Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of

A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s.  

Option: 1

1 m/s


Option: 2

2 m/s


Option: 3

0.5 m/s


Option: 4

1.5 m/s


Answers (1)

best_answer

f_{b}=f_{1}-f_{2}

2=680\left(\frac{340}{340-\nu_{s}}\right)-680\left(\frac{340}{340+\nu_{s}}\right)

=680\left(1-\frac{\nu_{s}}{340}\right)^{-1}-680\left(1+\frac{\nu_{s}}{340}\right)^{-1}

\begin{array}{l}{\text { Using Binomial expansion, we have }} \\ \\ {2=680\left[\frac{2 . \nu_{s}}{340}\right] \Rightarrow \nu_{s}=0.5 \mathrm{m} / \mathrm{s}}\end{array}

Posted by

manish

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 Session 2 Result 2026 tomorrow; steps to download, how to calculate marks
anam-khan

JEE Main 2026 Result LIVE: Session 2 scores tomorrow at jeemain.nta.nic.in; cutoff marks, toppers list
anam-khan

JEE Main Marks vs Rank 2026: Category-wise qualifying cut-offs show rise over years

Latest Question