Question

The tension of a stretched string is increased by $69\%$ . In order to keep its frequency of vibration constant, its length must be increased by-
(A) $30\%$
(B) $20\%$
(C) $69\%$
(D) $\sqrt {69} \%$

Answer 1

Hint
The frequency of vibration is proportional to the root of tension in the string. And inversely proportional to its length. When the tension in the string is increased by $x$ percent, the new value of tension becomes $\dfrac{{100 + x}}{{100}}$ times the original tension.
$f = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{\mu }}$
Where f is the frequency of vibration
L is the length of string
$\mu$ is the linear density of the material
T is the tension in the given string.

Complete step by step answer
Let the initial tension in the string be $T$ , when it increases by $69\%$ , it becomes-
$\dfrac{{169}}{{100}}T$ which is also equal to $1.69T$ .
If the initial frequency was-
${f_1} = \dfrac{1}{{2L}}\sqrt {\dfrac{{100T}}{\mu }}$
Then the new frequency,
${f_2} = \dfrac{1}{{2L}}\sqrt {\dfrac{{169T}}{\mu }}$
By taking the ratios we have:
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\sqrt {100T} }}{{\sqrt {169T} }}$
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{10}}{{13}}$
We know that,
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{{L_2}}}{{{L_1}}}$
$\dfrac{{{L_2}}}{{{L_1}}} = \dfrac{{10}}{{13}}$
$\dfrac{{{L_1}}}{{{L_2}}} = \dfrac{{13}}{{10}}$
${L_2}$ is $30\%$ less than ${L_1}$
Therefore when the tension is increased, it corresponds to a $30\%$ decrease in the length of the string, which means that if the tension were to be kept constant, a decrease of $30\%$ length would also give the frequency ${f_2}$ .
But in the question it is asked to keep the frequency equal to ${f_1}$ , so the length should be increased by $30\%$ to compensate for the effects caused by the change in tension $T$ .
Hence, option (A) is correct.

Note
It is important to understand whether the length should be increased or decreased to get the desired frequency. In this question it was already mentioned that the length should increase to cancel out the effect caused by increase in tension. But if it isn’t, then following points should be kept in mind-
-To produce the same effect that is produced by increase in tension, the length would have to be decreased.
-To keep the previous frequency constant, that is, to cancel the effect produced by increase in tension, the length would have to be decreased.