Question

Two children are at opposite ends of an aluminum rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in the air and in aluminum to reach the second child (velocity of sound in rod$V_R=6420m/s$).

Answer 1

Step 1: Given and assume

The velocity of sound in air, $v_a=340\frac{\text{m}}{\text{s}}$​.

The velocity of sound in aluminum rod $v_R=6420\frac{\text{m}}{\text{s}}$

Let,

The distance between the two children be $d$.

The time is taken by sound to reach another child via air be $t_1$​ and via aluminum rod be $t_2$​.

Step 2: To Find

The ratio of times taken by the sound wave in the air and in aluminum to reach the second child.

Step 3: Calculate the ratio of times taken by the sound wave in the air and in aluminum to reach the second child

The time taken by the sound to travel through air is,

$$\begin{gathered} t_1=\frac{d}{v_a} \\ =\frac{d}{340{\displaystyle \frac{\text{m}}{\text{s}}}} \end{gathered}$$ …(1)

The time taken by the sound to travel through rod is,

$$\begin{gathered} t_2=\frac{d}{v_R} \\ =\frac{d}{6420{\displaystyle \frac{\text{m}}{\text{s}}}} \end{gathered}$$ …(2)

By dividing equations (1) by (2), get

$$\begin{gathered} \frac{t_1}{t_2}=\frac{{\displaystyle \frac{d}{340}}}{{\displaystyle \frac{{\displaystyle d}}{{\displaystyle 6420}}}} \\ \frac{t_1}{t_2}=\frac{6420}{340} \\ =18.88:1 \end{gathered}$$

Therefore, the ratio will be $18.88:1$.