Question 253: Waves

Question

Two uniform strings A and B made of steel are made to vibrate under the same tension. If first overtone of A is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the length of string $B$ to that of $A$ is

Accepted Answer

To determine the ratio of the lengths of strings $ A $ and $ B $, we have the following information:
Radius of string $ A $ is twice that of $ B $: $ R_2 = 2R_1 $.
Both strings are under the same tension: $ T_1 = T_2 $.
The first overtone of string $ A $ equals the second overtone of string $ B $: $ 2n_A = 3n_B $.
Using the relationship for the frequency of harmonics:
$ n = \frac{1}{2l} \sqrt{\frac{T}{m}} $
where $ m $ is the mass per unit length given by:
$ m = \frac{\pi R^2 l ho}{l} = \pi R^2 ho $
Now, substituting into the frequency equation and comparing the frequencies:
$ 2 	imes \frac{1}{2 l_A} \sqrt{\frac{T}{m_A}} = 3 	imes \frac{1}{2 l_B} \sqrt{\frac{T}{m_B}} $
Simplifying from the above relation:
$ \frac{l_A}{l_B} = \frac{2}{3} \sqrt{\frac{m_B}{m_A}} = \frac{2}{3} \sqrt{\frac{\pi r_B^2 ho}{\pi r_A^2 ho}} $
Simplifying further using the relation $ R_A = 2R_B $:
$ \frac{l_A}{l_B} = \frac{2}{3} \sqrt{\frac{r_B^2}{(2r_B)^2}} = \frac{2}{3} \sqrt{\frac{1}{4}} = \frac{1}{3} $
Thus, the ratio of the length of string $ B $ to $ A $ is:
$ \frac{l_B}{l_A} = \frac{3}{1} $

Answer

$1: 2$

Answer

$4: 3$

Answer

$2: 3$

Answer

$3: 1$

Waves

Two uniform strings A and B made of steel are made to vibrate under the same tension. If first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B, the ratio of the length of string B to that of A is

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