Question

A hammer taps on the end of a 3.4-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 9.00 ms .What is the speed of sound in this metal?

Answer 1

Answer:

The speed of sound in this metal is 3726 m/s

Explanation:

Length Δx = 3.4 m, Δt = time of pulses separation = 9.0 ms = 9 × 10⁻³ s = 0.009 s

Velocity of air ($V_{air}$) = 343 m/s

The time interval of the sound pulse in air (Δ$t_{air}$) is given as:

Δ$t_{air}$ =  Δx / $V_{air}$ = 3.4 / 343 = 0.0099125 s

Δ$t_{air}$ = 0.0099 s

The time of pulse travelling through metal (Δ$t_{metal}$) = Δt - Δ$t_{air}$

Δ$t_{metal}$ = Δt - Δ$t_{air}$ =  0.0099125  - 0.009 = 0.0009125 s

Since the length of the metal is 3.4 m, the speed of sound in metal ($V_{air}$) is given as:

$V_{air}$ = Δx / Δ$t_{metal}$ = 3.4 / 0.0009125 = 3726 m/s

The speed of sound in this metal is 3726 m/s