When only \(π_2\) is emitting sound and it is at \(π\), the frequency of sound measured by the detector in Hz is _________.
Correct Answer: 648
Solution and Explanation
When only \(π_2\) is emitting sound and it is at \(π\), 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:
\(f'=f\times\frac{v+v_0}{v+v_s}\)ββ
Where:
- \('f\)β² is the observed frequency,
- \(f\) is the emitted frequency,
- \(v\) is the speed of sound in the medium (given as 324 msβ1),
- \(v_o\)β is the speed of the observer (in this case, at point π),
- \(v_s\)β is the speed of the source.
Initially, only π2 emits sound at π. To calculate the frequency observed at π, let's determine the relative speed of π2 towards the detector at π.
The observer's speed \((v_oβ)\) is 0 since the detector is stationary at π.
Now, the speed of π2 concerning the detector at π will be the component of the speed of π2 perpendicular to the line ππ.
Given that π2 moves at a uniform speed of 4\(\sqrt{2}\) \(ms^{β1}\) on a circular path around π, and π is equidistant from π and π
, which are diametrically opposite, the speed of π2 towards the detector at π is 4\(\sqrt{2}\)β \(ms^{β1}\)
Now, let's use the Doppler effect formula:
\(f'=f\times\frac{v+v_0}{v+v_s}\)
Given that \(f\)= \(656\) Hz, \(v\) = \(324 \;ms^{β1}\), \(v_o\)β=\(0\) \(ms^{β1}\) and \(v_s\)β = 4\(\sqrt{2}\) msβ1, let's calculate the observed frequency at \(π\) when only \(π_2\)emits sound
at \(π\).\(f'=656\times\frac{324+0}{324+4\sqrt{2}}\)
\(f'=656\times\frac{324}{324+4\sqrt{2}}\)
\(f' = 648.00\)
Consider both sources emitting sound. When π2 is at π
and π1 approaches the detector with a speed 4 m sβ1 , the beat frequency measured by the detector is _______Hz
Correct Answer: 8.2
Solution and Explanation
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
β΄ Beat frequency \(Ξ f = f_1 β f_2 \)
\(=656(\frac{324}{320}-1)\)
\(Ξ f = 8.2\)
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