Question

A string of length 36cm was in unison with a fork of frequency 256Hz. It was in unison with another fork when the vibrating length was 48cm, the tension being unaltered. The frequency of second fork is
A. 212Hz
B. 320Hz
C. 384Hz
D. 192Hz

Answer 1

Hint: You could deal with the question in two cases. Before that, recall the expression for fundamental frequency. Now by substituting the given quantities for the case of the first fork you could find the wave velocity. Then for the second fork under the condition of unaltered tension, you could substitute the same wave velocity to find the frequency.
Formula used:
Fundamental frequency,
$f=\dfrac{v}{2L}$

Complete answer:
In the question, we are given a string of length 36cm that is observed to vibrate in unison with a fork of frequency 256Hz. Now when the length was 48cm, it was found to be in unison with another fork of some frequency f. We are supposed to find the frequency of the second fork if the tension in the string was unaltered.
In order to answer this, let us recall the relation for fundamental frequency which is the lowest frequency mode known for a stretched string. This frequency is given by,
$f=\dfrac{\sqrt{\dfrac{TL}{m}}}{2L}$
Where, $\sqrt{\dfrac{TL}{m}}=v$ is the wave velocity.
$f=\dfrac{v}{2L}$
For the first case,
$v=2Lf$
$\Rightarrow v=2\times 36\times 256$
$\therefore v=18432$
As the tension is the same so will be the wave velocity.
So, for the second case,
$f=\dfrac{v}{2L}$
$\Rightarrow f=\dfrac{18432}{2\times 48}$
$\therefore f=192Hz$
Therefore, we found the frequency of the second fork to be 192Hz.

Hence, option D is found to be the answer.

Note:
From the expression for wave velocity, we see that it is dependent on the tension and mass per unit length of the string. Here, in the second case, the mass and length will obviously be the same and tension is also mentioned to be unaltered. So, we could thereby conclude that the wave velocity remains the same.