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(C) The given wave equation is $y = y_0 \sin 2 \pi (nt - \frac{x}{\lambda})$.Comparing this with the standard wave equation $y = A \sin (\omega t - kx

Vedclass · Accepted Answer

$\frac{\pi y_0}{4}$

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(C) The given wave equation is $y = y_0 \sin 2 \pi (nt - \frac{x}{\lambda})$.Comparing this with the standard wave equation $y = A \sin (\omega t - kx)$,we have $\omega = 2 \pi n$ and $k = \frac{2 \pi}{\lambda}$.The wave velocity $v = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / \lambda} = n \lambda$.The particle velocity $v_p = \frac{dy}{dt} = y_0 (2 \pi n) \cos 2 \pi (nt - \frac{x}{\lambda})$.The maximum particle velocity $v_{p, 	ext{max}} = 2 \pi n y_0$.According to the problem,$v = \frac{1}{8} v_{p, 	ext{max}}$.Substituting the values,$n \lambda = \frac{1}{8} (2 \pi n y_0)$.Solving for $\lambda$,we get $\lambda = \frac{2 \pi n y_0}{8 n} = \frac{\pi y_0}{4}$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

$A$ simple harmonic progressive wave is given by $y = y_0 \sin 2 \pi (nt - \frac{x}{\lambda})$. If the wave velocity is $\left(\frac{1}{8}\right)^{\text{th}}$ of the maximum particle velocity,then the wavelength is:

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