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Line PQ and RS are the interface of medium 1 and medium 2 respecively.A cart is running in between interface with velocity $\frac{3v}{4}$ towards right as shown in the attached image.'v' is the speed of sound in air.A man A is moving with velocity $\frac{v}{3}$ towards left and man B with velocity $\frac{v}{2}$ towards right in respective medium.

Given speed of sound in medium separted by PQ and medium separated by RS is 5v and 7v redpectively.Car is emitting sound of frequencu $f_0$.

Find frequency and wavelength of sound recieved by Man A and B respectively

i dont know how to find frequency recieved by them if medium changes.enter image description here

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    $\begingroup$ The frequency does not change across an interface. $\endgroup$
    – Farcher
    Apr 18 '17 at 13:05
  • $\begingroup$ So the wavelength changes but the frequency does not. $\endgroup$
    – Farcher
    Apr 18 '17 at 13:07
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So I finally made myself the solution and im posting it... As @Farcher told me Frequency doesn't change with medium..Hence,

$\lambda_A =\frac{vT + v_{s}T}{\frac{1}{5}} = \frac{35v}{4f_0}$

$f_A = \frac{5v -\frac{v}{3}}{\frac{35v}{4f_0}} = \frac{8f_0}{15}$

$\lambda_B=\frac{vT+v_{s}T}{\frac{1}{7}} = \frac{7v}{4f_0}$ And $f_B=\frac{7v-\frac{v}{2}}{\frac{7v}{4f_0}} =\frac{26f_0}{7}$

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Frequency only depends upon the source and not on the medium.

Velocity of A wrt cart = VAC = -v/3 - 3v/4 = -13v/12.
Using doppler's effect f' = fo(Vsource - VA) / Vsource
Therefore f' = fo(5V - 13V/12)/5V.

Similarly, You can find frequency for B

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  • $\begingroup$ Your answer is wrong ..according to my booklet the the frequency recieved by man A is $\frac{8}{15}f_0$ and manB is $\frac{26}{7}f_0$ $\endgroup$
    – Abhash Jha
    Apr 19 '17 at 1:55

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