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.
$f'=656\times\frac{324+0}{324+4\sqrt{2}}$
$f'=656\times\frac{324}{324+4\sqrt{2}}$