Question

Two identical piano wires kept under the same tension $$T$$ have a fundamental frequency of $$600\text{Hz}$$. The fractional increase in the tension of the wires which leads to occurrence of $$6\text{beats/s}$$ when both the wires oscillate together would be
A. 0.02
B. 0.03
C. 0.04
D. 0.01

Answer 1

Hint: Use the formula for the fundamental frequency in the stretched string given by the law of vibrations in stretched strings. This formula will give the relation between the fundamental frequency in the piano wire and the tension in the piano wires. Using this relation derives the relation for the fractional change in the fundamental frequency and tension.

Formula used:
The formula for the fundamental frequency is
$$\gamma ={\displaystyle \frac{1}{2L}}\sqrt{{\displaystyle \frac{T}{\mu }}}$$ …… (1)
Here, $$\gamma$$ is the fundamental frequency of the vibration, $$L$$ is the length of the stretched string, $$T$$ is the tension in the string and $$\mu$$ is the mass per unit length of the string.

Complete step by step answer:
We have given that the two piano wires have the same fundamental frequency $$600\text{Hz}$$ and are under the same tension $$T$$.
The change in the frequency of vibration of the piano wires is $$6\text{beats/s}$$.
$$\mathrm{\Delta }\gamma =6\text{beats/s}$$
From the equation (1), it can be concluded that the fundamental frequency $$\gamma$$ of the piano wires is directly proportional to the square root of the tension $$T$$ in the wires.
$$\gamma \propto \sqrt{T}$$
Let us take log and differentiate both the sides of the above relation.
$${\displaystyle \frac{\mathrm{\Delta }\gamma }{\gamma }}={\displaystyle \frac{1}{2}}{\displaystyle \frac{\mathrm{\Delta }T}{T}}$$
Rearrange the above equation for $${\displaystyle \frac{\mathrm{\Delta }T}{T}}$$.
$$\Rightarrow {\displaystyle \frac{\mathrm{\Delta }T}{T}}=2{\displaystyle \frac{\mathrm{\Delta }\gamma }{\gamma }}$$
Substitute $$6\text{beats/s}$$ for $$\mathrm{\Delta }\gamma$$ and $$600\text{Hz}$$ for $$\gamma$$ in the above equation.
$$\Rightarrow {\displaystyle \frac{\mathrm{\Delta }T}{T}}=2{\displaystyle \frac{6\text{beats/s}}{600\text{Hz}}}$$
$$\Rightarrow {\displaystyle \frac{\mathrm{\Delta }T}{T}}=0.02$$

Therefore, the fractional increase in the tension of the wire is $$0.02$$.

Hence, the correct option is A.

Note:
The students may wonder that the increase in the frequency of the piano wire is in beats and fundamental frequency of the wires is in hertz and change in frequency is converted from beats into hertz. But beat is also the SI unit used to measure tempo in music or heart beats in one minute. Hence, there is no need for unit conversion.