If the speed of the wave shown in the figure is 330m/s in the giv… - Vidyadip

MCQ

If the speed of the wave shown in the figure is $330m/s$ in the given medium, then the equation of the wave propagating in the positive $x-$direction will be (all quantities are in $M.K.S.$ units)

A
$y = 0.05\sin 2\pi (4000\,t - 12.5\,x)$
B
$y = 0.05\sin 2\pi (4000\,t - 122.5\,x)$
✓
$y = 0.05\sin 2\pi (3300\,t - 10\,x)$
D
$y = 0.05\sin 2\pi (3300\,x - 10\,t)$

✓

Answer

Correct option: C.

$y = 0.05\sin 2\pi (3300\,t - 10\,x)$

c
(c)Here $A = 0.05m,$ $\frac{{5\lambda }}{2} = 0.025$==> $\lambda = 0.1m$
Now standard equation of wave
$y = A\sin \frac{{2\pi }}{\lambda }(vt - x)$==> $y = 0.05\sin 2\pi (33t - 10x)$

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