Doppler shift near church pt2

DOPPLER SHIFT NEAR CHURCH PT2

INTRODUCTION

We know that Doppler Effect or Doppler shift occurs between a source and observer when they are in relative motion with respect to each other. In this case we’ll determine the Doppler shift that occurs when a car is moving away from a stationary sound source. Consider a car [observer] moving at a speed of 35 mph [56 kph] moving away from a Church [source] which continuously emits sound waves by ringing the bell. We’ll determine the apparent frequency as registered by the observer in car.

ASSUMPTIONS

1. The atmospheric air has standard temperature and pressure conditions

·         Temperature T = 298 K or 25°C or 77°F

·         Pressure = 1 bar = 10⁵ N/m²

2. The effect of humidity on sound is negligible

3. The amplitude of sound is unity

4. The air molecules do not move with respect to source and observer

CALCULATION

The equation for Doppler shift is given by,

f’ = f₀*{[V ± V_o]/[V ± V_s]} (Eq. 1)

f₀ – Original frequency (Hz)

f’ – Apparent or observed frequency (Hz)

V – Velocity of Sound in air at standard temperature and pressure conditions (m/s) {V = 343 m/s}

V_o – Velocity of observer [car] (m/s)

V_s – Velocity of Source [Church bell] (m/s)

Since the source is stationary,

V_s = 0 (Eq. 2)

Substitute equation (2) in equation (1),

f’ = f₀*{[V – V_o]/[V]} (Eq. 3)

The ‘–’ sign in the numerator of equation (3) indicates that the observer is moving away from the source.

The velocity of car V_o = 35 mph

                                    = 15.55 m/s (Eq. 4)

Frequency of Church bell f₀ = 1000 Hz (Eq. 5)

Speed of sound in air V = 343 m/s (Eq. 6)

Substitute equations (4), (5) and (6) in equation (3),

f’ = 1000*{[343–15.55]/[343]}

f’ = 954.66 Hz

This is the frequency of sound as registered by the observer in car when he recedes away from a church that rings the bell.

Difference in frequency = f’ – f₀

                                       = 954.66 – 1000

                                       = –45.34 Hz

The negative value indicates that apparent frequency is less than the original but magnitude is always positive.

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift.