Question

The natural frequency of a tuning fork P is 432 Hz. 3 beats per second are produced when tuning fork P and another fork Q are sounded together. If P is loaded with wax, the number of beats increases to 5 beats per second. The frequency of Q is
A. 429 Hz
B. 435 Hz
C. 437 Hz
D. 427 Hz

Answer 1

The beat frequency of two different frequencies (${{f}_{1}}$ and ${{f}_{2}}$) is given as ${{f}_{B}}=\left| {{f}_{1}}-{{f}_{2}} \right|$. Find the frequency of the tuning fork Q by this formula. However, we will get two values for the frequency. Then use the fact that the frequency of a tuning fork decreases when its mass increases and find the actual frequency of Q.

Formula used:
${{f}_{B}}=\left| {{f}_{1}}-{{f}_{2}} \right|$

Complete step by step answer:
When two forks of different frequencies are sounded together, we get a beat frequency. The beat frequency of two different frequencies (${{f}_{1}}$ and ${{f}_{2}}$) is given as ${{f}_{B}}=\left| {{f}_{1}}-{{f}_{2}} \right|$ …. (i), where ‘| |’ is a modulus function.
In this case, let frequency of the tuning P be ${{f}_{1}}$ and the frequency of the tuning Q be ${{f}_{2}}$.
It is given that ${{f}_{1}}=432$.
It is said when these two are sounded together, a beat frequency of 2 beats per second is obtained.
This means that ${{f}_{B}}=\left| 432-{{f}_{2}} \right|=3$.
$\Rightarrow 432-{{f}_{2}}=\pm 3$.
$\Rightarrow 432-{{f}_{2}}=3$ or $432-{{f}_{2}}=-3$.
Therefore,
$\Rightarrow {{f}_{2}}=432-3=429Hz$ or $\Rightarrow {{f}_{2}}=432+3=435Hz$
This means that the frequency of the tuning fork Q can be 429Hz or it can be 435 Hz.
However, it is given that when the tuning fork is loaded with Q the beat frequency increases to 5 beats per second. When a tuning fork is loaded with wax, its mass increases and its frequency decreases.
Let the new frequency of the Q be ${{f}_{2}}'$.
This means that $\left| {{f}_{1}}-{{f}_{2}}' \right|=5$.
$\Rightarrow 432-{{f}_{2}}'=\pm 5$
$\Rightarrow 432-{{f}_{2}}'=5$ or $\Rightarrow 432-{{f}_{2}}'=-5$
Therefore,
$\Rightarrow {{f}_{2}}'=432-5=427Hz$ or $\Rightarrow {{f}_{2}}'=432+5=437Hz$.
Since the frequency of Q has increased, its new frequency must be greater than 429Hz.
Therefore, ${{f}_{2}}'=427Hz$ is discarded.
Hence, the original frequency of Q is 435 Hz.

So, the correct answer is “Option B”.

Note:
The modulus function is a function which gives the positive value or absolute value of the variable inside the modulus sign.
Consider $y=|x|$.
This means that y is always positive.
Therefore, if x is positive then $y=x$ and if x is negative then $y=-x$.