Question
Explain Doppler effect in sound. Obtain an expression for apparent frequency of sound when source and listener are approaching each other.
✓
Answer
Whenever there is a relative motion between a source of sound and listener, the apparent frequency of sound heard by listener is different from actual frequency of sound emitted by the source.
Let S be a source of sound and L, the listener of sound, both initially at rest. Let v be the actual frequency of sound emitted by the source and $\lambda$ be the actual wavelength of sound emitted. If v is velocity of sound in still medium, then
$\lambda=\frac{v}{\text{v}}$
Let the distance between source and listener be V, so that v waves from the source reach listener in 1 second.
Vm = Velocity of medium, Vs = Velocity of source,
VL = Velocity of listener
Resultant velocity of sound along SL = (V + Vm)
SS' = Distance moved by source in 1 sec. = Vs along SL
$\therefore$ Relative velocity of sound w.r.t. source SS' = Vs' = [(V + Vm) - Vs]
As the frequency remains unchanged.
$\therefore$ V waves emitted in one second occupy the distance [(V + Vm) - Vs] apparent wavelength $\lambda'=\frac{[(\text{V}+\text{V}_\text{m})-\text{V}_\text{s}]}{\text{V}}$
Assuming both listener and source are moving in same direction, i.e. toward right.
LL' = VL’ relative velocity of sound wave w.r.t. listener (V + Vm) - VL
Apparent frequency of sound waves heard by listener is
$\text{v}'=\frac{(\text{V}+\text{V}_\text{m})-\text{V}_\text{L}}{\lambda'}$
$\text{v}'=\frac{[(\text{V}+\text{V}_\text{m})-\text{V}_\text{L}]\text{v}}{(\text{V}+\text{V}_\text{m})-\text{V}_\text{s}}$
When both approach each other, i.e. listener move towards the source:
Vs = Positive, VL = Negative.
$\text{v}'=\frac{\text{V}-(-\text{V}_\text{L})}{\text{V}-\text{v}_\text{s}}\text{v}=\Big(\frac{\text{V}+\text{V}_\text{L}}{\text{V}-\text{V}_\text{s}}\Big)\text{v}$
Apparent frequency (v') is greater than actual frequency (v).
Let S be a source of sound and L, the listener of sound, both initially at rest. Let v be the actual frequency of sound emitted by the source and $\lambda$ be the actual wavelength of sound emitted. If v is velocity of sound in still medium, then
$\lambda=\frac{v}{\text{v}}$
Let the distance between source and listener be V, so that v waves from the source reach listener in 1 second.
Vm = Velocity of medium, Vs = Velocity of source,
VL = Velocity of listener
Resultant velocity of sound along SL = (V + Vm)
SS' = Distance moved by source in 1 sec. = Vs along SL
$\therefore$ Relative velocity of sound w.r.t. source SS' = Vs' = [(V + Vm) - Vs]
As the frequency remains unchanged.
$\therefore$ V waves emitted in one second occupy the distance [(V + Vm) - Vs] apparent wavelength $\lambda'=\frac{[(\text{V}+\text{V}_\text{m})-\text{V}_\text{s}]}{\text{V}}$
Assuming both listener and source are moving in same direction, i.e. toward right.
LL' = VL’ relative velocity of sound wave w.r.t. listener (V + Vm) - VL
Apparent frequency of sound waves heard by listener is
$\text{v}'=\frac{(\text{V}+\text{V}_\text{m})-\text{V}_\text{L}}{\lambda'}$
$\text{v}'=\frac{[(\text{V}+\text{V}_\text{m})-\text{V}_\text{L}]\text{v}}{(\text{V}+\text{V}_\text{m})-\text{V}_\text{s}}$
When both approach each other, i.e. listener move towards the source:
Vs = Positive, VL = Negative.
$\text{v}'=\frac{\text{V}-(-\text{V}_\text{L})}{\text{V}-\text{v}_\text{s}}\text{v}=\Big(\frac{\text{V}+\text{V}_\text{L}}{\text{V}-\text{V}_\text{s}}\Big)\text{v}$
Apparent frequency (v') is greater than actual frequency (v).
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