Question

In a transverse progressive wave of amplitude A, the maximum particle velocity is four times its wave velocity. The wave- length of the wave is:

• A. $2\pi A$
• B. $\pi A$
• C. $\frac{\pi A}{2}$
• D. $\frac{\pi A}{4}$

Answer 1

Answer: (C)

Here: Amplitude of the wave = A Maximum velocity ${\upsilon }_1=4\upsilon$ (where $\upsilon$ is velocity of wave) Maximum velocity relation is given by as ${\upsilon }_1=a\omega$ or $\omega =\frac{4\upsilon }{A}$ ?(i) Hence, wavelength of the wave $\lambda =\frac{\upsilon }{f}=\frac{\upsilon }{\omega /2\pi }=\frac{2\pi \upsilon }{\omega }$ $\left(\because f=\frac{\omega }{2\pi }\right)$ $=\frac{2\pi \upsilon }{\frac{4\upsilon }{A}}=\frac{\pi A}{2}$ [from eq. (i)]