phys124s11-hw05

phys124s11-hw05 - HW 5 Solutions Energy of Harmonic...

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1 HW 5 Solutions Energy of Harmonic Oscillators As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horizontally moving block attached to a spring. Note that, since the gravitational potential energy is not changing in this case, it can be excluded from the calculations. For such a system, the potential energy is stored in the spring and is given by 2 1 2 Uk x = , where k is the force constant of the spring and x is the distance from the equilibrium position. The kinetic energy of the system is, as always, 2 1 2 Km v = , where m is the mass of the block and v is the speed of the block. We will also assume that there are no resistive forces; that is, E=const . Consider a harmonic oscillator at four different moments, labeled A, B, C, and D, as shown in the figure . Assume that the force constant k , the mass of the block, m , and the amplitude of vibrations, A , are given. Answer the following questions. a) Which moment corresponds to the maximum potential energy of the system? A B C D Note that the sign of x does not matter, just its magnitude, because U is a quadratic form of x . b) Which moment corresponds to the minimum kinetic energy of the system? Again, A.
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2 When the block is displaced a distance A from equilibrium, the spring is stretched (or compressed) the most, and the block is momentarily at rest ( K =0) Therefore, the total energy at this moment coincides with the maximum potential energy 2 1 2 Uk A = . In general, the mechanical energy of a harmonic oscillator equals its potential energy at the maximum or minimum displacement. c) Consider the block in the process of oscillating. If the kinetic energy of the block is increasing, the block must be at the equilibrium position. at the amplitude displacement. moving to the right. moving to the left. moving away from equilibrium. moving toward equilibrium. d) Which moment corresponds to the maximum kinetic energy of the system? A B C D e) Which moment corresponds to the minimum potential energy of the system? A B C D When the block is at the equilibrium position, the spring is not stretched (or compressed) at all. At that moment, of course, U =0. Meanwhile, the block is at its maximum speed ( v max ). The maximum kinetic energy can then be written as 2 max max 1 2 Km v = . Recall that E=K+U and that U =0 at the equilibrium position. Therefore, 2 max 1 2 E mv = . Recalling what we found out before, 2 1 2 E kA = ,
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3 we can now conclude that 22 max 11 kA mv = , or max k vA A m ω == . f) At which moment is K=U ? A B C D g) Find the kinetic energy K of the block at the moment labeled B. () () () 2 max 3 2 8 A Kt E Ut U Ut kA k ⎛⎞ =− = = = ⎜⎟ ⎝⎠ Position, Velocity, and Acceleration of an Oscillator Learning Goal: To learn to find kinematic variables from a graph of position vs. time.
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phys124s11-hw05 - HW 5 Solutions Energy of Harmonic...

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