Question 336: Waves

Question

A string A has twice the length, twice the diameter, twice the tension and twice the density of another string B. The overtone of A which will have the same fundamental frequency as that of $B$ is

Accepted Answer

The fundamental frequency of string B is given by the formula:
$ n_B = \frac{1}{2L_B} \sqrt{\frac{T_B}{\mu_B}} $
where $ \mu $ is the mass per unit length. For string B:
$ m_B = ho \pi r^2 l $
Thus, the mass per unit length becomes:
$ \mu_B = \frac{ho \pi r^2 l}{l} = ho \pi r^2  $
For string A, which has twice the diameter, the mass per unit length is:
$ \mu_A = 2 ho \pi (2r)^2 = 8 \mu_B $
The fundamental frequency of string A is calculated as follows:
$ n_A = \frac{1}{2(2L_B)} \sqrt{\frac{2T_B}{8\mu_B}} = \frac{1}{4} \left(\frac{1}{2L_B} \sqrt{\frac{T_B}{\mu_B}}ight) = \frac{1}{4} n_B $
The frequency of the $ p^{	ext{th}} $ overtone is given by:
$ (p+1)n  $
For string A to have the same fundamental frequency as string B:
$ (p+1) = 4 \Rightarrow p = 3 $
Therefore, the 3rd overtone of string A will have the same frequency as the fundamental frequency of string B.

Answer

first

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second

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third

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fourth

Waves

A string A has twice the length, twice the diameter, twice the tension and twice the density of another string B. The overtone of A which will have the same fundamental frequency as that of B is

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