Question

A hammer taps on the end of a 3.90-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 10.8ms .What is the speed of sound in this metal?

Vmetal =.........m/s

Answer 1

Answer 2

Time for the sound in air = x/v = 3.90m/343m/s = 0.01137s

So the time for the sound in the metal = 0.01137-0.0108 = 5.70x10^-4s

So v = x/t = 3.90/5.70x10^-4 = 6.84x10^3m/s

Answer 3

Time for sound in air = x/v = 3.90m/343m/s = 0.01137s

limit for sound of metal = 0.01137-0.0108 = 5.70x10^-4s

That means v = x/t = 3.90/5.70x10^-4 = 6.84x10^3m/s