Question 82: Waves

Question

A tuning fork resonates with a sonometer wire of length $1 \mathrm{~m}$ stretched with a tension of $6 \mathrm{~N}$. When the tension in the wire is changed to $54 \mathrm{~N}$, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is ________________ $\mathrm{Hz}$.

Accepted Answer

To solve this problem, we'll have to use the relationship between the frequency of a vibrating string and the tension applied to it. When a tuning fork resonates with a sonometer wire, their frequencies are equal.
The frequency of a vibrating string is given by the formula:
$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$
Where:
$ f $ is the frequency of the vibration
$ L $ is the length of the wire
$ T $ is the tension in the wire
$ \mu $ is the linear mass density of the wire
The linear mass density ($ \mu $) of the wire remains constant.
Initially, when the string resonates with the tuning fork, the frequency of both is given by:
$f_1 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}}$
Here $ T_1 = 6 \mathrm{~N} $ and $ L = 1 \mathrm{~m} $, so we have:
$f_1 = \frac{1}{2 \cdot 1} \sqrt{\frac{6}{\mu}} = \frac{1}{2} \sqrt{\frac{6}{\mu}}$
Now, when the tension is changed to $ T_2 = 54 \mathrm{~N} $, the frequency of the wire changes to $ f_2 $ and it is given by:
$f_2 = \frac{1}{2L} \sqrt{\frac{T_2}{\mu}}$
Since $ L $ and $ \mu $ remain the same, substituting $ T_2 $:
$f_2 = \frac{1}{2 \cdot 1} \sqrt{\frac{54}{\mu}} = \frac{1}{2} \sqrt{\frac{54}{\mu}}$
Notice that $ 54 = 6 	imes 9 $, therefore:
$f_2 = \frac{1}{2} \sqrt{\frac{6 	imes 9}{\mu}} = \frac{1}{2} \sqrt{9} \sqrt{\frac{6}{\mu}} = \frac{3}{2} \sqrt{\frac{6}{\mu}} = 3 f_1$
When the tension was increased, the frequency became thrice the original frequency.
Since the second instance of the string produces 12 beats per second this means that the frequency of the tuning fork (and original string) and the new frequency (of the string with higher tension) differ by 12 Hz. If $ f_F $ is the frequency of the tuning fork, then:
Either $ f_2 = f_F + 12 $ Hz or $ f_2 = f_F - 12 $ Hz.
Since we have determined $ f_2 = 3 f_1 $ and $ f_1 = f_F $, we can state:
Either $ 3f_F = f_F + 12 $ or $ 3f_F = f_F - 12 $.
If $ 3f_F = f_F - 12 $, then :
$3f_F - f_F = -12$
$2f_F = -12$
This result is not possible since frequency cannot be negative.
So, we must consider the correct equation which is:
$f_2 = f_F + 12$
Now, substituting $ f_2 = 3f_F $:
$3f_F = f_F + 12$
$2f_F = 12$
$f_F = 6 	ext{ Hz}$
Thus, the frequency of the tuning fork is 6 Hz.

Waves

A tuning fork resonates with a sonometer wire of length 1 \mathrm{~m} stretched with a tension of 6 \mathrm{~N}. When the tension in the wire is changed to 54 \mathrm{~N}, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is ________________ \mathrm{Hz}.

Choose the correct answer: