Solution - Maximizing Amplitude

[9:02 AM | 0 comments ]

Consider the block when it has distance x from equilibrium point and velocity v like in the picture below.Then the energy at this point is
$E_0 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \frac{1}{2}kA_0^2$

When point A starts moving to the left, consider the block from point A's frame. The block will have velocity of v+vA. Therefore, we can write the equation for its energy in A's frame.
$E = \frac{1}{2}kx^2 + \frac{1}{2}m(v+v_A)^2$
$E = \frac{1}{2}kx^2 + \frac{1}{2}m(v^2+v_A^2+2vv_A)$
$E = (\frac{1}{2}kx^2 + \frac{1}{2}mv^2)+\frac{1}{2}m(v_A^2+2vv_A)$

We can substitute $\frac{1}{2}kx^2 + \frac{1}{2}mv^2$ from the equation above. Hence,
$E = \frac{1}{2}kA_0^2 + \frac{1}{2}mv_A^2 + mvv_A$

In order to maximize its amplitude, we need to maximize its energy. And from the equation above, we can conclude that we have to maximize , because $\frac{1}{2}mv_A^2$ and $\frac{1}{2}kA_0$ is constant.
The variable would be maximum if the block has maximum velocity (i.e. at its equilibrium point).
So point A should start moving when the block at its equilibrium point and towards right.

And if point A is accelerated to the left, the equilibrium point of the block will be shifted to the right, because there is a fiction force towards right if we observe from A's frame.
Simply, if we want the new amplitude maximum, point A should start being accelerated when the block is at its leftmost position.