Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
0 votes

An observer is moving towards a stationary source of frequency 320 Hz with a velocity of 10 m/s . The speed of sound in air is 343 m/s . The new frequency heard by the observer is

$(a)\;310 \;Hz\qquad(b)\;336 \;Hz\qquad(c)\;340 \;Hz\qquad(d)\;329 \;Hz$

Can you answer this question?

1 Answer

0 votes
Answer : (d) $\;329 \;Hz$
Explanation :
New frequency $\;f^{|} =(\large\frac{v_{0}+v}{v})\;f$
$v_{0}\;$ velocity of observer
$v\;$ velocity of sound
By Doppler's effect
therefore , $\;f^{|}=(\large\frac{10+343}{343})\times320$
$=329\;Hz \;.$
answered Mar 8, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App