Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
An engine whestles at a frequency of 256Hz moves towards a wall with a velocity of 50m/s. How many number of beats will be sounded to the engine?
Doppler's fomula is n' = n (v -vo)/(v-vs) where n' = apprant frequency n = original frequency v = spped of sound in air that is 330 m/s vo = speed of observer vs = speed of the source Here n = 256 hz vo = 0 vs = 50m/s pluging in the values we will get n = 497 hz now with this frequency the sond will reflect from the wall so the source will be stationary we will again use the same formula n" = n (v -vo)/(v-vs) now vo = 50m/s , vs = 0 and v =330m/s n" = frequency heard by the observer now pluging in the values we will have n" = 497(330 - 50)/(330 - 0) = 497x 280/330 = 422hz so the no of beats heard by the observer is n" - n = 422 - 256= 166 beats per second
Doppler's fomula is
n' = n (v -vo)/(v-vs)
where n' = apprant frequency
n = original frequency
v = spped of sound in air that is 330 m/s
vo = speed of observer
vs = speed of the source
Here n = 256 hz vo = 0 vs = 50m/s pluging in the values we will get n = 497 hz
now with this frequency the sond will reflect from the wall so the source will be stationary
we will again use the same formula
n" = n (v -vo)/(v-vs)
now vo = 50m/s , vs = 0 and v =330m/s n" = frequency heard by the observer
now pluging in the values we will have
n" = 497(330 - 50)/(330 - 0)
= 497x 280/330
= 422hz so the no of beats heard by the observer is n" - n = 422 - 256= 166 beats per second
Find solution of question on doubt solving app Instasolv
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -