# Video: Relating the Speed of a Sound Wave Source to the Doppler Shift Produced

A hawk swoops down to catch a mouse, screeching as it does so. The hawk approaches the mouse at a speed of 25.0 m/s, and the frequency of the screech is 3500 Hz. What is the frequency of the screech as heard by the mouse? Use a value of 331 m/s for the speed of sound in air.

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### Video Transcript

A hawk swoops down to catch a mouse, screeching as it does so. The hawk approaches the mouse at a speed of 25.0 meters per second. And the frequency of the screech is 3500 hertz. What is the frequency of the screech as heard by the mouse? Use a value of 331 meters per second for the speed of sound in air.

Letโs highlight some of the important information weโve been given in this statement. First, weโre told that the speed of the hawk as it relates to the mouse is 25.0 meters per second. Weโll call that ๐ฃ sub โ. Weโre also told that when the hawk screeches, the screech frequency is 3500 hertz. Weโll call that ๐ sub โ. Weโre told that the speed of sound in air can be treated as exactly 331 meters per second. Weโll call that ๐ฃ sub ๐ . We want to know the frequency of the hawkโs screech as heard by the mouse. Weโll call that ๐ sub ๐.

Letโs begin our solution by drawing a diagram of this situation. In this situation, we have a hawk approaching a mouse at a relative speed of 25.0 meters per second. As it approaches, the hawk screeches loudly at a frequency of 3500 hertz. As weโll see, this is not exactly the frequency that is heard by the stationary mouse while the hawk approaches. That value that weโve called ๐ sub ๐ can be determined by considering the doppler effect on the sound waves emitted by the hawk.

Letโs recall the equation for the shift in frequency that occurs due to the motion of the source. This is called a Doppler shift. This equation says that ๐ sub ๐, the observed frequency, is equal to ๐ sub ๐ , the source frequency, multiplied by this fraction, where ๐ฃ is the speed of sound in air, ๐ฃ sub ๐  is the source velocity, and ๐ฃ sub ๐ is the observerโs velocity.

If we apply this Doppler shift equation to our scenario, then we can write that the frequency that the mouse observes is equal to the speed of sound, ๐ฃ sub ๐ , plus the speed of the mouse, ๐ฃ sub ๐, divided by the speed of sound, ๐ฃ sub ๐ , minus the speed of the hawk, ๐ฃ sub โ, all multiplied by ๐ sub โ, the original frequency emitted by the hawk.

In our problem, the mouse is stationary. So ๐ฃ sub ๐ is zero. When we substitute in for ๐ฃ sub ๐ , the speed of sound; ๐ฃ sub โ, the speed of the hawk; and ๐ sub โ, the original source frequency, when we calculate this value, we find the frequency of the hawkโs screech as observed by the mouse. To three significant figures, that frequency is 3.79 times 10 to the third hertz. So we see that because of the hawkโs incoming velocity, the mouse hears a higher frequency than the original source sound.