Harmonic Oscillator Acceleration Learning Goal: To understand the application of

Question

Harmonic Oscillator Acceleration Learning Goal: To understand the application of the general harmonic equation to finding the acceleration of a spring oscillator as a function of time. One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x=0. The length of the relaxed spring is L. (Figure 1) The block is slowly pulled from its equilibrium position to some position xinit>0 along the x axis. At time t=0 , the block is released with zero initial velocity. The goal of this problem is to determine the acceleration of the block a(t) as a function of time in terms of k, m, and xinit. It is known that a general solution for the position of a harmonic oscillator is x(t)=Ccos(t)+Ssin(t), where C, S, and are constants. (Figure 2) Your task, therefore, is to determine the values of C, S, and in terms of k, m,and xinit and then use the connection between x(t) and a(t) to find the acceleration. Part C Find the angular frequency . Express your answer in terms of k and m.

Explanation / Answer

x(t)=Ccos(t)+Ssin(t)

The velocity,

dx(t)/dt = v(t) = -C sin(t) + S cos(t)

At time t=0 , the block is released with zero initial velocity

v(0) = 0 = S

S = 0

Hence, x(t)=Ccos(t)

here C is the Amplitude of the motion

v(t) = -C sin(t)

a = dv(t)/dt = -C^2 cos(t)

For SHM, the oscillation frequency depends on the restoring force. For a mass on a spring, where the restoring force is F = -kx, this gives:

F = -kx = -kCcos(t)

This is the net force acting, so it equals ma:

F = -kCcos(t) = ma = m(-C^2 cos(t))

This gives a relationship between the angular velocity, the spring constant, and the mass:

^2 = k/m

= sqrt(k/m)

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