Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A source of sound of frequency 500 Hz situated between a stationary observer <img alt="O" src="https://entrancecorner.on

A source of sound S of frequency 500 Hz situated between a stationary observer O and a wall, moves towards the wall with a speed of 2 m/s. If the velocity of sound is 332 m/s, then the number of beats per second heard by the  observer is (approximately)-

 

Option: 1

8 Hz


Option: 2

6 Hz


Option: 3

4 Hz


Option: 4

2 Hz


Answers (1)

best_answer

 

\mathrm{\begin{aligned} n_1=n\left(\frac{v}{v+v_s}\right) & =500\left(\frac{332}{332+2}\right) \\ & =500\left(\frac{332}{334}\right)=497 \mathrm{~Hz} . \end{aligned}} ------------(1)

For sound reflected from the wall-

\mathrm{\begin{aligned} & n_2=n\left(\frac{v}{v-v_5}\right) \\ & n_2=500\left(\frac{3.32}{332-2}\right)=503 \mathrm{~Hz} . \end{aligned}} -------------(2)

Beat frequency \mathrm{\begin{aligned} =n_2-h_1 & =503-457 \\ & =6 \mathrm{HZ} \end{aligned}}

Posted by

Deependra Verma

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon at jeemain.nta.nic.in
anam-khan

Focus on modern physics, formulas to solve tricky problems; 5 tips to tackle JEE Main physics 2025

Latest Question