay approxiate cos 0 θ with 1 and sin θ with θ in radians. Then the length of the spring is given by 0 + Lθ / and its elongation is x = Lθ/. The force it exerts on the rod has agnitude F = k x = klθ/. Since θ is sall we ay approxiate the torque exerted by the spring on the rod by τ = FL/, where the pivot point was taken as the origin. Thus τ = (kl /4)θ. The constant of proportionality C that relates the torque and angle of rotation is C = kl /4. The rotational inertia for a rod pivoted at its center is I = L /1, where is its ass. See Table 10-. Thus the period of oscillation is I L / 1 T = π = π = π C kl / 4. 3k With = kg and k = 1850 N/, we obtain T = s.

50 50. (a) For the physical pendulu we have T = π I gh = π I co + h gh. If we substitute r for h and use ite (i) in Table 10-, we have π a + b T = + r g 1r In the figure below, we plot T as a function of r, for a = 0.35 and b = (b) The iniu of T can be located by setting its derivative to zero, dt / dr = 0. This yields a + b (0.35 ) + (0.45 ) r = = = (c) The direction fro the center does not atter, so the locus of points is a circle around the center, of radius [(a + b )/1] 1/.

51 51. This is siilar to the situation treated in Saple Proble 15-5, except that O is no longer at the end of the stick. Referring to the center of ass as C (assued to be the geoetric center of the stick), we see that the distance between O and C is h = x. The parallel axis theore (see Eq ) leads to And Eq gives F HG I KJ = 1 + = I L h L x. c h c h. L I 1 + x T = π = π = π gh gx L + 1x 1gx (a) Miniizing T by graphing (or special calculator functions) is straightforward, but the standard calculus ethod (setting the derivative equal to zero and solving) is soewhat awkward. We pursue the calculus ethod but choose to work with 1gT /π instead of T (it should be clear that 1gT /π is a iniu whenever T is a iniu). 1gT d L e d x x π j d + 1 i L = 0 = = + 1 dx dx x which yields x= L/ 1 = (1.85 )/ 1 = 0.53 as the value of x which should produce the sallest possible value of T. (b) With L = 1.85 and x = 0.53, we obtain T =.1 s fro the expression derived in part (a).

52 5. Consider that the length of the spring as shown in the figure (with one of the block s corners lying directly above the block s center) is soe value L (its rest length). If the (constant) distance between the block s center and the point on the wall where the spring attaches is a distance r, then rcosθ = d/ and rcosθ = L defines the angle θ easured fro a line on the block drawn fro the center to the top corner to the line of r (a straight line fro the center of the block to the point of attachent of the spring on the wall). In ters of this angle, then, the proble asks us to consider the dynaics that results fro increasing θ fro its original value θ o to θ o + 3º and then releasing the syste and letting it oscillate. If the new (stretched) length of spring is L (when θ = θ o + 3º), then it is a straightforward trigonoetric exercise to show that (L ) = r + (d/ ) r(d/ )cos(θ o + 3º) = L + d d cos(3º)+ Ldsin(3º). since θ o = 45º. The difference between L (as deterined by this expression) and the original spring length L is the aount the spring has been stretched (denoted here as x ). If one plots x versus L over a range that sees reasonable considering the figure shown in the proble (say, fro L = 0.03 to L = 0.10 ) one quickly sees that x 0.00 is an excellent approxiation (and is very close to what one would get by approxiating x as the arc length of the path ade by that upper block corner as the block is turned through 3º, even though this latter procedure should in principle overestiate x ). Using this value of x with the given spring constant leads to a potential energy of 1 k x = J. Setting this equal to the kinetic energy the block has as it passes back through the initial position, we have J = 1 I ω where ω is the axiu angular speed of the block (and is not to be confused with the angular frequency ω of the oscillation, though they are related by ω = θ o ω if θ o is expressed in radians). The rotational inertia of the block is I = 1 6 Md = kg. Thus, we can solve the above relation for the axiu angular speed of the block: ω = (0.0096)/ = 1.81 rad/s. Therefore the angular frequency of the oscillation is ω = ω /θ o = 34.6 rad/s. Using Eq. 15-5, then, the period is T = 0.18 s.

