A transverse sinusoidal wave moves along a st | vedclass

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Question

(A) The velocity of a particle in a transverse wave is given by $v_p = \omega \sqrt{A^2 - y^2}$,where $A$ is the amplitude and $y$ is the displacement

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The velocity of a particle in a transverse wave is given by $v_p = \omega \sqrt{A^2 - y^2}$,where $A$ is the amplitude and $y$ is the displacement.Given: wave speed $v = 10 \, cm/s = 0.1 \, m/s$,wavelength $\lambda = 0.5 \, m$,amplitude $A = 10 \, cm = 0.1 \, m$,and displacement $y = 5 \, cm = 0.05 \, m$.The angular frequency $\omega = \frac{2\pi v}{\lambda} = \frac{2\pi 	imes 0.1}{0.5} = 0.4\pi \, rad/s$.Substituting the values into the velocity formula:$v_p = 0.4\pi \sqrt{(0.1)^2 - (0.05)^2} = 0.4\pi \sqrt{0.01 - 0.0025} = 0.4\pi \sqrt{0.0075} = 0.4\pi 	imes 0.05\sqrt{3} = 0.02\sqrt{3}\pi \, m/s$.Converting $0.02\sqrt{3}\pi$ to fraction form: $\frac{2\sqrt{3}\pi}{100} = \frac{\sqrt{3}\pi}{50} \, m/s$.From the figure,as the wave travels in the positive $x-$ direction,point $P$ is on the downward slope of the crest,meaning it is moving upwards in the positive $y-$ direction. Thus,the velocity is $\frac{\sqrt{3}\pi}{50} \hat{j} \, m/s$.

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