A transverse wave is described by the equatio | vedclass

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(B) Comparing the given equation $Y = Y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$ with the standard wave equation $y = a \sin(\omega t - kx)$

Vedclass · Accepted Answer

$\lambda = \frac{\pi Y_0}{2}$

Vedclass · Answer

(B) Comparing the given equation $Y = Y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} ight)$ with the standard wave equation $y = a \sin(\omega t - kx)$:We identify the amplitude $a = Y_0$,angular frequency $\omega = 2\pi f$,and wave number $k = \frac{2\pi}{\lambda}$.The maximum particle velocity is given by $(v_{\max})_{	ext{particle}} = a\omega = Y_0 	imes 2\pi f$.The wave velocity (phase velocity) is given by $v_{	ext{wave}} = \frac{\omega}{k} = \frac{2\pi f}{2\pi / \lambda} = f\lambda$.According to the problem,the maximum particle velocity is four times the wave velocity:$(v_{\max})_{	ext{particle}} = 4 v_{	ext{wave}}$$Y_0 	imes 2\pi f = 4 f\lambda$Dividing both sides by $4f$,we get:$\lambda = \frac{2\pi Y_0}{4} = \frac{\pi Y_0}{2}$.

$A$ transverse wave is described by the equation $Y = Y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$. The maximum particle velocity is four times the wave velocity if:

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