MCQ
A transverse wave is given by $y = A\sin 2\pi \left( {\frac{t}{T} - \frac{x}{\lambda }} \right)$. The maximum particle velocity is equal to $4$ times the wave velocity when
  • A
    $\lambda = 2\pi A$
  • $\lambda = \frac{1}{2}\pi A$
  • C
    $\lambda = \pi A$
  • D
    $\lambda = \frac{1}{4}\pi A$

Answer

Correct option: B.
$\lambda = \frac{1}{2}\pi A$
b
(b) Given $A\omega = 4v \Rightarrow A2\pi n = 4n\lambda \Rightarrow \lambda = \frac{{\pi A}}{2}$

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