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)

Question

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) (A) y=0.05 sin 2 π(4000 t-12.5 x) (B) y=0.05 sin 2 π(4000 t-122.5 x) (C) y=0.05 sin 2 π(3300 t-10 x) (D) y=0.05 sin 2 π(3300 x-10 t)

Tardigrade Admin · Accepted Answer

Here A=0.05 m , (5 λ/2)=0.025 ⇒ λ=0.1 m Now standard equation of wave y=A sin (2 π/λ)(v t-x) ⇒ y=0.05 sin 2 π(3300 t-10 x)

Q. 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)

$y = 0.05 sin 2 π (4000 t - 12.5 x)$

$y = 0.05 sin 2 π (4000 t - 122.5 x)$

$y = 0.05 sin 2 π (3300 t - 10 x)$

$y = 0.05 sin 2 π (3300 x - 10 t)$

Here $A = 0.05 m, \frac{5 λ}{2} = 0.025$
$\Rightarrow λ = 0.1 m$
Now standard equation of wave
$y = A sin \frac{2 π}{λ} (v t - x)$
$\Rightarrow y = 0.05 sin 2 π (3300 t - 10 x)$

Q. 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)

y=0.05sin2π(4000t−12.5x)

y=0.05sin2π(4000t−122.5x)

y=0.05sin2π(3300t−10x)

y=0.05sin2π(3300x−10t)