53 53. Replacing x and v in Eq and Eq with θ and dθ/dt, respectively, we identify 4.44 rad/s as the angular frequency ω. Then we evaluate the expressions at t = 0 and divide the second by the first: dθ/dt θ = ω tanφ. at t = 0 (a) The value of θ at t = 0 is rad, and the value of dθ/dt then is 0.00 rad/s, so we are able to solve for the phase constant: φ = tan 1 [0.00/( x 4.44)] = rad. (b) Once φ is deterined we can plug back in to θ o = θ cosφ to solve for the angular aplitude. We find θ = rad.

54 54. We note that the initial angle is θ o = 7º = 0.1 rad (though it turns out this value will cancel in later calculations). If we approxiate the initial stretch of the spring as the arclength that the corresponding point on the plate has oved through (x = rθ o where r = 0.05 ) then the initial potential energy is approxiately 1 kx = J. This should equal to the kinetic energy of the plate ( 1 Iω where this ω is the axiu angular speed of the plate, not the angular frequency ω). Noting that the axiu angular speed of the plate is ω = ωθ o where ω = π/t with T = 0 s = 0.0 s as deterined fro the graph, then we can find the rotational inertial fro 1 I ω = J. Thus, I = kg 5.

55 55. (a) The period of the pendulu is given by T = π I / gd, where I is its rotational inertia, =.1 g is its ass, and d is the distance fro the center of ass to the pivot point. The rotational inertia of a rod pivoted at its center is L /1 with L =.0. According to the parallel-axis theore, its rotational inertia when it is pivoted a distance d fro the center is I = L /1 + d. Thus, T = π L ( / 1 + d) = π gd L + 1d 1gd. Miniizing T with respect to d, dt/d(d)=0, we obtain d = L/ 1. Therefore, the iniu period T is T L + 1( L/ 1) L (.0 ) in π π π = = = =.6 s. 1 gl ( / 1) 1g 1(9.80 /s ) (b) If d is chosen to iniize the period, then as L is increased the period will increase as well. (c) The period does not depend on the ass of the pendulu, so T does not change when increases.

56 56. The table of oents of inertia in Chapter 11, plus the parallel axis theore found in that chapter, leads to I P = 1 MR + Mh = 1 (.5 kg)(0.1 ) + (.5 kg)(0.97 ) =.41 kg ² where P is the hinge pin shown in the figure (the point of support for the physical pendulu), which is a distance h = away fro the center of the disk. (a) Without the torsion spring connected, the period is T = π I P Mgh =.00 s. (b) Now we have two restoring torques acting in tande to pull the pendulu back to the vertical position when it is displaced. The agnitude of the torque-su is (Mgh + κ)θ = I P α, where the sall angle approxiation (sinθ θ in radians) and Newton s second law (for rotational dynaics) have been used. Making the appropriate adjustent to the period forula, we have T = π I P Mgh + κ. The proble stateent requires T = T s. Thus, T = ( )s = 1.50 s. Consequently, κ = 4π T I P Mgh = 18.5 N /rad.

57 57. Referring to the nubers in Saple Proble 15-7, we have = 0.5 kg, b = kg/s and T = 0.34 s. Thus, when t = 0T, the daping factor becoes e = e b. gb gb. g/ b. g = bt

58 58. Since the energy is proportional to the aplitude squared (see Eq. 15-1), we find the fractional change (assued sall) is E E E de E dx = = xdx dx =. x x x Thus, if we approxiate the fractional change in x as dx /x, then the above calculation shows that ultiplying this by should give the fractional energy change. Therefore, if x decreases by 3%, then E ust decrease by 6.0 %.

59 59. (a) We want to solve e bt/ = 1/3 for t. We take the natural logarith of both sides to obtain bt/ = ln(1/3). Therefore, t = (/b) ln(1/3) = (/b) ln 3. Thus, (b) The angular frequency is a f = 1.50 kg t 3= kg / s ln s k b 800. N/ kg / s ω = = = 31. rad / s kg kg The period is T = π/ω = (π)/(.31 rad/s) =.7 s and the nuber of oscillations is a a f f t/t = (14.3 s)/(.7 s) = 5.7.

