Energy in simple Harmonic motion




We can learn even more about simple harmonic motion by using energy considerations. The only horizontal force on the body in SHM in Figs. 14.2 and 14.13 is the conservative force exerted by an ideal spring. The vertical forces do no work, so the total mechanical energy of the system is conserved. We also assume that the mass of the spring itself is negligible.

The kinetic energy of the body is K = 1 /2 mv2 and the potential energy of the spring is U = 1 /2 kx2 , just as in Section 7.2. There are no nonconservative forces that do work, so the total mechanical energy E = K + U is conserved:

Energy in simple Harmonic motion

 

(Since the motion is one-dimensional, v2 = vx 2.) The total mechanical energy E is also directly related to the amplitude A of the motion. When the body reaches the point x = A, its maximum displacement from equilibrium, it momentarily stops as it turns back toward the equilibrium position. That is, when x = A (or -A), vx = 0. At this point the energy is entirely potential, and E = 1 2 kA2 . Because E is constant, it is equal to 1/ 2 kA2 at any other point. Combining this expression with Eq. (14.20), we get

Energy in simple Harmonic motion

 

 

Energy in simple Harmonic motion

 

 

 

 

 

 

 

We can verify this equation by substituting x and vx from Eqs. (14.13) and (14.15) and using w2 = k/m from Eq. (14.9):

Energy in simple Harmonic motion

 

 

Hence our expressions for displacement and velocity in SHM are consistent with energy conservation, as they must be.

We can use Eq. (14.21) to solve for the velocity vx of the body at a given displacement x:

Energy in simple Harmonic motion

 

 

The ± sign means that at a given value of x the body can be moving in either direction. For example, when x = ±A/2,

Energy in simple Harmonic motion

 

 

Equation (14.22) also shows that the maximum speed vmax occurs at x = 0. Using Eq. (14.10), v = (k)1/2/m, we find that

Energy in simple Harmonic motion

 

 

This agrees with Eq. (14.15): wx oscillates between -wA and +wA

Energy in simple Harmonic motion



Frequently Asked Questions

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Ans: We still need to find the displacement x as a function of time for a harmonic oscillator. Equation (14.4) for a body in SHM along the x-axis is identical to Eq. (14.8) for the x-coordinate of the reference point in uniform circular motion with constant angular speed v = 2k/m view more..
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Ans: the period and frequency of simple harmonic motion are completely determined by the mass m and the force constant k. In simple harmonic motion the period and frequency do not depend on the amplitude A. view more..
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Ans: To explore the properties of simple harmonic motion, we must express the displacement x of the oscillating body as a function of time, x1t2. view more..
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Ans: We can learn even more about simple harmonic motion by using energy considerations. The only horizontal force on the body in SHM in Figs. 14.2 and 14.13 is the conservative force exerted by an ideal spring. The vertical forces do no work, so the total mechanical energy of the system is conserved. We also assume that the mass of the spring itself is negligible. view more..
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Ans: the energy quantities E, K, and U at x = 0, x = ±A/2, and x = ±A. Figure 14.15 is a graphical display of Eq. (14.21); energy (kinetic, potential, and total) is plotted vertically and the coordinate x is plotted horizontally. The parabolic curve in Fig. 14.15a represents the potential energy U = 1/2 kx2 . The horizontal line represents the total mechanical energy E, which is constant and does not vary with x. view more..
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Ans: PROBLEM SOLVING STRATEGY ON ENERGY MOMENTUM OF SHM view more..
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Ans: So far, we’ve looked at a grand total of one situation in which simple harmonic motion (SHM) occurs: a body attached to an ideal horizontal spring. But SHM can occur in any system in which there is a restoring force that is directly proportional to the displacement from equilibrium, as given by Eq. (14.3), Fx = -kx view more..
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Ans: A mechanical watch keeps time based on the oscillations of a balance wheel (Fig. 14.19). The wheel has a moment of inertia I about its axis. A coil spring exerts a restoring torque tz that is proportional to the angular displacement u from the equilibrium position. We write tz = -ku, where k (the Greek letter kappa) is a constant called the torsion constant. Using the rotational analog of Newton’s second law for a rigid body, gtz = Iaz = I d2 u>dt2 view more..
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Ans: The following discussion of the vibrations of molecules uses the binomial theorem. If you aren’t familiar with this theorem, you should read about it in the appropriate section of a math textbook. view more..
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Ans: A simple pendulum is an idealized model consisting of a point mass suspended by a massless, unstretchable string. When the point mass is pulled to one side of its straight-down equilibrium position and released, it oscillates about the equilibrium position. view more..
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Ans: A physical pendulum is any real pendulum that uses an extended body, as contrasted to the idealized simple pendulum with all of its mass concentrated at a point. F view more..




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