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Unformatted text preview: Physics 1240 Homework 6 solutions Brief solutions to homework set #6. Your numbers will be different, but the idea is the same. I’m just using sample numbers here. 1. A sailor strikes the side of her ship just below the waterline. She hears the echo of the sound reflected from the ocean floor directly below 1.99 s later. How deep is the ocean at this point? Assume the velocity of sound in water is 1560 m/s. The sound travels from the boat, down to the bottom of the ocean, then back to the sailor. This means that if the depth of the ocean is d , she hears the sound after it travels a distance of 2 d (down and then back). This is the trick in this question! We can then use that velocity is distance divided by time: normally we write this as v = d/t . But here the distance the sound travels is 2 d . Rearranging, 2 d = vt , so d = 1 2 vt . d = 1 2 vt = 0 . 5 × 1560 m/s × 1 . 99 s = 1552 . 2 m . 2. Change the previous problem to consider sound traveling the same distance, but through air. Suppose the sailor strikes the side of her ship, and the sound travels the same distance as the depth of the ocean you calculated in the previous problem. Now the sound travels through air, reflects (echoes) off a cliff, and returns to the sailor. How much time elapses between when the sailor produces the sound, and when she hears its echo? Assume the velocity of sound in air is 344 m/s. What physical property of liquids and solids do you think causes the speed of sound in them to be to be so much greater than the speed of sound in air?...
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This note was uploaded on 08/24/2010 for the course PHYS 1240 taught by Professor Holland,murray during the Spring '08 term at Colorado.
- Spring '08