34. An ambulance is emitting siren of frequency (fs) (100 Hz) , and moving towar

Question

34. An ambulance is emitting siren of frequency (fs) (100 Hz) , and moving towards you at velocity 10m/s , while you are also running toward the ambulance with 0.1 m/s velocity . what will be frequency you will hear ? (assume sound velocity = 340 m/s) (a) 100 Hz (b) 103 Hz (c) 121 Hz (d) none of them 35. You can hear sound waves in space (a) True (b)Flase 36. what will be the wavelength of third harmonics of one end open and one end close pipe of length L? (a) (2/3)L (b) (3/4)L (c) (4/3)L (d) none of them

Explanation / Answer

Given

34.

ambulance frequency fs = 100 Hz

velocityof the ambulace towards the observer is vs = 10 m/s

velocity of the observer moving towards the source is vo = 0.1 m/s

from the given situation the frequency heard by the observer will be grater than the original freqency fs

from Doppler's effect  

the apparent chagen in frequency of the sound due to relative motion between the source and observer

the apparent frequency is given by f' = fs(v+vo/v-vs)

f' = 100 ((340+0.1) /(340-10)) Hz

f' = 103.06 Hz

the answer is option is b 103 Hz------------ which is greater than fs

35.

Faslse  

sound is a longitudinal wave, to propagate the wave from one point to other we require medium , so in space there is empty space whern the sound waves can not move so we can not hear sound in sspace

36.

closed pipe

L = Lambda/4 ==> lambda = 4*L

v = Lambda*f ==> f = V/lambda

fundamental frequency or first harmonic is f1 = v/ 4*L  

the second harmonic is  

L = 3Lambda/4 ==> lambda = 4*L/3

v = Lambda*f ==> f = V/lambda

fundamental frequency or first harmonic is f1 = 3v/ 4*L

so the wavelength of the third harmonic of closed pipe is lambda = (4/3)L

answer is option c <<----- Answer

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