Question

A source emitting a sound of frequency \(v\) is placed at a large distance from an observer. The source starts moving towards the observer with uniform acceleration \(a\). The speed of sound in the medium is \(v\). The frequency the observer hears, corresponding to the wave emitted just after the source starts, is:

• A. \(\left(\dfrac{2vv^{2}}{2vv-a}\right)\)
• B. \(\dfrac{2vv}{2vv-a}\)
• C. \(\dfrac{2vv^{2}}{2vv^{2}-a}\)
• D. \(\dfrac{vv^{2}}{2vv-a}\)

Hint:

Over the first period the source speed averages \(\tfrac12 aT=a/(2\nu)\); feed that into \(\nu'=\nu c/(c-u)\).

Answer 1

The Correct Option is A

Use the standard Doppler shift but with the source speed evaluated over the first period, because the source is accelerating.

Let frequency be $\nu$, speed of sound $c$. For a source approaching at speed $u$, \[\nu'=\nu\,\frac{c}{c-u}\] Just after the start, the relevant speed is the average speed built up while the first wavelength is being laid down. Over one period $T=1/\nu$ the source accelerates from rest, so its effective (average) speed is \[u=\tfrac12 aT=\frac{a}{2\nu}\] Insert this into the Doppler formula: \[\nu'=\nu\,\frac{c}{c-\dfrac{a}{2\nu}}=\nu\,\frac{2c\nu}{2c\nu-a}=\frac{2c\nu^{2}}{2c\nu-a}\] With both symbols shown as $v$, this is $\dfrac{2vv^{2}}{2vv-a}$, option (A). Since the denominator is smaller than $2c\nu$, the heard pitch is higher than $\nu$, consistent with an approaching source. \[\boxed{\nu'=\frac{2c\nu^{2}}{2c\nu-a}}\]