91Ó°ÊÓ

The frequency of the sound emitted by the speakers is ${\mathit{f}}_{\mathbf{s}}\mathbf{=}\frac{\mathbf{v}}{\mathbf{λ}}$and the angular frequency of the turntable is ${\mathit{v}}_{\mathbf{s}}\mathbf{=}\mathit{r}\mathit{ϖ}$and the frequency detected by the listener is ${\mathit{f}}_{\mathbf{L}}\mathbf{=}{\mathit{f}}_{\mathbf{s}}\left(\frac{v}{v∓v_s}\right)$where, v is the speed of sound in the air, vs is the speed of speaker and speaker2 as well, ${\mathit{f}}_{\mathbf{s}}$ is the frequency of the sound emitted by each speaker

The speaker that will be moving toward the listener (speaker) and (speaker2) for the speaker that moves away from the listener as the turntable rotates.

So, let's find the frequency of the sound emitted by the speakers, which can be determined as follows

${\mathrm{f}}_{\mathrm{s}}=\frac{\mathrm{v}}{\mathrm{λ}}=\frac{344}{31.3×{10}^{-2}}=1100\mathrm{Hz}$

Now, the other thing we need to determine is the velocity of each speaker, where the velocity of the speakers is the same as the velocity of the table's rims, so we need to convert it to (rad/s)

$v_s=r\mathrm{Ï–}$where (r= 0.75 m) is the radius of the turntable and (w =75 rpm) is the angular frequency of the turntable and convert into rad/s

$$\begin{gathered} {\mathrm{v}}_{\mathrm{s}}=0.75\mathrm{m}×75\mathrm{rpm}×\frac{2\mathrm{Ï€°ù²¹»å}}{1\mathrm{rev}}×\frac{1\min }{60\mathrm{s}} \\ {\mathrm{v}}_{\mathrm{s}}=5.9\mathrm{m}/\mathrm{s} \end{gathered}$$

The frequency detected by the listener from speaker is $f_L=f_s\left(\frac{v}{v∓v_s}\right)$

Substitute the values in the above equation we get,

Use the minus sign in the denominator because the source "speaker " is moving toward the listener,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{L}1}=1100×\left(\frac{344}{344-5.9}\right) \\ {\mathrm{f}}_{\mathrm{L}1}=1119\mathrm{Hz} \end{gathered}$$

Use the plus sign in denominator because the source "speaker2" is moving away from the listener:

$$\begin{gathered} {\mathrm{f}}_{\mathrm{L}2}=1100×\left(\frac{344}{344-5.9}\right) \\ {\mathrm{f}}_{\mathrm{L}2}=1081\mathrm{Hz} \end{gathered}$$

A beat frequency of 38 Hz means that there will be a 38 pulse per second, so, no, the listener will not be able to distinguish individual beats.

$$\begin{gathered} {\mathrm{f}}_{\mathrm{beat}}={\mathrm{f}}_{\mathrm{L}1}-{\mathrm{f}}_{\mathrm{L}2} \\ =1119\mathrm{Hz}-1081\mathrm{Hz} \\ =38\mathrm{Hz} \end{gathered}$$

Therefore, the beat frequency of the listener is 38Hz