Another Method to get a Sine Wave. X = A cos θ V = Acc =

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1 LAST NAME FIRST NAME DATE PER CJ Wave Assignment 10.3 Energy & Simple Harmonic Motion Conceptual Questions 3, 4, 6, 7, 9 page 313 6, 7, 33, 34 page Tracing the movement of the mass on the end of the spring with the moving paper gives a sine wave. Another Method to get a Sine Wave X = A cos θ V = Acc = Trace displacement, velocity and acceleration.

2 Conceptual Questions 3, 4, 6, 7, 9 page The shadow moves in a simple horizontal motion. Where on the shadow s path is (a) the velocity equal to zero and (b) the acceleration equal to zero? Figures and show the velocity and the acceleration, respectively, of the shadow of a ball that undergoes uniform circular motion. The shadow undergoes simple harmonic motion. a. The velocity of the shadow is given by Equation 10.7: v A sin. The velocity of the shadow will be zero when = 0 or rad. From Figure 10.11, we see that the velocity equals zero each time the shadow reaches the right and left endpoints of its motion (that is, when the ball crosses the x axis). x b. The acceleration of the shadow is given by Equation 10.9: a A cos. The acceleration of the shadow will be zero when is / or 3 /. From Figure 10.14, we see that the acceleration equals zero each time that the shadow passes through the point x = 0 m (that is, when the ball crosses the y axis). x

3 4. A steel ball is dropped onto a concrete floor. Over and over again, it rebounds to its original height. Is this simple motion harmonic motion? Justify your answer. Simple harmonic motion is the oscillatory motion that occurs when a restoring force of the form of Equation 10., F kx, acts on an object. The force changes continually as the displacement x changes. x A steel ball is dropped onto a concrete floor. Over and over again, it rebounds to its original height. During the time when the ball is in the air, either falling down or rebounding up, the only force acting on the ball is its weight, which is nearly constant, to the extent that the ball remains near the earth s surface. Thus, the motion of the bouncing ball is not simple harmonic motion. 6. A block is attached to a horizontal spring that slides back and forth in simple harmonic motion on a frictionless horizontal surface. A second identical block is suddenly attached to the first block. The attachment is accomplished by joining the blocks at one extreme end of the oscillation cycle. The velocities of the blocks are exactly matched at the instant of joining. Explain how (a) the amplitude, (b) the frequency, and (c) the maximum speed of the oscillation change. A block is attached to a horizontal spring and slides back and forth in simple harmonic motion on a frictionless horizontal surface. A second identical block is suddenly attached to the first block when the first block is at one extreme end of the oscillation cycle. a. Since the attachment is made at one extreme end of the oscillation cycle, where the velocity is zero, the extreme end of the oscillation cycle will remain at the same point; in other words, the amplitude remains the same. b. The angular frequency of an object of mass m in simple harmonic motion at the end of a spring of force constant k is given by Equation 10.11: k/ m. Since the mass m is doubled while the force constant k remains the same, the angular frequency decreases by a factor of. The vibrational frequency f is related to by f = /( ); the vibrational frequency f will also decrease by a factor of. c. The maximum speed of oscillation is given by Equation 10.8: v max A. Since the amplitude, A, remains the same and the angular frequency,, decreases by a factor of, the maximum speed of oscillation also decreases by a factor of.

4 7. A particle is oscillating in simple harmonic motion. The time required for the particle to travel through one complete cycle is equal to the period of motion, no matter what the amplitude is. But how can this be, since larger amplitudes mean that the particle travels further? Explain. The time required for a particle in simple harmonic motion to travel through one complete cycle (the period) is independent of the amplitude of the motion, even though at larger amplitudes the particle travels further. This is possible because, at larger amplitudes, the maximum speed of the particle is greater. Thus, even though the particle must cover larger distances at larger amplitudes, it does so with greater speeds. 9. Is more potential energy stored in a spring when the spring is compressed by one centimeter than when it is stretched by the same amount? Explain. The potential energy that a spring has by virtue of being stretched or compressed is given by Equation 10.13: PE (1/)kx, where x is the amount by which the spring is stretched or compressed relative to its unstrained length. The amount of stretch or compression appears squared, so that the potential energy is positive and independent of the sign of x. Therefore, the amount of potential energy stored in a spring when it is compressed by one centimeter is the same as when it is stretched by the same amount.

