Question: A man strike one end of a thin rod with a hammer. The speed of sound in the rod is 15times the speed of sound in the air. A woman, at the other end with her ear close to the rod, hears the sound of the blow twice with a 0.12 sinterval between; one sound comes through the rod and the other comes through the air alongside the rod. If the speed of sound in air is 343 m/swhat is the length of the rod?

1. The speed of the sound in the rod is$v_r=15v_s$
2. The time delayis$∆t=0.12s$.
3. The speed of the sound in air is$v_s=3\text{43m/s}$ .

By finding the equation for time delay$∆t$and using the speed vof sound, find the lengthof the rod.

Formulae are as follow:

1. The time of travel for sound,$\mathbf{t}\mathbf{=}\frac{\mathbf{l}}{\mathbf{v}}$
2. The time delay is,$\mathbf{\Delta t}\mathbf{=}{\mathbf{t}}_{\mathbf{s}}\mathbf{-}{\mathbf{t}}_{\mathbf{r}}$

Let, I be the length of the rod. Then, the time of travel for sound in the air will be,

$t_s=\frac{l}{v_s}$

And the time of travel for the compression waves in the rod will be,

$t_s=\frac{l}{v_r}$

Hence, the time delayis$∆t$ given by,

$$\begin{gathered} ∆t=t_s-t_r \\ =l\left(\frac{1}{v_s}-\frac{1}{v_r}\right) \\ =0.12\text{s} \end{gathered}$$

Thus, the length of the rod is given by,

$$\begin{gathered} ∆t=l\left(\frac{v_r-v_s}{v_s{v}_r}\right) \\ ∆t=0.12\text{s} \\ l=\frac{\left(0.12\text{s}\right)\times v_s{v}_r}{\left(v_r-v_s\right)} \end{gathered}$$

But,

$$\begin{gathered} v_r=15v_s \\ =15\times 343\text{m/s} \\ =514\text{5m/s} \end{gathered}$$

Substituting value,

$$\begin{gathered} l=\frac{0.12\times 343m/s\times 5145m/s}{\left(5145m/s-343m/s\right)} \\ I=44m \end{gathered}$$

Hence, the length of the rod is I = 44 m.