Question

The frequency of 1st harmonic of a sonometer wire is $ 160Hz $ . If the length of the wire is increased by 50% and the tension in the wire is decreased by 19%, the frequency of its first overtone is:
(Assume linear mass density to be constant)
(A) $ 180Hz $
(B) $ 192Hz $
(C) $ 220Hz $
(D) $ 232Hz $

Answer 1

Hint
The frequency of the nth mode of a sonometer wire is given by, $ f = \dfrac{n}{{2L}}\sqrt {\dfrac{T}{\mu }} $ . For the two cases given in the question, we substitute the values of the variables and find the ratio of the frequency of the two cases. Since the frequency in the first case is given from there we can find the frequency of the second case.
In the solution to this question, we will be using the following formula,
$\Rightarrow f = \dfrac{n}{{2L}}\sqrt {\dfrac{T}{\mu }} $
where $ f $ is the frequency of the wire, $ n $ is the harmonic, $ L $ is the length of the wire, $ T $ is the tension in the wire and $ \mu $ is the mass per unit length of the wire.

Complete step by step answer
In this question, we are provided with the frequency of the first harmonic of a sonometer wire. The frequency of a sonometer wire is given by the formula,
 $\Rightarrow f = \dfrac{n}{{2L}}\sqrt {\dfrac{T}{\mu }} $
here the harmonic is 1 so we can substitute 1 in place of $ n $ . Hence we get,
 $\Rightarrow f = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{\mu }} $
This is the first case.
In the second situation, the length and the tension in the wire are increased. The length is increased by 50%. So we get the new length as,
 $\Rightarrow L' = L + \dfrac{{50}}{{100}}L $
So on adding we get,
 $\Rightarrow L' = 1.5L $
Similarly, the tension in the wire is decreased by 19%. So we get the new tension as,
 $\Rightarrow T' = T - \dfrac{{19}}{{100}}T $
So on adding we get,
 $\Rightarrow T' = 0.81T $
The value $ \mu $ remains unchanged in both cases as the material of the wire remains the same.
So using these new values of the length and tension we get the new frequency as,
 $\Rightarrow f' = \dfrac{1}{{2L'}}\sqrt {\dfrac{{T'}}{\mu }} $
By substituting the values of $ L' $ and $ T' $ in the equation we get,
 $\Rightarrow f' = \dfrac{1}{{2 \times 1.5L}}\sqrt {\dfrac{{0.81T}}{\mu }} $
Now we can take the ratio of the frequencies in the first and second cases. Therefore we get,
 $\Rightarrow \dfrac{{f'}}{f} = \dfrac{{\dfrac{1}{{2 \times 1.5L}}\sqrt {\dfrac{{0.81T}}{\mu }} }}{{\dfrac{1}{{2L}}\sqrt {\dfrac{T}{\mu }} }} $
by arranging the equation we get,
 $\Rightarrow \dfrac{{f'}}{f} = \dfrac{1}{{2 \times 1.5L}}\sqrt {\dfrac{{0.81T}}{\mu }} \times 2L\sqrt {\dfrac{\mu }{T}} $
Therefore we can cancel all the variables from the numerator and the denominator we get,
 $\Rightarrow \dfrac{{f'}}{f} = \dfrac{{\sqrt {0.81} \times 2}}{{2 \times 1.5}} $
The 2 gets cancelled from the numerator and denominator. On doing the further calculation we get,
 $\Rightarrow \dfrac{{f'}}{f} = 0.6 $
We are given that the frequency of the first harmonic in the first case, is $ f = 160Hz $ . So from the above equation, $ f' = 0.6 \times f $
Substituting the value of $ f = 160Hz $ we get,
 $\Rightarrow f' = 0.6 \times 160Hz $
This gives us the frequency of the first harmonic in the second case as,
 $\Rightarrow f' = 96Hz $
In the question, we are asked the frequency of the first overtone. The frequency of the first overtone is given by $ 2 \times $ the frequency of the first harmonic.
Therefore, the frequency of the first overtone $ = 2 \times 96 = 192Hz $
So the correct answer is option (B); $ 192Hz $ .

Note
A sonometer is a device that is used to demonstrate the relation between the frequency of the sound produced by a plucked string and the length, tension, and the mass per unit length of that string. In the question, we are asked to find the frequency of the first overtone which is also known as the second harmonic given by $ n = 2 $ .