Question

A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire B produces 6 beats per second with it when the tension in A is slightly increased. Find the the ratio of the tension in A to the tension in B.

Answer 1

Mass per unit length of both the wires​ = m

Fundamental frequency of wire of length $\left(l\right)$ and tension $\left(T\right)$ is given by:

$n=\frac{1}{2I}\sqrt{\frac{T}{m}}$

It is clear from the above relation that as the tension increases, the frequency increases.

Fundamental frequency of wire A is given by:

$n_{\mathrm{A}}=\frac{1}{2I}\sqrt{\frac{T_{\mathrm{A}}}{m}}$

Fundamental frequency of wire B is given by:

$n_{\mathrm{B}}=\frac{1}{2I}\sqrt{\frac{T_{\mathrm{B}}}{m}}$

It is given that 6 beats are produced when the tension in A is increased.

⇒​ $n_A=606=\frac{1}{2l}\sqrt{\frac{T_A}{m}}$
Therefore, the ratio can be obtained as:

$$\begin{gathered} \frac{n_{\mathrm{A}}}{n_{\mathrm{B}}}=\frac{606}{600}=\frac{\left(1/2I\right)\sqrt{\left(T_{\mathrm{A}}/m\right)}}{\left(1/2I\right)\sqrt{T_{\mathrm{B}}/m}} \\ \Rightarrow \frac{606}{600}=\frac{\sqrt{T_{\mathrm{A}}}}{\sqrt{T_{\mathrm{B}}}} \\ \Rightarrow \frac{\sqrt{T_{\mathrm{A}}}}{\sqrt{T_{\mathrm{B}}}}=\frac{606}{600}=1.01 \\ \Rightarrow \frac{T_{\mathrm{A}}}{T_{\mathrm{B}}}=1.02 \end{gathered}$$