A transverse wave is described by the equation y=y_ 6 2 (f t- x ) The maximum particle…

Physics · JEE Main · NTA ExamsOscillations and Waves

A transverse wave is described by the equation \(y=y_{6} \sin 2 \pi\left(f t-\frac{x}{\lambda}\right)\)
 The maximum particle velocity is equal to four times the wave velocity if
  1. \(\lambda=\frac{\pi y_{0}}{4}\)
  2. \(\lambda=\frac{\pi y_{0}}{2}\)
  3. \(\lambda=\pi y_{0}\)
  4. \(\lambda=2 \pi y_{0}\)

Answer

(B) = y_ 0 2

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Related practice questions

More Oscillations and Waves questions · Browse all practice questions