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STD 11 - 14. waves and sound — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 14. waves and sound4 Marks
MCQ
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
  • A
    $\lambda = \frac{{\pi {Y_0}}}{4}$
  • ✓
    $\lambda = \frac{{\pi {Y_0}}}{2}$
  • C
    $\lambda = \pi {Y_0}$
  • D
    $\lambda = 2\pi {Y_0}$
✓

Answer

Correct option: B.
$\lambda = \frac{{\pi {Y_0}}}{2}$
b
(b) Comparing the given equation with $y = a\sin (\omega t - kx)$,

We get $a = Y_0, \, \omega  = 2\ pi  f, k = \frac{{2\pi }}{\lambda }$.

Hence maximum particle velocity ${({v_{\max }})_{particle}} = a\omega = {Y_0} \times 2\pi f$

and wave velocity ${(v)_{wave}} = \frac{\omega }{k} = \frac{{2\pi f}}{{2\pi /\lambda }} = f\lambda $

$\because \,\,\,{({v_{\max }})_{Particle}} = 4{v_{Wave}}$==> ${Y_0} \times 2\pi f = 4f\lambda $ ==> $\lambda = \frac{{\pi {Y_0}}}{2}$.

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