Question

A small airplane is approaching a stationary crowd at 100 meters per second. The engine of the plane is emitting a sound of 485 Hz. What is the apparent frequency as it approaches the crowd?

Answer 1

695.87 Hz

Explanation:

Using the equation for Doppler shift, the frequency of the apparent sound is f' = f(v ± v')/(v ± v") where f = frequency of sound emitted by the plane's engine = 485 Hz, v = speed of sound in air = 330 m/s, v' = speed of crowd(detector) = 0 m/s (since the crowd are stationary) and v" = speed of plane(source) = 100 m/s

So, f' = f(v ± v')/(v ± v")

f' = f(v ± 0)/(v ± v")

f' = fv/(v ± v")

Since the source moves towards the detector, we use a minus a minus sign in the denominator to get a shift upwards in frequency since a shift upwards means the objects are moving closer to each other.

So,

f' = fv/(v - v")

Substituting the values of the variables into the equation, we have

f' = 485 Hz × 330 m/s/(330 m/s - 100 m/s)

f' = 160050 Hzm/s ÷ 230 m/s

f' = 695.87 Hz