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„Ž. Weโ€™re also told that when the hawk screeches, the screech frequency is 3500 hertz. Weโ€™ll call that ๐‘“ sub โ„Ž. 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‘š.

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‘š 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‘œ, 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‘œ 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‘š, divided by the speed of sound, ๐‘ฃ sub ๐‘ , minus the speed of the hawk, ๐‘ฃ sub โ„Ž, all multiplied by ๐‘“ sub โ„Ž, the original frequency emitted by the hawk.

In our problem, the mouse is stationary. So ๐‘ฃ sub ๐‘š is zero. When we substitute in for ๐‘ฃ sub ๐‘ , the speed of sound; ๐‘ฃ sub โ„Ž, the speed of the hawk; and ๐‘“ sub โ„Ž, 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.

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