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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:
$y=y_0 \sin 2 \pi f t$
For what value of $\lambda$, maximum particle velocity equals to 4 times the wave velocity?
Solution:
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Verified Answer
The correct answer is:
$y_0 \frac{\pi}{2}$
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 $\lambda=\frac{\pi y_0}{2}$
$y=y_0 \sin 2 \pi f t$
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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