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 ______

Options

• 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 length = nλ

Maximum particle velocity = 2πnY_0​

It is given that: 

`nlambda = 1/8 * 2pinY_0`

Now, cancel n from both sides:

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

`lambda = (piY_0)/4`