Question

The maximum potential energy / length increases with:

• A. Amplitude
• B. Wavelength
• C. Frequency
• D. Velocity

Answer 1

The correct option is C Frequency

If $y=A\sin (\omega t-kx)$ is the equation of a wave through a string, then the slope of the wave is $\frac{dy}{dx}=-Ak\cos (\omega t-kx)$.
The maximum potential energy will be $T\times \Delta x\times (\frac{dy}{dx}{)}^2{A}^2{k}^2=A^2{k}^2{\cos }^2(\omega t-kx)T\Delta x$
The maximum potential energy will be obtained if cos (\omega t - kx)=1. Thus, maximum potential energy = $4{\pi }^2{A}^{2}Tf\times T\times \Delta x$; T is the tension in the string
We also know that $T=\mu v^2$. Substituting, we get,
Maximum potential energy = $4{\pi }^2{f}^2{A}^2\mu$
Thus maximum potential energy depends on frequency and as frequency increases, potential energy also increases
The correct option is (c)