Harmonic Oscillator Kinematics?

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.

The block is slowly pulled from its equilibrium position to some position x_init > 0 along the x axis. At time t=0 , the block is released with zero initial velocity.

The goal is to determine the position of the block x(t) as a function of time in terms of omega and x_init.

It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is

x(t) = C*cos(omega*t) S*sin(omega*t)

where C, S, and omega are constants.

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What is the equation x(t) for the block?

Express your answer in terms of t, omega, and x_init.

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Now, imagine that we have exactly the same physical situation but that the x axis is translated, so that the position of the wall is now defined to be x = 0 .

The initial position of the block is the same as before, but in the new coordinate system, the block’s starting position is given by x_new(t=0) = L x_init

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Find the equation for the block’s position x_new(t) in the new coordinate system.

Express your answer in terms of L, x_init, omega, and t.

Answer:

old: x(t) = x_init* cos(w*t), where w=√(k/m);

new: x(t) = x_init* cos(w*t) +L;

L + x_init*cos(omega*t)

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