A transverse wave is expressed as, y=y_0 2 π f t For what value of λ , maximum particle velocity equals to 4 times the wave velocity?

Correct answer: A. y_0 (π)/(2) Given equation is y=y_0 2 π f t where A=y_0, k=(2 π)/(λ) According to question, array rlrl v & =4 v_p & A ω & =4 (ω)/(k) & y_0 ω & =(4 ω)/(k) & & y_0=(4)/(k) & =(4)/(2 π) × λ array Therefore, wavelength λ=(π y_0)/(2)

JIPMER 2011

Exam: JIPMER
Year: 2011
Subject: Physics
Chapter: Waves and Sound

JIPMER 2011 · Physics · Waves and Sound

A transverse wave is expressed as, y=y_0 2 π f t For what value of...

A transverse wave is expressed as,

y=y_0 \sin 2 \pi f t

For what value of

\lambda

, maximum particle velocity equals to 4 times the wave velocity?

Options

A $y_0 \frac{\pi}{2}$
B $2 y_0 \pi$
C $y_0 \pi$
D $y_0 \frac{\pi}{4}$

Correct answer

A. $y_0 \frac{\pi}{2}$

Step-by-step solution

Given equation is

y=y_0 \sin 2 \pi f t

where

A=y_0, k=\frac{2 \pi}{\lambda}

According to question,

\begin{array}{rlrl}v & =4 v_p \\ \therefore & A \omega & =4 \frac{\omega}{k} \\ \therefore & y_0 \omega & =\frac{4 \omega}{k} \\ \Rightarrow & & y_0=\frac{4}{k} & =\frac{4}{2 \pi} \times \lambda\end{array}

Therefore, wavelength

\lambda=\frac{\pi y_0}{2}

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