Question

A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire Bproduces 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 : \[n_A  = \frac{1}{2I}\sqrt{\frac{T_A}{m}}\]

Fundamental frequency of wire B is given by:

\[n_B  = \frac{1}{2I}\sqrt{\frac{T_B}{m}}\]

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:

\[  \frac{n_A}{n_B} = \frac{606}{600} = \frac{\left( 1/2I \right)\sqrt{\left( T_A /m \right)}}{\left( 1/2I \right)\sqrt{T_B /m}}\] 

\[ \Rightarrow \frac{606}{600} = \frac{\sqrt{T_A}}{\sqrt{T_B}}\] 

\[ \Rightarrow   \frac{\sqrt{T_A}}{\sqrt{T_B}} = \frac{606}{600} = 1 . 01\] 

\[ \Rightarrow   \frac{T_A}{T_B} = 1 . 02\]