Question

A simple harmonic progressive wave is given by Y = Y₀ sin 2π `(nt - x/lambda)`. If the wave velocity is `(1/8)^{th}` of the maximum particle velocity, then the wavelength is ______

विकल्प

• πY₀/2

• πY₀/16

• πY₀/8

• πY₀/4

Answer 1

A simple harmonic progressive wave is given by Y = Y₀ sin 2π `(nt - x/lambda)`. If the wave velocity is `(1/8)^{th}` of the maximum particle velocity, then the wavelength is πY₀/4.

Explanation:

Y = Y₀ sin 2π `(nt - x/lambda)`

Wave velocity = nλ

Particle velocity = `dy/dt = Y_0 2pin cos2pi(nt - x/lambda)`

`dy/dt|_{max} = Y_0 2pin`

According to condition given

8nλ = Y₀2πn

`lambda = (2piY_0)/8 = (nY_0)/4`