A transverse sinusoidal wave moves along a st | vedclass

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(B) Given: Wave speed $v = 10 	ext{ cm/s} = 0.1 	ext{ m/s}$,wavelength $\lambda = 0.5 	ext{ m}$,amplitude $A = 10 	ext{ cm} = 0.1 	ext{ m}$.Angul

Vedclass · Accepted Answer

$-\frac{\sqrt{3} \pi}{50} \hat{j} 	ext{ m/s}$

Vedclass · Answer

(B) Given: Wave speed $v = 10 	ext{ cm/s} = 0.1 	ext{ m/s}$,wavelength $\lambda = 0.5 	ext{ m}$,amplitude $A = 10 	ext{ cm} = 0.1 	ext{ m}$.Angular frequency $\omega = \frac{2\pi v}{\lambda} = \frac{2\pi 	imes 0.1}{0.5} = 0.4\pi 	ext{ rad/s} = \frac{2\pi}{5} 	ext{ rad/s}$.The displacement equation is $y = A \sin(kx - \omega t + \phi)$.At displacement $y = 5 	ext{ cm} = 0.05 	ext{ m}$,we have $0.05 = 0.1 \sin(	heta)$,so $\sin(	heta) = 0.5$,which means $	heta = 30^{\circ}$ or $150^{\circ}$.From the figure,point $P$ is on the downward slope,so its velocity $v_y = \frac{dy}{dt}$ must be negative.The particle velocity is $v_y = A\omega \cos(	heta)$.Since the wave moves in the positive $x$-direction,a particle on the downward slope has a negative vertical velocity.Thus,$v_y = -(0.1) 	imes (0.4\pi) 	imes \cos(30^{\circ}) = -0.04\pi 	imes \frac{\sqrt{3}}{2} = -0.02\pi\sqrt{3} 	ext{ m/s} = -\frac{\sqrt{3}\pi}{50} \hat{j} 	ext{ m/s}$.

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