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

The Minimum Length will be 0.391 cm

Explanation:

• We can calculate the frequency from,

f = $\frac {1}{2L} \sqrt \frac {T}{\rho}$

• When the length was L=25 cm. Then,

256 - 4 = $\frac {1}{2L} \sqrt \frac {T}{\rho}$

252 = $\frac {1}{2L} \sqrt \frac {T}{\rho}$

• When the length is shortened, we assume it to be L’.

256 = $\frac {1}{2L} \sqrt \frac {T}{\rho}$

256 = $\frac {1}{2L'} \sqrt \frac {T}{\rho}$

• Now,

$\frac {L'}{L} = \frac {252}{256}$

$L' = 24.609 cm$

Hence the decrease in length will be,

$\Delta L = (25 - 24.609)$ cm

$\Delta L$ = 0.391 cm