1. Tardigrade
  2. Question
  3. Physics
  4. 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:

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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:

JIPMERJIPMER 1997

Solution:

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)]