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(B) The given wave equation is $Y = Y_0 \sin(2 \pi n t - \frac{2 \pi x}{\lambda})$.Comparing this with the standard form $Y = Y_0 \sin(\omega t - kx)$

Vedclass · Accepted Answer

$\frac{\pi Y_0}{4}$

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(B) The given wave equation is $Y = Y_0 \sin(2 \pi n t - \frac{2 \pi x}{\lambda})$.Comparing this with the standard form $Y = Y_0 \sin(\omega t - kx)$,we get angular frequency $\omega = 2 \pi n$.The maximum particle velocity $v_{p, 	ext{max}} = Y_0 \omega = Y_0 (2 \pi n) = 2 \pi n Y_0$.The wave velocity $v = n \lambda$.According to the problem,the wave velocity is $(1/8)^{	ext{th}}$ of the maximum particle velocity:$v = \frac{1}{8} v_{p, 	ext{max}}$$n \lambda = \frac{1}{8} (2 \pi n Y_0)$$n \lambda = \frac{\pi n Y_0}{4}$$\lambda = \frac{\pi Y_0}{4}$.

$A$ simple harmonic progressive wave is given by $Y = Y_0 \sin 2 \pi (nt - \frac{x}{\lambda})$. If the wave velocity is $(1/8)^{\text{th}}$ of the maximum particle velocity,then the wavelength is

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