5 6. A rifle fires a kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by m from its unstrained length. The pellet rises to a maximum height of 6.10 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant. Since air resistance is negligible, we can apply the principle of conservation of mechanical energy, which indicates that the total mechanical energy of the pellet/spring system is the same when the pellet comes to a momentary halt at the top of its trajectory as it is when the pellet is resting on the compressed spring. The fact that the total mechanical energy is conserved will allow us to determine the spring constant. SOLUTION The conservation of mechanical energy states that the final total mechanical energy E f is equal to the initial total mechanical energy E 0. The expression for the total mechanical energy for an object on a spring is given by Equation 10.14, so that we have mvf I f mghf ky f mv 0 I 0 mgh0 ky 0 (1) E f 0 The pellet does not rotate, so the angular speeds f and 0 are zero. Since the pellet is at rest as it sits on the spring and since the pellet comes to a momentary halt at the top of its trajectory, the translational speeds v 0 and v f are also zero. Because the spring is unstrained when the pellet reaches its maximum height, the final displacement y f of the spring is likewise zero. Thus, Equation (1) simplifies to : E 1 f 0 0 mgh mgh ky Solving this simplified energy-conservation expression for the spring constant k and noting that the pellet rises to distance of hf hi 6.10 m above its position on the compressed spring, we find that f h 0 y mg h kg 9.80 m/s 6.10 m k m 303 N/m

6 7. A.00-kg object is hanging from the end of a vertical spring. The spring constant is 50.0 N/m. The object is pulled 0.00 m downward and released from rest. Complete the table below by calculating the transitional kinetic energy, the gravitational potential energy, the potential energy, and the total mechanical energy E for each vertical position indicated. If we neglect air resistance, only the conservative forces of the spring and act on the ball. Therefore, the principle of conservation of mechanical energy applies. When the.00 kg object is hung on the end of the vertical spring, it stretches the spring by an amount y, where F mg (.00 kg)(9.80 m/s ) y 0.39 m k k 50.0 N/m (10.1) This position represents the equilibrium position of the system with the.00-kg object suspended from the spring. The object is then pulled down another 0.00 m and released from rest ( v0 0 m/s). At this point the spring is stretched by an amount of 0.39 m m = 0.59 m. This point represents the zero reference level ( h 0 m) for the gravitational potential energy. h = 0 m: The kinetic energy, the gravitational potential energy, and the potential energy at the point of release are: 1 1 mv 0 m KE (0 m/s) 0 J PE mgh mg(0 m) 0 J 1 1 ky 0 PE (50.0 N/m)(0.59 m) 8.76 J The total mechanical energy E 0 at the point of release is the sum of the three energies above: E J. h = 0.00 m: When the object has risen a distance of h 0.00 m above the release point, the spring is stretched by an amount of 0.59 m 0.00 m 0.39 m. Since the total mechanical energy is conserved, its value at this point is still E 8.76 J. The gravitational and potential energies are: PE mgh (.00 kg)(9.80 m/s )(0.00 m) 3.9 J 1 1 ky PE (50.0 N/m)(0.39 m) 3.84 J

7 Since KE PE PE E, KE E PE PE 8.76 J 3.9 J 3.84 J 1.00 J h = m: When the object has risen a distance of h m above the release point, the spring is stretched by an amount of 0.59 m m 0.19 m. At this point, the total mechanical energy is still E 8.76 J. The gravitational and potential energies are: PE mgh (.00 kg)(9.80 m/s )(0.400 m) 7.84 J 1 1 ky PE (50.0 N/m)(0.19 m) 0.9 J The kinetic energy is KE E PE PE 8.76 J 7.84 J 0.9 J 0 J The results are summarized in the table below:

8 33. A 1.1-kg object is suspended from a vertical spring whose spring constant is 10 N/m. (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight downward by an additional distance of 0.0 m and released from rest. Find the speed with which the object passes through its original position on the way up. The amount by which the spring stretches due to the weight of the kg object can be calculated using Equation 10.1 (with the variable x replaced by y), where the force F y Applied is equal to the weight of the object. The position of the object when the spring is stretched is the equilibrium position for the vertical harmonic motion of the object-spring system. a. Solving Equation 10.1 for y with F y Applied equal to the weight of the object gives Applied y mg F (1.1 kg)(9.80 m/s ) y m k k 10 N/m b. The object is then pulled down another 0.0 m and released from rest ( v0 0 m/s). At this point the spring is stretched by an amount of m m = 0.9 m. We will let this point be the zero reference level ( h 0 m) for the gravitational potential energy. are: The kinetic energy, the gravitational potential energy, and the potential energy at the point of release 1 1 mv 0 m KE (0 m/s) 0 J PE mgh mg(0 m) 0 J 1 1 ky 0 PE (10 N/m)(0.9 m) 5.0 J The total mechanical energy E 0 is the sum of these three energies, so E0 5.0 J. When the object has risen a distance of h 0.0 m above the release point, the spring is stretched by an amount of 0.9 m 0.0 m m. Since the total mechanical energy is conserved, its value at this point is still 5.0 J. Thus, E KE PE PE 1 1 E mv mgh ky 1 1 v 5.0 J (1.1 kg) (1.1 kg)(9.80 m/s )(0.0 m) (10 N/m)(0.090 m) Solving for v yields v.1 m/s.

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