Question

A tuning fork of frequency 256 Hz produces 4 beats per second with a wire of length 25 cm vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire we shortened so that it produces no beats with the tuning fork?

Answer 1

Given:
Length of the wire l = 25 cm = 25 × 10⁻² m
Frequency of tuning fork \[f\] = 256 Hz
Let T be the tension and m the mass per unit length of the wire.

Frequency of the fundamental note in the wire is given by : \[f = \frac{1}{2l}\sqrt{\frac{T}{m}}\]

It is clear from the above relation that by shortening the length of the wire, the frequency of the vibrations increases.
In the first case :

\[256 = \frac{1}{2 \times 25 \times {10}^{- 2}}\sqrt{\left( \frac{T}{m} \right)}         .  .  . (1)\]

Let the length of the wire be l₁, after it is slightly shortened.

As the vibrating wire produces 4 beats with 256 Hz, its frequency must be 252 Hz or 260 Hz. Again, its frequency must be 252 Hz, as the beat frequency decreases on shortening the wire.

In the second case :

\[\Rightarrow   252 = \frac{1}{2 \times I_1}\sqrt{\frac{T}{m}}         .  .  . (2)\]

Dividing (2) by (1), we have:

\[\frac{252}{256} = \frac{I_1}{25 \times {10}^{- 2}}\] 

\[ \Rightarrow    I_1  = \frac{252 \times 25 \times {10}^{- 2}}{256}\] 

\[               =   0 . 24609 \text{ m }\]

So, it must be shortened by (25 − 24.61)
 = 0.39 cm.