MCQ
A standing wave pattern of amplitude $A$ in a string of length $L$ shows $2$ nodes (plus those at two ends). If one end of the string corresponds to the origin and $v$ is the speed of progressive wave, the disturbance in the string, could be represented (with appropriate phase) as: 
  • A
    $y(x,t) = A\sin \left( {\frac{{2\pi x}}{L}} \right)\cos \left( {\frac{{2\pi vt}}{L}} \right)$
  • B
    $y(x,t) = A\cos \left( {\frac{{3\pi x}}{L}} \right)\sin \left( {\frac{{3\pi vt}}{L}} \right)$
  • C
    $y(x,t) = A\cos \left( {\frac{{4\pi x}}{L}} \right)\cos \left( {\frac{{4\pi vt}}{L}} \right)$
  • $y(x,t) = A\sin \left( {\frac{{3\pi x}}{L}} \right)\cos \left( {\frac{{3\pi vt}}{L}} \right)$

Answer

Correct option: D.
$y(x,t) = A\sin \left( {\frac{{3\pi x}}{L}} \right)\cos \left( {\frac{{3\pi vt}}{L}} \right)$
d
From the given information, there are four nodes of the wave( $n=4$ ), including the two at the ends. Clearly, it forms a sine wave. With amplitude $A,$ wave number, $k=$ $(n-1) \pi / L=3 \pi / L .$ Writing this in the standard sinusoidal standing wave form, $y=$

$A \sin (k x) \cos (\omega t)$

$\omega=k v$

$y=A \sin (3 \pi x / L) \cos (3 \pi v t / L)$

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