If the speed of the wave shown in the figure is 330 m/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)

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$y = 0 .05 \sin 2 π (3300 x - 10 t)$

$y = 0 .05 \sin 2 π (3300 t - 10 x)$

$y = 0 .05 \sin 2 π (4000 t - 122 .5 x)$

$y = 0 .05 \sin 2 π (4000 t - 12 .5 x)$

answer is C.

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Detailed Solution

Here $A = 0 .05 m, \frac{5 λ}{2} = 0 .25 \Rightarrow λ = 0 .1 m$
Now standard equation of wave
$y = A \sin \frac{2 π}{λ} (vt - x) \Rightarrow y = 0 .05 \sin 2 π (3300 t - 10 x)$

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