The preceding problems in this chapter have assumed that the springs had negligi

Question

The preceding problems in this chapter have assumed that the springs had negligible mass. But of course no spring is completely massless. To find the effect of the spring's mass, consider a spring with mass M, equilibrium length L0, and spring constant k. When stretched or compressed to a length L, the potential energy is 1/2kx^2, where x=L-L0.

Take the time derivative of the conservation of energy equation, for a mass m moving on the end of a massless spring. By comparing your results ax=-w^2x, find the angular frequency of oscillation.

Explanation / Answer

Potential energy:

U = (1/2) k x^2

Force:

F = -dU/dx = -k x

acceleration:

a = F/m = -(k/m) x

And we have: a = -w^2 x

So:

-(k/m) x = -w^2 x

==> w^2 = k/m

==> w = sqrt(k/m)

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