Question

A motorcycle and a police car are moving toward one another. The police car emits sound with a frequency of 523 Hz and has a speed of 32.6 m/s. The motorcycle has a speed of 13.6 m/s. Part A What frequency does the motorcyclist hear

Answer 1

Answer:

The frequency the motorcyclist hears is 601.60 Hz

Explanation:

The Doppler effect is a physical phenomenon where an apparent change in wave frequency is presented by a sound source with respect to its observer.

The following expression is considered the general case of the Doppler effect:

$f'=f*\frac{v+-vR}{v-+vE}$

Where:

f ’, f: Frequency perceived by the receiver and frequency emitted by the emitter respectively. Its unit of measurement in the International System (S.I.) is hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s-1)

v: propagation speed of the wave in the medium. It is constant and depends on the characteristics of the medium. In this case, the speed of sound in air is considered to be 34o m / s

vR, vE: Receiver and emitter speed respectively. Your unit of measure in the S.I. is the m / s

±, ∓:

The + sign is used:

• In the numerator if the receiver approaches the sender .
• In the denominator if the emitter moves away from the receiver .

The sign - is used:

• In the numerator if the receiver moves away from the sender .
• In the denominator if the emitter approaches the receiver.

In this case:

• f: 523 Hz:
• v= 340 m/s
• vR= 13.6 m/s
• vE= 32.6 m/s
• The emitter approaches the receiver, so the + sign is used in the numerator and the - sign is used in the denominator.

Replacing:

$f'=523 Hz*\frac{340 m/s+13.6 m/s}{340 m/s-32.6 m/s}$

Solving:

$f'=523 Hz*\frac{353.6 m/s}{307.4 m/s}$

f' = 601.60 Hz

The frequency the motorcyclist hears is 601.60 Hz