Question

Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.

Answer 1

Let's consider the length of the aluminum rod as d.
The speed of the sound wave at 25°C in aluminum (V_Al​) is 6420 m/s.
The time taken for the sound wave to reach the other end of the aluminum rod is given by:

T_Al ​= \(\frac{d}{V_{Al}​}\)​ = \(\frac{d}{6420}\)​

The speed of sound in air (V_air​) is 346 m/s.
The time taken by sound to reach the other end through air is given by:

T_air ​= \(\frac{d}{V_{air}}​\)​ = \(\frac{d}{346}\)​

Therefore, the ratio of the time taken by sound in aluminum to the time taken in air is:

\(\frac{T_{Al}}{​T_{air}}\)​​ = \(\frac{6420}{346}\) ​= 18.55

Answer 2

The distance between the two children is denoted as d.
Also, let the time it takes for sound to reach the other child through the air be t₁ and via aluminium rod be t₂.
Velocity of sound in air v_a = 340 m/s
Therefore, \(t_1=\frac{d}{v_a}=\frac{d}{340}\ \ ....(i)\)
Velocity of sound in aluminium rod v_R = 6420 m/s
Therefore, \(t_2=\frac{d}{v_R}=\frac{d}{6420}\ \ .....(ii)\)
Now, by dividing the equation (i) and (ii), we get
\(\frac{t_1}{t_2}=\frac{6420}{340}=18.88\)