Question

Question #2148b

Answer 1

Given that the velocity of stationary wave produced in a stretched string depending on its tension (T)and its mass per unit length (m) is `(V=270.9"m/s")`

As shown in above figure the wave lengths `lambda_i` of first three harmonics are related with the length `L` of the stretched string as follows.

For first i th harmonic `lambda_i=(2L)/p_i`,

where `p_i` is the number of loops produced in the given length L of the stretched string

For 1st harmonic wave length `lambda_1=2L`

For 2nd harmonic wave length `lambda_2=(2L)/2`

For 3rdharmonic wave length `lambda_2=(2L)/3`

So

For 1st harmonic, frequency

`n_1=V/lambda_1=V/(2L)=(270.9xx10^2cms^-1)/(32.4cm)=836.1Hz`

For 2nd harmonic, frequency

`n_2=V/lambda_2=(2V)/(2L)=2n_1=1672.2Hz`

For 3rd harmonic, frequency

`n_3=V/lambda_2=(3V)/(2L)=3n_1=2508.3Hz`