Doppler Effect
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Doppler Effect
Wave Motion of Class 11
Doppler Effect
The speed of sound wave in still air is about 340 m/s. If an observer moves toward or away from the sound source, the apparent speed of sound waves must increase or decrease with respect to the observer. Consequently, the rate at which sound wavefronts strike at the observer’s ear increases if the observer is in motion toward the source or decreases if he moves away from the source. The rate at which wavefronts reach the observer’s ear is just the frequency he notes for that sound wave. Therefore, a moving observer notices a change in the frequency of the sound source.
Similarly, a stationary observer measures a change in the frequency if the source is in motion. The change in frequency due to the motion of the source, observer, or both is called the Doppler effect.
Source Stationary & Observer Moving
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Let us consider the case when the observer moves toward or away from the stationary source. If the observer moves toward the source with speed vo, the wavefronts have a relative speed of (v + vo) toward the observer. And, if the observer moves away from the source, the wavefronts have a relative speed of (v – vo) toward the observer.
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Fig. A stationary sources produces spherical waves.The speed of the waves relative to the observer O, who is moving towards the source, is( v + vo) |
If the relative speed of the sound waves with respect to observer is v ± vo, then the number of waves reaching the observer in time t is
N =( v ± vo) t/λ
and the frequency of the sound waves detected by the observer is
f ′ = N/t = (( v ± vo) t/λ)/t = ( v ± vo) /λ
Since λ = v/ f, therefore
f ′ = f ( v ± vo/v) (14.45)
Use: + for motion toward source, and – for motion away from source
Observer Stationary & Source Moving
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Let us suppose that the observer is at rest and the source is in motion with speed vs. It is obvious from Fig.(14.23) that the wavelength of the sound waves decreases for motion toward an observer and increases for motion away from an observer. |
Fig. - The wavelength in front of the moving source is less than normal, whereas in the rear, the wavelength is larger than normal |
During the time between wavefront emissions the source has moved to the right, decreasing the distance between wavefronts by just the distance it travels during this time.
Therefore,λ′ = λ - (Distance the source moves between wavefront emission)
λ′ = λ - (vs × Time between emissions)
orλ′ = λ - vsT = λ - Vs/ f
or v/f' = v/f - vs/ f
Thus,f ′ = f (v/v + vs) (source moving toward observer)(14.46)
If the source moves away from the observer, then the distance between wavefronts is increased vsT.
Therefore,λ′ = λ + vsT orλ′ = λ + Vs/ f
or v/f' = v/f + vs/ f = v + vs/f
Thus,f ′ = f (v/v + vs) (source moving away from the source)(14.47)
Combining equations (14.46) and (14.47) we get,
f ′ = f (v/v + vs) (14.48)
Use:+ for motion away from observer, - for motion towards observer
If both source and observer are in motion, the general expression for the observed frequency given by
f ′ = f (v ± vo /v + vs)(14.46)
Table: Doppler Frequencies (f′) for Different Situations
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Source Stationary |
Source Towards Observer |
Source Away from Observer |
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Observer Stationary |
f |
f(v/v + vs) |
f(v/v - vs) |
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Observer towards source |
f(v + vo /v ) |
f (v + vo / v - vs) |
f (v + vo / v + vs) |
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Observer away from source |
f (v - vo /v ) |
f (v - vo / v - vs) |
f(v - vo / v + vs) |
Wind Effect
The above formulae can be modified by taking the wind effects into account. The velocity of sound should be taken as
v + v w or v - vw if the wind is blowing in the same or opposite direction as SO (source to observer).
Example 14.20
A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity of 5 m/s. How many beats per second will be heard by an observer O standing in such a position that the source S is between observer O and the wall? v = 330 m/s.
Solution
Frequency of sound directly heard
f1 = fo (v - vo / v - vs) = 256 (330/330+5) = 252.2 Hz......(1)
Frequency received by the wall
f ′ = fo(v/v - vs) = 256 x 330/330 - 5 Hz(....2)Frequency received by the observer directly
f2 = f ′ (Because the wall is not moving)
Thus, f2 = 256 x 330/325 = 259.9 Hz
Beats frequency = f2 – f1 = (259.9 – 252.2) Hz = 7.7 Hz
7 beats per second will be heard by the observer.
Example 14.21
A stationary source emits a sound towards a wall moving towards it with a velocity u. Speed of sound in air = v. Find the fractional change in wavelength of the sound sent and the reflected sound.
Solution
Let the frequency of the sound sent by the source be fo.
Frequency of sound as observed by the wall is
∴ f1 = (v + u/v) fo
Frequency of the sound reflected by the wall is equal that received by it.
Now, the frequency f2 received by the source is
f2 = (v/v - u)f1
f2 = (v/v - u x v + u/v) fo
f2 = (v + u/v - u) fo
Initial wavelength = λi = v / fo
Final wavelength = λf = v / f2 = v/fo (v + u/v - u)
Fractional change in wavelength = Δλ /λi = -2u/v + u
