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

Step 1 (notation): The paper uses the same letter for two things. Let the emitted frequency be \(\nu\) and the speed of sound be \(c\) (these are the \(v\) and \(v\) in the options). The source starts from rest and moves toward the observer with acceleration \(a\).

Step 2: Track two crests emitted right after the start. Crest 1 leaves at \(t=0\) from distance \(D\); crest 2 leaves one period \(T=1/\nu\) later. In that time the source has moved forward by \(\tfrac{1}{2}aT^{2}\), so crest 2 starts from distance \(D-\tfrac{1}{2}aT^{2}\).

Step 3: Arrival times at the observer: \[t_1=\frac{D}{c},\qquad t_2=T+\frac{D-\tfrac{1}{2}aT^{2}}{c}\] The observed period is \[T'=t_2-t_1=T-\frac{aT^{2}}{2c}\]
Step 4: The observed frequency is \[\nu'=\frac{1}{T'}=\frac{1}{T\left(1-\dfrac{aT}{2c}\right)}=\frac{\nu}{1-\dfrac{a}{2c\nu}}\]
Step 5: Multiply top and bottom by \(2c\nu\): \[\nu'=\frac{2c\nu^{2}}{2c\nu-a}\] Writing \(\nu\) and \(c\) both as \(v\) gives \(\dfrac{2vv^{2}}{2vv-a}\), which is option (A). Note \(2c\nu-a<2c\nu\), so \(\nu'>\nu\), as expected for an approaching source. \[\boxed{\nu'=\frac{2c\nu^{2}}{2c\nu-a}}\]