Question

A tuning fork unknown frequency makes three beats per second with a standard fork of 384Hz.The beat frequency decreases when a small piece of wax is put on a prong of the first fork. What is the frequency of this fork (this fork)

Answer 1

Hint:A tuning fork is a useful device for how vibrating objects can produce sound. When it is hit with a rubber hammer, then the tuning fork starts to vibrate. The rear vibration of the times produces disturbances of surrounding air molecules. When two tuning forks A and B are sounded together, a number of beats per second are heard. The frequency of A is f1. when prong of B puts a small piece of wax on a prong of the first fork A, the number of beats per second decreases.

Complete step by step answer:
The beat frequency ($\beta $) is the no. of beats per unit time. It is given by
$\beta \, = \left| {{{\rm{f}}_1} - \;{{\rm{f}}_2}} \right|$
Here f1 and f2 are the frequencies.
If we add wax to a new tuning fork, the frequency ${{\rm{f}}_1}$​reduced and also beats reduced this means that${{\rm{f}}_1}\; > {{\rm{f}}_2}$.
Here, $\beta \; = \;\dfrac{3}{1}{s^{ - 1}}\; = \;3{\rm{Hz}}$
A standard fork (B) frequency ${{\rm{f}}_2}\; = \;384{\rm{Hz}}$
Therefore from the above beat and frequency formula, we get,
${{\rm{f}}_1} - \;{{\rm{f}}_2}\; = \,3$
$ \Rightarrow {{\rm{f}}_1}\; = \;{{\rm{f}}_2} + \;3{\rm{Hz}}$
$ \Rightarrow \;{{\rm{f}}_1}\; = \;384{\rm{Hz}} + \;3{\rm{Hz}}$
$ \Rightarrow \;{{\rm{f}}_1}\; = \;387{\rm{Hz}}$
 Thus, the frequency of the fork = 387Hz.

Note:remember that loading of any tuning fork decreases its frequency. Contrary to this, the ringing of a tuning fork increases its frequency. Because of the change in the inertia of the tuning fork, this is a property of mass. As the tuning fork is loaded with wax, inertia increases resulting in a decrease of frequency. In case of filing, the inertia decreases, so the frequency increases.
i) The echo will have the same speed as a sound.
ii) The beats frequency has the same units as frequency.