60 60. (a) Fro Hooke s law, we have ( 500 kg)( 9.8 /s ) k = 10c = N/c. (b) The aplitude decreasing by 50% during one period of the otion iplies e bt 1 π = where T = ω. Since the proble asks us to estiate, we let ω ω = k/. That is, we let ω N/ 500 kg 9.9 rad / s, so that T 0.63 s. Taking the (natural) log of both sides of the above equation, and rearranging, we find b g b = b ln 0.69 = T kg / s 0.63 Note: if one worries about the ω ω approxiation, it is quite possible (though essy) to use Eq in its full for and solve for b. The result would be (quoting ore figures than are significant) g b = ln k ( ln ) +4π = 1086 kg / s which is in good agreeent with the value gotten the easy way above.

61 61. With ω = π/t then Eq can be used to calculate the angular frequencies for the given pendulus. For the given range of.00 < ω < 4.00 (in rad/s), we find only two of the given pendulus have appropriate values of ω: pendulu (d) with length of 0.80 (for which ω = 3.5 rad/s) and pendulu (e) with length of 1. (for which ω =.86 rad/s).

62 6. (a) We set ω = ω d and find that the given expression reduces to x = F /bω at resonance. (b) In the discussion iediately after Eq. 15-6, the book introduces the velocity aplitude v = ωx. Thus, at resonance, we have v = ωf /bω = F /b.

63 63. With M = 1000 kg and = 8 kg, we adapt Eq to this situation by writing π k ω = = T M + 4. If d = 4.0 is the distance traveled (at constant car speed v) between ipulses, then we ay write T = v/d, in which case the above equation ay be solved for the spring constant: πv k πv = k= ( M + 4 ). d M + 4 d Before the people got out, the equilibriu copression is x i = (M + 4)g/k, and afterward it is x f = Mg/k. Therefore, with v = 16000/3600 = 4.44 /s, we find the rise of the car body on its suspension is = 4 = 4 g g d xi xf = k M +4 πv F H I K

64 64. We note (fro the graph) that a = ω x = 4.00 c/s. Also the value at t = 0 is a o = 1.00 c/s. Then Eq leads to φ = cos 1 ( 1.00/4.00) = +1.8 rad or 4.46 rad. The other root (+4.46 rad) can be rejected on the grounds that it would lead to a negative slope at t = 0.

65 65. (a) Fro the graph, we find x = 7.0 c = 0.070, and T = 40 s = s. Thus, the angular frequency is ω = π/t = 157 rad/s. Using = 0.00 kg, the axiu kinetic energy is then 1 v = 1 ω x = 1. J. (b) Using Eq. 15-5, we have f = ω/π = 50 oscillations per second. Of course, Eq. 15- can also be used for this.

66 66. (a) Fro the graph we see that x = 7.0 c = and T = 40 s = s. The axiu speed is x ω = x π/t = 11 /s. (b) The axiu acceleration is x ω = x (π/t) = /s.

67 67. Setting 15 J (0.015 J) equal to the axiu kinetic energy leads to v ax = /s. Then one can use either an exact approach using v ax = gl(1 cos(θ ax )) or the SHM approach where v ax = Lω ax = Lωθ ax = L g/l θ ax to find L. Both approaches lead to L = 1.53.

68 68. Its total echanical energy is equal to its axiu potential energy 1 kx, and its potential energy at t = 0 is 1 kx o where x o = x cos(π/5) in this proble. The ratio is therefore cos (π/5) = = 65.5%.

69 69. (a) We note that ω = k/ = 1500/0.055 = rad/s. We consider the ost direct path in each part of this proble. That is, we consider in part (a) the otion directly fro x 1 = x at tie t 1 to x = x at tie t (as opposed to, say, the block oving fro x 1 = x through x = x, through x = 0, reaching x = x and after returning back through x = 0 then getting to x = x ). Eq leads to ωt 1 + φ = cos 1 (0.800) = rad ωt + φ = cos 1 (0.600) = 0.97 rad. Subtracting the first of these equations fro the second leads to ω(t t 1 ) = = rad. Using the value for ω coputed earlier, we find t t 1 = s. (b) Let t 3 be when the block reaches x = 0.800x in the direct sense discussed above. Then the reasoning used in part (a) leads here to and thus to t 3 t 1 = s. ω(t 3 t 1 ) = ( ) rad = rad

70 70. Since ω = πf where f =. Hz, we find that the angular frequency is ω = 13.8 rad/s. Thus, with x = 0.010, the acceleration aplitude is a = x ω = 1.91 /s. We set up a ratio: a F HG I F I KJ H K a = = 1.91 =0.19. g g 9.8 g g

