Question

A bug on the surface of a pond is observed to move up and down a total vertical distance of 0.12 m, lowest to highest point, as a wave passes. (a) What is the amplitude of the wave? (b) If the amplitude increases to 0.16 m, by what factor does the bug’s maximum kinetic energy change?

Short answer

(a) The amplitude of the wave is 0.06 m.

(b) The maximum kinetic energy changes by a factor of 7.11.

Step-by-step solution

Data given in the question

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 \({A_2} = \;0.16\;{\rm{m}}\).

Concepts used in the solution

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.

Calculation for determining the desired results

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.