Question #2148b

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Question #2148b

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Answer 1 · 2017-02-24T03:31:19

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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 $(V = 270.9 m/s)$ $(V=270.9"m/s")$

As shown in above figure the wave lengths $λ_{i}$ $lambda_i$ of first three harmonics are related with the length $L$ $L$ of the stretched string as follows.

For first i th harmonic $λ_{i} = \frac{2 L}{p_{i}}$ $lambda_i=(2L)/p_i$ ,

where $p_{i}$ $p_i$ is the number of loops produced in the given length L of the stretched string

For 1st harmonic wave length $λ_{1} = 2 L$ $lambda_1=2L$

For 2nd harmonic wave length $λ_{2} = \frac{2 L}{2}$ $lambda_2=(2L)/2$

For 3rdharmonic wave length $λ_{2} = \frac{2 L}{3}$ $lambda_2=(2L)/3$

So

For 1st harmonic, frequency

$n_{1} = \frac{V}{λ_{1}} = \frac{V}{2 L} = \frac{270.9 \times 10^{2} c m s^{- 1}}{32.4 c m} = 836.1 H z$ $n_1=V/lambda_1=V/(2L)=(270.9xx10^2cms^-1)/(32.4cm)=836.1Hz$

For 2nd harmonic, frequency

$n_{2} = \frac{V}{λ_{2}} = \frac{2 V}{2 L} = 2 n_{1} = 1672.2 H z$ $n_2=V/lambda_2=(2V)/(2L)=2n_1=1672.2Hz$

For 3rd harmonic, frequency

$n_{3} = \frac{V}{λ_{2}} = \frac{3 V}{2 L} = 3 n_{1} = 2508.3 H z$ $n_3=V/lambda_2=(3V)/(2L)=3n_1=2508.3Hz$

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Question #2148b

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