A source, approaching with speed u towards the open end of a...

Consider a tuning fork as a source of sound which is moving towards the stationary pipe with velocity $u$.

According to the Doppler effect ,

$f ' = {f}_{s}\left(\dfrac{v - {v}_{o}}{v - {v}_{s}}\right)</p><p>Where,</p><p>$v =$is speed of sound</p><p>${v}_{s} =$is velocity of source</p><p>${v}_{o} =$is velocity of observer</p><p>${f}^{'} =$is appeared frequency</p><p>${f}_{s} =$is actual frequency</p><p>So the appeared frequency of the sound will be,</p><p>$f ' = {f}_{s}\left(\dfrac{v - 0}{v - u}\right)$</p><p>Since pipe is closed at one end so it will be like a close organ pipe and we know that close organ pipe has only odd harmonics.</p><p>And for resonance, appeared frequency should match with any of the natural frequency of the closed organ pipe.If fundamental frequency of the closed organ pipe is${f}_{0}$, then its natural frequencies will be${f}_{0}, 3{f}_{0}, 5{f}_{0} . . . . \left(2n - 1\right){f}_{0}$</p><p>So for resonance,</p><p>${f}_{s}\left(\dfrac{v}{v - u}\right) = \left(2n - 1\right){f}_{0}$</p><p>${f}_{s} = \left(1 - \dfrac{u}{v}\right)\left(2n - 1\right){f}_{0}$</p><p>Now from options\left(A, B, C\right)$

$∵ u = 0 . 8v$

So,

${f}_{s} = \left(1 - 0 . 8\right)\left(2n - 1\right){f}_{0}$

${f}_{s} = \dfrac{2n - 1}{5}{f}_{0}$

$n = 1, {f}_{s} = \dfrac{{f}_{0}}{5}$

$n = 2, {f}_{s} = \dfrac{3{f}_{0}}{5}$

$n = 3, {f}_{s} = \dfrac{5{f}_{0}}{5} = {f}_{0}$

$n = 4, {f}_{s} = \dfrac{7{f}_{0}}{5}$

$n = 5, {f}_{s} = \dfrac{9{f}_{0}}{5}$

$n = 6, {f}_{s} = \dfrac{11{f}_{0}}{5}$

So option (A) is correct, (B) and (C) are incorrect

Now from option\left(D\right)$</p><p>$∵ u = 0 . 5v$</p><p>So,</p><p>${f}_{s} = \left(1 - 0 . 5\right)\left(2n - 1\right){f}_{0}$</p><p>${f}_{s} = \left(\dfrac{2n - 1}{2}\right){f}_{0}$</p><p>$n = 1, {f}_{s} = \dfrac{{f}_{0}}{2}$</p><p>$n = 2, {f}_{s} = \dfrac{3{f}_{0}}{2} = 1 . 5{f}_{0}$</p><p>$n = 3, {f}_{s} = \dfrac{5{f}_{0}}{2}$</p><p>Hence option \left(D\right)$ is correct.