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

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

TestHub · Accepted Answer

The fundamental frequency of a vibrating string is given by:

$n = \frac{1}{2 L} \frac{T}{m}$

where $m$ is the linear mass density, $m = π r^{2} ρ$ .

Thus, $n = \frac{1}{2 L r} \frac{T}{π ρ}$ .

Since both strings are made of steel and are under the same tension, $T$ and $ρ$ are constant. Therefore, $n \propto \frac{1}{L r}$ .

The first overtone of string A is its second harmonic, which is $2 n_{A}$ .
The second overtone of string B is its third harmonic, which is $3 n_{B}$ .

Given that the first overtone of A is equal to the second overtone of B:

$2 n_{A} = 3 n_{B}$

Substituting the proportionality $n \propto \frac{1}{L r}$ :

$\frac{2}{L _{A} r _{A}} = \frac{3}{L _{B} r _{B}}$

Rearranging to find the ratio of the length of string B to that of A:

$\frac{L _{B}}{L _{A}} = \frac{3}{2} \times \frac{r _{A}}{r _{B}}$

Given that the radius of A is twice that of B, $r_{A} = 2 r_{B}$ , which means $\frac{r _{A}}{r _{B}} = 2$ .

Substituting this value:

$\frac{L _{B}}{L _{A}} = \frac{3}{2} \times 2 = \frac{3}{1}$

The ratio of the length of string B to that of A is $3 : 1$ .

Answer: $3 : 1$

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