Question

A hammer taps on the end of a 4.10-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 11.1 ms .

Answer 1

Answer:

Speed of sound inside metal is ≅ 8200 $\frac{m}{s}$

Explanation:

Given :

Length of metal bar $x = 4.10$ m

From general velocity equation,

 $v = \frac{x}{t}$

Where $v =$ speed of sound in air = 343 $\frac{m}{s}$

For finding time from above equation,

  $t = \frac{x}{v}$

 $t = \frac{4}{343}$

$t = 0.01166$ sec

Since pulses are separated by  $t_{o} =$  $11.1 \times 10^{-3} = 0.0111$ sec

So we take time difference,

$\Delta t = t_{} -t_{o} = 0.0005$

So speed of sound in metal is,

 $v = \frac{x}{\Delta t }$

 $v = \frac{4.10}{0.0005}$

 $v = 8200$ $\frac{m}{s}$