Question

A hammer taps on the end of a 4.9 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 10.4 ms. what is the speed of sound in this metal

Answer 1

The speed of sound in air is

`c_a=343 m/s`
The pulse moving in air covers a distance of
`L=4.9 m` with this velocity, so the time it takes is

`t_a= (L)/(c_a)= (4.9 m)/(343 m/s)=0.0143 s`

The problem says that the second pulse (moving in the metal bar) arrives with a delay of
`\delta t = 10.4 ms = 0.0104 s`. Therefore, the time
`t_m` at which the pulse in the metal arrives at the microphone is

`t_m = t_a + \Delta t=0.0143 s+0.0104 s=0.0247 s`

The distance covered by this pulse is still
`L=4.9 m`, therefore we can find the speed of the pusle in metal (so, the speed of sound in this metal):

`c_m= (L)/(t_m)= (4.9 m)/(0.0247 s) =198 m/s`