91Ó°ÊÓ

The vertical distance traveled by the bug in the first case is \({D_1}\; = \;0.12\;{\rm{m}}\).

The amplitude of the wave in the second case is \({A_2} = \;0.16\;{\rm{m}}\).

The maximum kinetic energy of a wave is given by,

\(\begin{aligned}{c}K{E_{\max }} = \;\frac{1}{2}m{\omega ^2}{A^2}\\K{E_{\max }} \propto \;{A^2}\end{aligned}\)

Here \(A\) is the amplitude, \(m\) is the mass, a constant and \(\omega \)is the angular frequency, which is also a constant.

Part (a)

The amplitude of the wave is half of the distance from the lowest to the highest point. So, the amplitude in the first case is,

\(\begin{aligned}{c}{A_1} &= \frac{{{D_1}}}{2}\\ &= \;\frac{{0.12\;{\rm{m}}}}{2}\\ &= \;0.06\;{\rm{m}}\end{aligned}\)

Therefore, the amplitude of the wave is 0.06 m.

Part (b)

From the relationship of the maximum kinetic energy and amplitude, we get

\(\begin{aligned}\frac{{K{E_{\max - 2}}}}{{K{E_{\max - 1}}}} &= \;\frac{{A_2^2}}{{A_1^2}}\\ &= \;\frac{{{{\left( {0.16\;{\rm{m}}} \right)}^2}}}{{{{\left( {0.06\;{\rm{m}}} \right)}^2}}}\\ &= \;7.11\end{aligned}\)

Therefore, maximum kinetic energy changes by a factor of 7.11.