91Ó°ÊÓ

Constructive interference occurs at a point where the superposing waves are in phase and destructive interference occurs at points where the two waves are out of phase.

Since the frequency of sound changes and the speed of sound is constant, the wavelength of sound must change too to keep the speed of sound same

$$\begin{gathered} v_s={\mathrm{λ}}_{new}f \\ {\mathrm{λ}}_{new}=\frac{v_s}{f} \end{gathered}$$

Since the new frequency is half the original ie$$\begin{gathered} f=\frac{f_0}{2} \\ =\frac{500HZ}{2} \\ =250Hz \end{gathered}$$

$$\begin{gathered} {\mathrm{λ}}_{new}=\frac{340m/s}{250Hz} \\ =1.36m \end{gathered}$$

The ratio is given by

${\mathrm{λ}}_{new}=2{\mathrm{λ}}_0$

Earlier when the wavelength was unchanged then the students at the following positons were able to hear the sound

$\mathrm{Δ}x=m{\mathrm{λ}}_0$

Where m is an integer,

Students at$\mathrm{Δ}x={\mathrm{λ}}_0,2{\mathrm{λ}}_0,3{\mathrm{λ}}_0.....$would hear the maximum sound.

When the wavelength is changed, path difference is given by

$\mathrm{Δ}{x}_{new}=2m{\mathrm{λ}}_0$

This means that at the position in which sound was initially maximum, the interference in the new case for students at${\mathrm{λ}}_0,3{\mathrm{λ}}_0,5{\mathrm{λ}}_0.....$would be the destructive one

But the interference in the new case for the students at $2{\mathrm{λ}}_0,4{\mathrm{λ}}_0,6{\mathrm{λ}}_0.....$ will stay constructive.

Therefore, some students who initially heard a loud tone will hear again a loud tone, but others win this group will now hear nothing. Hence the correct option is C