Asked by Anonymous
A hammer taps on the end of a 3.32 m long metal bar at room temperature. A microphone at the other end of the bar picks up two pulses of sound, one that travels through the metal and one that travels through the air. The pulses are separated in time by 5.71 ms. What is the speed of sound in this metal?(the speed of sound in air is 343 m/s)
Answers
Answered by
drwls
Let L = 33.2 m be the length of the bar.
Use a procedure similar to the P and S wave earthquake example I solved earlier.
L/Vair - L/Vmetal = 5.71*10^-3 s
= L [1/343 - 1/Vmetal]
You know L and Vair. Solve for Vmetal
Use a procedure similar to the P and S wave earthquake example I solved earlier.
L/Vair - L/Vmetal = 5.71*10^-3 s
= L [1/343 - 1/Vmetal]
You know L and Vair. Solve for Vmetal
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