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  4. A transverse wave is represented by y = A sin(ω t - kx) For what value of the wavelength is the wave velocity equal to the maximum particle velocity?

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Q. A transverse wave is represented by $y = A \sin(\omega t - kx)$ For what value of the wavelength is the wave velocity equal to the maximum particle velocity?

AIPMTAIPMT 2010Waves

Solution:

The given equation is, $y = A \sin (\omega t - kx )$
Wave velocity, $v=\frac{\omega}{ k }$.....(i)
Particle velocity, $v_{ p }=\frac{ dy }{ dt }= A \omega \cos (\omega t - kx )$
Maximum particle velocity, $\left(v_{ p }\right)_{\max }= A \omega$........(ii)
According to the given question, $v=\left(v_{ p }\right)_{\max }$
$\frac{\omega}{ k }= A \omega$
(Using (i) and (ii))
$\frac{1}{ k }= A$ or $\frac{\lambda}{2 \pi}= A \left(\because k =\frac{2 \pi}{\lambda}\right)$