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A bat locates insects by emitting ultrasonic “chirps” and then listening for echoes from the bugs. Suppose a bat chirp has a frequency of 25 kHz. How fast would the bat have to fly, and in what direction, for you to just barely be able to hear the chirp at 20 kHz?

This problem aims to find the speed of a bat flying near the observer at a particular frequency. The concept required to solve this problem is entirely related to doppler’s effect.

Suppose that a sound or a wave of some frequency is being produced by a moving source at some distance from the observer, such that any change in the frequency of that sound or wave generated by that moving source with reference to the observer is known as Doppler’s effect.

In physics terms, the Doppler effect is the noticeable change in the frequency of sound waves due to the comparable motion between the source and the observer. We can extrapolate the obvious frequency in the Doppler effect using the equation:

\[f’=\dfrac{(v \pm v_0)}{(v \pm v_s)} f_s\]

Where:

$f’=\text{frequency observed by the observer,}$

$f_s=\text{frequency of the source of the sound,}$

$v=\text{velocity of sound waves or speed of sound,}$

$v_0=\text{velocity of the observer is positive when it is from listener to source,}$

$v_s=\text{velocity of the source is positive when is it from source to listener.}$

This equation can be changed in different situations relying on the velocities of the observer or the source of the sound waves.

Expert Answer

When the sound-generating source and the observer are moving with respect to each other, the frequency of the sound listened by the observer is not equal in magnitude to the source frequency. For instance, when a car comes near you with its horn blowing, the pitch appears to decline as the car perishes.

In this problem, we are requested to find the speed with which the source of the sound passes by the observer so that the observer hears a sound of frequency $20kHz$. The toughest part is deciding the direction for each velocity.
Since the source moves away from the observer to make a frequency less than its actual frequency, a sound of less frequency is heard rather than the actual frequency from the source. Using the doppler’s equation:

\[f’=\dfrac{(v \pm v_0)}{(v \pm v_s)} f_s\]

Since the observer is stationary:

$v_0=0$,

$v_s$ is positive as the source is moving away from the listener,

Plugging them in:

\[f’=\dfrac{(v + 0)}{(v + v_s)} f_s\]

\[v+v_s=\dfrac{(v\times f_s)}{f’}\]

\[v_s=\dfrac{(v\times f_s)}{f’} – v \]

We have speed of sound $v = 343 m/s$, the frequency of source $f_s = 25000 Hz$, and the frequency of the sound heard by the listener $f’ = 20000 Hz$, plugging them in:

\[v_s=\dfrac{((343)\times (25000 ))}{20000 } – 343\]

\[v_s=(343)\times (1.25) – 343 \]

\[v_s=428.75 – 343\]

\[v_s=85.75 m/s \]

Numerical Result

The speed of the source is $v_s = 85.75 m/s$.

Example

Two cars are moving toward each other at a speed of $432 km/h$. If the frequency of the horn blown by the first car is $800Hz$, find the frequency heard by the person in the other car.

The observer and the source are moving toward each other, therefore,

\[f’=\dfrac{(v + 0)}{(v – v_s)} f_s \]

Changing $432 km/h$ into $m/s$ we get $120 m/s$.

Substituting the values:

\[f’=\dfrac{(360 + 120)}{(360 – 120)} 800=1600\space Hz\]

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