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Labels 1,2,3,4,51,2,3,4,5 show wavefronts emitted successively at regular interval T=1fT=\frac{1}{f} , ff is frequency, when the source is located at S1,S2,S3,S4,S5S_{1},S_{2},S_{3},S_{4},S_{5} respectively. Each wavefront will have its centre at the position where the source was situated while emitting the wavefront. The radius of preceding wavefront will exceed the next one by an amount VtVt. The wavefront for position S5S_{5} is just being emitted and therefore it is a point

Consider the diagram above, AB=43mAB=\frac{4}{3}m. Speed of sound and frequency are 330 m/s330~m/s and 330 Hz330~Hz respectively. The velocity of source is

In the above diagram value of CDCD will be

Note by Tanishq Varshney
4 years, 1 month ago

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@Raghav Vaidyanathan @Ronak Agarwal @Nishant Rai , Answers are 110m/s and CD=23mCD=\frac{2}{3}m

Tanishq Varshney - 4 years, 1 month ago

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This is easy. The difference of the radii of the two different circles is given by Wavelength Wavelength since you see every wavefront is being released at time

1/f ( Distance=Speed Time , v1/f )

But the extra distance between A and B is coming because of the speed of the source

The extra distance is vSourcef \dfrac{{v}_{Source}}{f}

With this difference you can calculate the speed.

Find CD is also easy since you can subtract the extra difference from the wavelength to get the answer.

Ronak Agarwal - 4 years, 1 month ago

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can u elaborate the CD one, in the first one u mean to say 43330330=vsourcef\frac{4}{3}-\frac{330}{330}=\frac{v_{source}}{f}

Tanishq Varshney - 4 years, 1 month ago

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ok i got its CD1=13CD-1=\frac{1}{3}

Tanishq Varshney - 4 years, 1 month ago

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