Question

refer to thr attachment for the question

Answer 1

Let the equation of the wave be ,

⇒  y( x , t) = A sin(kx ± ωt + ∅ )

At any point x = x₀

The velocity is given by ,

v(t) = d y(  x₀ , t ) / dt

⇒ v (t) = Aω cos( kx₀ - ωt + ∅ )   ₋₋₋₋₋₋₋₋₋ (1)

Now acceleration can be given by,

⇒ a(t) = -Aω² sin ( kx₀ - ωt + Φ )  ₋₋₋₋₋₋₋₋₋(2)

But given that velocity of particle = 3 m/s ; acceleration = 90 m/s²

Taking the ratio of  (1) & (2), we get :

1 / ω = 3 / 90

⇒ ω = 30 s⁻¹

Maximum  velocity can be given by , V = Aω

⇒ 3 = A x 30

⇒ A = 0.1 m or 10 cm.

Now speed of the wave = υ x λ = ω / k = 20 m/s

⇒ k = 30 / 20 = 1.5 m⁻¹

so the final equation is :

y (x,t) = 10 sin(1.5X ± 30t + ∅)

( ± is because the direction of the wave is not given)

HOPE THIS HELPS YOU !!