Question

When only \(𝑆_2\) is emitting sound and it is at \(𝑄\), the frequency of sound measured by the detector in Hz is _________.

Answer 1

Correct Answer: 648

When only \(𝑆_2\) is emitting sound and it is at \(𝑄\), the frequency of sound measured by the detector in Hz is \(\underline{648.00}.\)

Explanation:  

The 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, \(v\) = \(324 \;ms^{−1}\), \(v_o\)​=\(0\) \(ms^{−1}\) and \(v_s\)_​ = 4\(\sqrt{2}\) ms⁻¹, let's calculate the observed frequency at \(𝑃\) when only \(𝑆_2\)emits sound

at \(𝑄\).\(f'=656\times\frac{324+0}{324+4\sqrt{2}}\)

\(f'=656\times\frac{324}{324+4\sqrt{2}}\)

\(f' = 648.00\)

Answer 2

f₀ = 656 Hz 
v = 324 m/s Frequency heard due to movement of (S₁)
\(f_1=(\frac{v}{v-u_s})f_0\)

\(f_1=\frac{324}{3520}\times656\)

And frequency heard due to movement of (S₂) 
f₂ = 656 Hz   
∴ Beat frequency \(Δ f = f_1 – f_2 \)
\(=656(\frac{324}{320}-1)\)

\(Δ f = 8.2\)