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 MKS units)

y = 0.05 sin 2 π (4000 t - 12.5 x)

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y = 0.05 sin 2 π (4000 t - 122.5 x)

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y = 0.05 sin 2 π (3300 t - 10 x)

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y = 0.05 sin 2 π (3300 x - 10 t)

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Solution

The correct option is C $y = 0.05 sin 2 π (3300 t - 10 x)$
The length $0.25 m$ corresponds to $2.5 λ$
$0.25 = 2.5 λ$
$λ = 0.1 m$
$k = \frac{2 π}{0.1}$
we know
$V = λ f$
here $v =$ speed of wave,
$λ$ = wavelength
$f =$ frequency
$330 = 0.1 f$ $\Rightarrow f = 3300 \Rightarrow ω = 2 π \times 3300$

$y = A sin (ω t - k x)$
$y = 0.05 sin (3300 \times 2 π t - \frac{2 π}{0.1} x)$
$y = 0.05 sin 2 π (3300 t - 10 x)$

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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 MKS units)

The correct option is C y=0.05 sin 2π(3300t−10x)The length 0.25 m corresponds to 2.5λ 0.25=2.5λ λ=0.1 m k=2π0.1 we know V=λf here v= speed of wave, λ = wavelength f= frequency 330=0.1f⇒f=3300⇒ω=2π×3300 y=A sin(ωt−kx) y=0.05sin(3300×2πt−2π0.1x) y=0.05 sin2π(3300t−10x)