Waves

To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of $\pi $. The equation of the reflected wave will be

$y = a\sin \left( {\omega t + kx + \pi } \right)$

$ \Rightarrow y = - a\sin \left( {\omega t + kx} \right)$

KEY CONCEPT : The frequency of a tuning fork is given by the expression

$f = {{{m^2}k} \over {4\sqrt 3 \pi {\ell ^2}}}\sqrt {{Y \over \rho }} $

As temperature increases, $\ell $ increases and therefore $f$ decreases.

KEY CONCEPT : The fundamental frequency for closed organ pipe is given by ${\upsilon _c} = {v \over {4\ell }}$ and

For open organ pipe is given by ${\upsilon _0} = {v \over {2\ell }}$

$\therefore$ ${{{\upsilon _0}} \over {{\upsilon _c}}} = {v \over {2\ell }} \times {{4\ell } \over v} = {2 \over 1}$

A tuning fork produces $4$ beats/sec with another tuning fork of frequency $288$ cps. From this information we can conclude that the frequency of unknown fork is $288+4$ $cps$ or $288-4$ $cps$ i.e. $292$ $cps$ or $284$ $cps.$

Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork. Here also beats per second decreases to 2 from 4. So the difference between frequency decreases.

This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork.

So the frequency of the unknown tuninh fork is = 292 Hz