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Question: A transverse wave is described by the equation \(y = A\sin \left[ {2\pi \left( {nt - \dfrac{x}{{\lam...

A transverse wave is described by the equation y=Asin[2π(ntxλ0)]y = A\sin \left[ {2\pi \left( {nt - \dfrac{x}{{\lambda_0}}} \right)} \right]. The maximum particle velocity is equal to 33 times the wave velocity if:

Explanation

Solution

Here the concept of simple harmonic motion is applied to the particle.Now using the transverse wave general wave equation we can complete it with the given wave equation. With this comparison we can find the wave velocity of the particle. Now equating 33 times of this wave velocity with maximum velocity of the particle. In this way we can calculate the solution.

Complete step by step answer:
As per the given problem there is a transverse wave is described by the equation y=Asin[2π(ntxλ0)]y = A\sin \left[ {2\pi \left( {nt - \dfrac{x}{{\lambda_0}}} \right)} \right].
We need to find the condition when the maximum particle velocity is equal to 33 times the wave velocity.The general equation of transverse wave is represented as,
y=Asin(ωt±kx)y = A\sin \left( {\omega t \pm kx} \right)
Comparing with our given wave equation the general equation will be,
y=Asin(ωtkx)y = A\sin \left( {\omega t - kx} \right)

Know transverse wave equation is,
y=Asin[2π(ntxλ0)]y = A\sin \left[ {2\pi \left( {nt - \dfrac{x}{{\lambda_0}}} \right)} \right]
Taking 2π2\pi inside the bracket we will get,
y=Asin[(2πnt2πxλ0)]y = A\sin \left[ {\left( {2\pi nt - \dfrac{{2\pi x}}{{\lambda_0}}} \right)} \right]
On comparing with the general equation we will get,
ω=2πn\omega = 2\pi n
k=2πλ0k = \dfrac{{2\pi }}{{\lambda_0}}
We know wave velocity as,
v=ωkv = \dfrac{\omega }{k}

On putting the values we will get,
v=2πn2πλ0 v=λ0nv = \dfrac{{2\pi n}}{{\dfrac{{2\pi }}{{\lambda_0}}}} \\\ \Rightarrow v = \lambda_0n
The maximum velocity of a particle in simple harmonic motion is ±Aω\pm A\omega.So from the given condition we can write,
vmax=vv\max = v
A×ω=3λ0n A×2πn=3λ0n\Rightarrow A \times \omega = 3\lambda_0n \\\ \Rightarrow A \times 2\pi n = 3\lambda_0n
On rearranging we will get,
λ0=2πA3\therefore \lambda_0 = \dfrac{{2\pi A}}{3}

Hence λ0\lambda_0 is the condition for the maximum particle velocity is equal to 33 times the wave velocity.

Note: Remember in a plane progressive transverse wave particles of the medium oscillate simply harmonically about their mean position. Here the maximum velocity of the particle is about the mean position. Also note that in the wave equation both position and negative term can be possible which depends upon the direction of the wave.