### 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.

Letโs highlight some of the important information weโve been given in this statement. First, weโre told that the speed of the hawk as it relates to the mouse is 25.0 meters per second. Weโll call that ๐ฃ sub โ. Weโre also told that when the hawk screeches, the screech frequency is 3500 hertz. Weโll call that ๐ sub โ. Weโre told that the speed of sound in air can be treated as exactly 331 meters per second. Weโll call that ๐ฃ sub ๐ . We want to know the frequency of the hawkโs screech as heard by the mouse. Weโll call that ๐ sub ๐.

Letโs 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 weโll see, this is not exactly the frequency that is heard by the stationary mouse while the hawk approaches. That value that weโve called ๐ sub ๐ can be determined by considering the doppler effect on the sound waves emitted by the hawk.

Letโs 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 ๐ฃ is the speed of sound in air, ๐ฃ sub ๐ is the source velocity, and ๐ฃ sub ๐ is the observerโs 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, ๐ฃ sub ๐ , plus the speed of the mouse, ๐ฃ sub ๐, divided by the speed of sound, ๐ฃ sub ๐ , minus the speed of the hawk, ๐ฃ sub โ, all multiplied by ๐ sub โ, the original frequency emitted by the hawk.

In our problem, the mouse is stationary. So ๐ฃ sub ๐ is zero. When we substitute in for ๐ฃ sub ๐ , the speed of sound; ๐ฃ 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 hawkโs 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 hawkโs incoming velocity, the mouse hears a higher frequency than the original source sound.