Physics · Sound Waves

JEE Advanced 2023 — Paper 2 — Question 40

S1S_{1} and S2S_{2} are two identical sound sources of frequency 656 Hz . The source S1S_{1} is located at OO and S2S_{2} moves anticlockwise with a uniform speed 42 m s−14 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1} on a circular path around O , as shown in the figure. There are three points P,Q\mathrm{P}, \mathrm{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 PP. The source S1\mathrm{S}_{1} can move along direction OP. [Given: The speed of sound in air is 324 m s−1324 \mathrm{~m} \mathrm{~s}^{-1} ]

Consider both sources emitting sound. When S2\mathrm{S}_{2} is at R and S1\mathrm{S}_{1} approaches the detector with a speed 4 m s−14 \mathrm{~m} \mathrm{~s}^{-1}, the beat frequency measured by the detector is ____\_\_\_\_ Hz.

Answer: 8.2

Numerical answer — enter this value.

Step-by-step solution

f1=(vv−vs′)f0(vs′=4m/s)f_{1}=\left(\frac{v}{v-v_{s}^{\prime}}\right) f_{0} \left(v_{s}^{\prime}=4 m / s\right) f1=(324324−4)656=324320×656f_{1}=\left(\frac{324}{324-4}\right) 656=\frac{324}{320} \times 656 f1=664.2 Hz\mathrm{f}_{1}=664.2 \mathrm{~Hz} Now, f2=(vv−vscos⁡90)f0=f0f_{2}=\left(\frac{v}{v-v_{s} \cos 90}\right) f_{0}=f_{0} f2=656 Hz\mathrm{f}_{2}=656 \mathrm{~Hz}. Hence, the beats frequency measured by the detector. fb=∣f1−f2∣=664.2−656=8.2 Hz\mathrm{f}_{\mathrm{b}}=\left|\mathrm{f}_{1}-\mathrm{f}_{2}\right|=664.2-656=8.2 \mathrm{~Hz}.

Solution figure

Answer key and solution verified before publishing.

Practise Sound Waves

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Advanced 2023
Paper
Paper 2
Subject
Physics
Chapter
Sound Waves
Topic
Doppler Effect of Sound
S 1 and S 2 are two identical sound sources of frequency 656 Hz . The… | JEE Advanced 2023 PYQ with Solution · DhiX AI