71 71. (a) Assue the bullet becoes ebedded and oves with the block before the block oves a significant distance. Then the oentu of the bullet-block syste is conserved during the collision. Let be the ass of the bullet, M be the ass of the block, v 0 be the initial speed of the bullet, and v be the final speed of the block and bullet. Conservation of oentu yields v 0 = ( + M)v, so v0 v = = + M a fa f kg 150 / s kg kg =1.85 /s. When the block is in its initial position the spring and gravitational forces balance, so the spring is elongated by Mg/k. After the collision, however, the block oscillates with siple haronic otion about the point where the spring and gravitational forces balance with the bullet ebedded. At this point the spring is elongated a distance = am+ f g/ k, soewhat different fro the initial elongation. Mechanical energy is conserved during the oscillation. At the initial position, just after the bullet is ebedded, the kinetic energy is 1 ( M+ ) v and the elastic potential energy is 1 kmg k potential energy to be zero at this point. When the block and bullet reach the highest point in their otion the kinetic energy is zero. The block is then a distance y above the position where the spring and gravitational forces balance. Note that y is the aplitude of the otion. The spring is copressed by y, so the elastic potential energy is ( / ). We take the gravitational 1 ky ( ). The gravitational potential energy is (M + )gy. Conservation of echanical energy yields 1 F H a f b g a f M v k Mg k We substitute = M+ g k y = I K = 1 ky + M+ gy. a f /. Algebraic anipulation leads to a+ Mfv g am+ f k k a kg kgfa185. / sf a kgfc9. 8 / s h = 40 a. kgf kg 500 N/ ( 500 N/) = (b) The original energy of the bullet is E0 = v0 = ( kg)( 150 / s) = 563 J. The kinetic energy of the bullet-block syste just after the collision is a f a fa f E= 1 + M v = kg kg 1.85 / s = 6.94 J.

72 Since the block does not ove significantly during the collision, the elastic and gravitational potential energies do not change. Thus, E is the energy that is transferred. The ratio is E/E 0 = (6.94 J)/(563 J) = or 1.3%.

73 7. (a) The rotational inertia of a hoop is I = R, and the energy of the syste becoes 1 1 E = Iω + kx and θ is in radians. We note that rω = v (where v = dx/dt). Thus, the energy becoes F R E = v kx H G 1 I 1 r K J + which looks like the energy of the siple haronic oscillator discussed in 15-4 if we identify the ass in that section with the ter R /r appearing in this proble. Making this identification, Eq yields k ω = = R / r r R k. (b) If r = R the result of part (a) reduces to ω = k/. (c) And if r = 0 then ω = 0 (the spring exerts no restoring torque on the wheel so that it is not brought back towards its equilibriu position).

74 73. (a) The graphs suggest that T = 0.40 s and κ = 4/0. = 0.0 N /rad. With these values, Eq can be used to deterine the rotational inertia: I = κt /4π = kg.. (b) We note (fro the graph) that θ ax = 0.0 rad. Setting the axiu kinetic energy ( 1 Iω ax ) equal to the axiu potential energy (see the hint in the proble) leads to ω ax = θ ax κ/i = 3.14 rad/s.

75 74. (a) Let v ax be the axiu speed attained during the first oscillation. By taking the derivative of Eq and using the approxiations available to us fro the fact that b «k (see section 15-8), then we have v ax ωx e bt/ where ω = k/. The axiu x occurs at a different tie than the axiu speed so that when we consider the ratio bv ax /kx ax = (b/k)ωe b t/ we ust account for that tie difference through the t ter (corresponding to a quarter-period) in the exponential. Thus, this expression can be reduced to bv ax kx ax = b k exp( πb/(4 k )). Using the data fro that Saple Proble (converted to SI units) we get for this ratio. (b) Due to the sall level of daping in this proble, the answer is no.

76 75. (a) Hooke s law readily yields k = (15 kg)(9.8 /s )/(0.1 ) = 15 N/. Rounding to three significant figures, the spring constant is therefore 1.3 kn/. (b) We are told f =.00 Hz =.00 cycles/sec. Since a cycle is equivalent to π radians, we have ω = π(.00) = 4π rad/s (understood to be valid to three significant figures). Using Eq. 15-1, we find k 15 N/ ω = = = 7.76kg. ( 4 π rad/s) Consequently, the weight of the package is g = 76.0 N.

