Question #2148b

1 Answer
P dilip_k
Feb 24, 2017

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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

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