Question

𝑆₁ and 𝑆₂ are two identical sound sources of frequency 656 Hz. The source 𝑆₁ is located at 𝑂 and 𝑆₂ moves anti-clockwise with a uniform speed of 4\(\sqrt{2}\) ms^-1 on a circular path around 𝑂, as shown in the figure. There are three points 𝑃, 𝑄 and 𝑅 on this path such that 𝑃 and 𝑅 are diametrically opposite while 𝑄 is equidistant from them. A sound detector is placed at point 𝑃. The source 𝑆₁ can move along the direction 𝑂𝑃. [Given: The speed of sound in air is 324 ms^-1]. When only 𝑆₂ is emitting sound and it is at 𝑄, the frequency of sound measured by the detector in Hz is _________

Answer 1

he Doppler effect formula for sound when the source is moving towards the observer is given by:
\(f'=f\times\frac{v+v_0}{v+v_s}\)​​
Where:

• ′f′ is the observed frequency,
• f is the emitted frequency,
• v is the speed of sound in the medium (given as 324 ms⁻¹),
• v_o​ is the speed of the observer (in this case, at point 𝑃),
• v_s​ is the speed of the source.

Initially, only 𝑆₂ emits sound at 𝑄. To calculate the frequency observed at 𝑃, let's determine the relative speed of 𝑆₂ towards the detector at 𝑃.
The observer's speed (v_o​) is 0 since the detector is stationary at 𝑃.
Now, the speed of 𝑆₂ concerning the detector at 𝑃 will be the component of the speed of 𝑆₂ perpendicular to the line 𝑄𝑃.
Given that 𝑆₂ moves at a uniform speed of 4\(\sqrt{2}\) ms⁻¹ on a circular path around 𝑂, and 𝑄 is equidistant from 𝑃 and 𝑅, which are diametrically opposite, the speed of 𝑆₂ towards the detector at 𝑃 is 4\(\sqrt{2}\)​ ms⁻¹
Now, let's use the Doppler effect formula:
\(f'=f\times\frac{v+v_0}{v+v_s}\)
Given that f=656 Hz, v=324 ms⁻¹,v_o​=0 ms⁻¹and v_s​=4\(\sqrt{2}\) ms⁻¹, let's calculate the observed frequency at 𝑃 when only 𝑆₂ emits sound at 𝑄.
\(f'=656\times\frac{324+0}{324+4\sqrt{2}}\)
\(f'=656\times\frac{324}{324+4\sqrt{2}}\)