IIT JEE , AIEEE Forum - Physics , Mechanics

sachin_gupta1991

Q. A receiver and a source of sonic oscillations of frequency 2000 Hz are located on the x-axis.The source swings harmonically on the x-axis with circular frequency

and amplitude 50 cm.At what value of

will the frequency bandwidth(difference of max. and min. frequency) registered by stationary receiver be equal to 200 Hz?The velocity of sound in air is equal to 340 m/s.

Decoder

lookk u want a beat frequency of 200Hz..

draw the diagram...with both source & detector on +ve side of x-axis..

now .think when is the max. frequency detected..

by doppler effect it will be max. wen source moves with max. velocity ...( in s.h.m tht is at mean pos.)..

now the ques. comes source moves towards observer or away frm it..

in case of away the frequency detected will be least..

so wat is tht velocity... ==>> wA..
where both of them r given to u..
solve for both..

take the diffrence and equate it to 200..

cheers.!

sachin_gupta1991

I am not getting the correct answer by this method.

The.CHOSEN.ONE

me too not getting it...........more codes pleese DECODER. : )

harsha_27

The answer is

= 680(

101 - 10)

33.9 rad/sec

Decoder

hey harsha how u got tht root..

harsha_27

see let the velocity be v.

velocity of sound in air = c = 340m/s

now f1 = f { c / (c-v)}

f2 = f { c/(c+v)}

now given f1-f2 = 200Hz

=> 2fcv/(c²-v²) = 200

divide and multiply by v²....

2f (c/v)/[(c²/v²)-1] = 200

let c/v = x........

now 2fx/(x²-1) = 200

substituting f=2000Hz and simplifying.......

x²-20x-1 = 0

solving for x ......

x = 10 +

101

=> c/v = 10 +

101

=> v = 340(

101 - 10)

now v = Aw

=> w = v/A

=> w = 340(

101-10)/(50x10^-2)

=> w = 680(

101-10)

=> w

33.9 rad/sec

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