Labels \(1,2,3,4,5\) show wavefronts emitted successively at regular interval \(T=\frac{1}{f}\) , \(f\) is frequency, when the source is located at \(S_{1},S_{2},S_{3},S_{4},S_{5}\) 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 \(Vt\). The wavefront for position \(S_{5}\) is just being emitted and therefore it is a point

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

In the above diagram value of \(CD\) will be

## Comments

Sort by:

TopNewest@Raghav Vaidyanathan @Ronak Agarwal @Nishant Rai , Answers are 110m/s and \(CD=\frac{2}{3}m\) – Tanishq Varshney · 2 years ago

Log in to reply

This is easy. The difference of the radii of the two different circles is given by \( Wavelength\) 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 \( \dfrac{{v}_{Source}}{f} \)

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 · 2 years ago

Log in to reply

– Tanishq Varshney · 2 years ago

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

– Tanishq Varshney · 2 years ago

ok i got its \(CD-1=\frac{1}{3}\)Log in to reply