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 be shortened so that it produces no beats with the tuning fork ?

Answer 1

Given that,

I=25 cm $=25\times {10}^{-2}m$

By shortening the wire the frequency increases $[f=(\frac{1}{2I}\left)\sqrt{\left(\frac{T}{m}\right)}\right]$

As the vibrating wire proudces 4 betas with 256 Hz, its frequency must be 252 Hz or 260 Hz. Its frequency decreases by shortening the wire.

SO, $252=\frac{1}{2\times 25\times {10}^{-2}}\sqrt{\left(\frac{T}{M}\right)}.....\left(1\right)$

Let length of the wire will be I, after it is slightly shortened,

$\Rightarrow 252=\frac{1}{2\times I_1}\sqrt{\left(\frac{T}{M}\right)}........\left(2\right)$

Dividing (1)by (2), we get,

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

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

=0.24609 m

So it should be shorten by (25-24.61)

=0.39 cm.