DOPPLER EFFECT

The apparent shift in frequency of the wave motion when the source of sound or light moves with reect to the observer, is called Doppler Effect.

Calculation of Apparent Frequency

Suppose v is the velocity of sound in air, v_s is the velocity of the source of sound(s), v_o is the velocity of the observer (O), and f is the frequency of the source.

(i) Source moves towards stationary observer.

If the source were stationary the f waves sent out in one second towards the observer O would occupy a distance v, and the wavelength would be v/f.

If S moves with a velocity v_s towards O, the f waves sent out occupy a distance (v - v_s) because S has moved a distance v_s towards O in 1s. So the apparent wavelength would be $\lambda ’$ = $\frac{v-v_s}{f}$

Thus, apparent frequency f ' = $\frac{VelocityofsoundrelativetoO}{WavelengthofwavereachingO}$

f' = $\frac{v}{{\lambda }^{\prime }}=f\left(\frac{v}{v-v_s}\right)$

(ii) Source moves away from stationary observer.

Now, apparent wavelength

$\lambda ’$ = $\frac{v+v_s}{f}$

$\therefore$ Apparent frequency f ' = $v/\lambda ’$

f' = f$\left(\frac{v}{v+v_s}\right)$

(iii) Observer moves towards stationary source.

f' = $\frac{VelocityofsoundrelativetoO}{WavelengthofwavereachingO}$

Here, velocity of sound relative to O = v + v_O and wavelength of waves reaching O = v/f

$\therefore$ f' = $\frac{v+v_o}{v/f}=f\left(\frac{v+v_o}{v}\right)$

(iv) Observer moves away from the stationary source.

f' = $\frac{v-v_o}{v/f}=f\left(\frac{v-v_o}{v}\right)$

(v) Source and observer both moves toward each other.

f' = $\frac{v+v_o}{\frac{v-v_s}{f}}=f\left(\frac{v+v_o}{v-v_s}\right)$

(vi) Both moves away from each other.

f' = $f\left(\frac{v-v_o}{v+v_s}\right)$

(vii) Source moves towards observer but observer moves away from source

f = f' $\left(\frac{v-v_o}{v-v_s}\right)$

(viii) Source moves away from observer but observer moves towards source

f = f' $\left(\frac{v+v_o}{v+v_s}\right)$

Wind's 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 - v_w if the wind is blowing in the same or opposite direction as SO (source to observer).

Illustration 1: Two trains are moving with speeds 30m/s and 25m/s towards each other. The frequency of the whistle of the faster train is 500 Hz. What is the apparent frequency of this whistle if heard by an observer on the slower train? What will be the apparent frequency after the trains have crossed each other?

Solution: (a) The faster train's speed is the source's speed and the other is the observed speed

Taking the direction from S to O as +ve,

V_s = +30 m/s, V₀ = –25 m/s

v' = $v\left(\frac{c-V_0}{c-V_s}\right)$ v' = $v\left(\frac{c-V_0}{c-V_s}\right)$

v' = $v\left(\frac{c-\left(-25\right)}{c-\left(+30\right)}\right)=500\left(\frac{345+25}{345-30}\right)$

= 547.62 Hz

(b) Behind the train: After train have crossed each other, they will move away from each other as shown

Now V₀ = + 25 m/s, V_s = –30 m/s

v' = v $\left(\frac{c--25}{c-\left(-30\right)}\right)$ = 500$\left(\frac{345-25}{345+30}\right)$

= 426.67 Hz.