Let’s begin with a straightforward example of simple harmonic motion (SHM). A sp

Question

Let’s begin with a straightforward example of simple harmonic motion (SHM). A spring is mounted horizontally on an air track as in (Figure 1) , with the left end held stationary. We attach a spring balance to the free end of the spring, pull toward the right, and measure the elongation. We determine that the stretching force is proportional to the displacement and that a force of 6.0 N causes an elongation of 0.030 m. We remove the spring balance and attach a 0.50 kg object to the end, pull it a distance of 0.040 m, release it, and watch it oscillate in SHM as in (Figure 2) . Find the following quantities:

The force constant of the spring

The maximum and minimum velocities attained by the vibrating object

The maximum and minimum accelerations

The velocity and acceleration when the object has moved halfway to the center from its initial position

The kinetic energy, potential energy, and total energy in the halfway position

SOLUTION

SET UP AND SOLVE For this glider, the amplitude is A=0.040m.

Part (a): When x=0.030m, the force the spring exerts on the object is Fx=?6.0N. From Fx=?kx, the force constant k is

k=?Fxx=??6.0N0.030m=200N/m

Part (b): From vmax=k/m?????A, the maximum velocity occurs at the equilibrium position, x=0:

vmax=km????A=200N/m0.50kg??????????(0.040m)=0.80m/s

The minimum (most negative) velocity also occurs at x=0 and is the negative of the maximum value. The minimum magnitude of velocity, occurring at x=±A, is zero.

Part (c): The acceleration for any x is given by

ax=?kmx

The maximum (most positive) acceleration amax occurs at the most negative value of x—that is,x=?A; therefore,

amax=?km(?A)=?200N/m0.50kg(?0.040m)=16.0m/s2

The minimum (most negative) acceleration is ?16.0 m/s2, occurring at x=+A=+0.040m.

Part (d): The velocity vx at any position x is given by:

vx=±km????A2?x2?????????

At the halfway point, x=A/2=0.020m. We find that

vx=?200N/m0.50kg??????????(0.040m)2?(0.020m)2???????????????????????=?0.69m/s

We’ve chosen the negative square root because the object is moving in the ?x direction, from x=A toward x=0. From ax=?kmx,

ax=?kmx=?200N/m0.50kg(0.020m)=?8.0m/s2

Part (e): The energy quantities are the terms in the equation E=12kA2=12mv2x+12kx2. From part (d), when x=A/2=(0.040m)/2, vx=±0.69m/s. At this point, the kinetic energy is

K=12mv2x=12(0.50kg)(0.69m/s)2=0.12J

the potential energy is

U=12kx2=12(200N/m)(0.020m)2=0.040J

and the total energy is E=K+U=0.12J+0.040J=0.16J.

Alternatively,

E=12kA2=12(200N/m)(0.040m)2=0.16J

REFLECT The equations we've used in this solution enable us to find relationships among position, velocity, acceleration, and force. But they don’t help us to find directly how these quantities vary with time. In the next section, we’ll work out relationships that enable us to do just that and also to relate the period and frequency of the motion to the mass and force constant of the system.

Part A - Practice Problem:

If you had pulled the object out a distance of 0.056 m before releasing it, how much kinetic energy would it have at the 0.022 m mark?

Express your answer in joules to two significant figures.

Explanation / Answer

change in kinetic energy = change in potential energy

initial kinetic energy = 0 it is in the extreme position

final kineti cenergy = 1/2kx1^2-1/2kx2^2 = 1/2*200*(0.056^2-0.022^2) = 0.266 J

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