77 76. (a) The proble gives the frequency f = 440 Hz, where the SI unit abbreviation Hz stands for Hertz, which eans a cycle-per-second. The angular frequency ω is siilar to frequency except that ω is in radians-per-second. Recalling that π radians are equivalent to a cycle, we have ω = πf rad/s. (b) In the discussion iediately after Eq. 15-6, the book introduces the velocity aplitude v = ωx. With x = and the above value for ω, this expression yields v =.1 /s. (c) In the discussion iediately after Eq. 15-7, the book introduces the acceleration aplitude a = ω x, which (if the ore precise value ω = 765 rad/s is used) yields a = 5.7 k/s.

78 77. We use v = ωx = πfx, where the frequency is 180/(60 s) = 3.0 Hz and the aplitude is half the stroke, or x = Thus, v = π(3.0 Hz)(0.38 ) = 7. /s.

79 78. (a) The textbook notes (in the discussion iediately after Eq. 15-7) that the acceleration aplitude is a = ω x, where ω is the angular frequency (ω = π f since there are π radians in one cycle). Therefore, in this circustance, we obtain b a fga f 4 a = π 1000 Hz = / s. (b) Siilarly, in the discussion after Eq. 15-6, we find v = ωx so that c b ghb g v = π 1000 Hz =.5 / s. (c) Fro Eq. 15-8, we have (in absolute value) c b ghb g 3 a = π 1000 Hz = / s. (d) This can be approached with the energy ethods of 15-4, but here we will use trigonoetric relations along with Eq and Eq Thus, allowing for both roots steing fro the square root, v x sin ( ωt+ φ) =± 1 cos ( ωt+ φ) =± 1. ωx x Taking absolute values and siplifying, we obtain ( ) v = π f x x = π =. /s.

80 79. The agnitude of the downhill coponent of the gravitational force acting on each ore car is b gc h θ w x = kg 9.8 / s sin where θ = 30 (and it is iportant to have the calculator in degrees ode during this proble). We are told that a downhill pull of 3ω x causes the cable to stretch x = Since the cable is expected to obey Hooke s law, its spring constant is = 3 wx 5 k = N/. x (a) Noting that the oscillating ass is that of two of the cars, we apply Eq (divided by π). f = π =. N/ 0000 kg =1.1 Hz. (b) The difference between the equilibriu positions of the end of the cable when supporting two as opposed to three cars is = 3 w x wx x = k

81 80. (a) First consider a single spring with spring constant k and unstretched length L. One end is attached to a wall and the other is attached to an object. If it is elongated by x the agnitude of the force it exerts on the object is F = k x. Now consider it to be two springs, with spring constants k 1 and k, arranged so spring 1 is attached to the object. If spring 1 is elongated by x 1 then the agnitude of the force exerted on the object is F = k 1 x 1. This ust be the sae as the force of the single spring, so k x = k 1 x 1. We ust deterine the relationship between x and x 1. The springs are unifor so equal unstretched lengths are elongated by the sae aount and the elongation of any portion of the spring is proportional to its unstretched length. This eans spring 1 is elongated by x 1 = CL 1 and spring is elongated by x = CL, where C is a constant of proportionality. The total elongation is x = x 1 + x = C(L 1 + L ) = CL (n + 1), where L 1 = nl was used to obtain the last for. Since L = L 1 /n, this can also be written x = CL 1 (n + 1)/n. We substitute x 1 = CL 1 and x = CL 1 (n + 1)/n into k x = k 1 x 1 and solve for k 1. With k = 8600 N/ and n = L 1 /L = 0.70, we obtain (8600 N/) 0886 N/ N/ n k + 1 k + = = = n 0.70 (b) Now suppose the object is placed at the other end of the coposite spring, so spring exerts a force on it. Now k x = k x. We use x = CL and x = CL (n + 1), then solve for k. The result is k = k(n + 1). k n k 4 = ( + 1) = ( )(8600 N/) = 1460 N/ N/ (c) To find the frequency when spring 1 is attached to ass, we replace k in a1/ πf k/ with k(n + 1)/n. With f = a1/ πf k/, we obtain, for f = 00 Hz and n = 0.70 f 1 ( n+ 1) k n π n n = = f = (00 Hz) = Hz. (d) To find the frequency when spring is attached to the ass, we replace k with k(n + 1) to obtain 1 ( n+ 1) k f n f π = = + 1 = (00 Hz) =.6 10 Hz.