Question

Two children are at opposite ends of metal rods. One strikes the end of the rod with a big pebble. Find the ratio of time taken by the sound waves in air and in metal to reach the second child:
(Speed of sound in rod $= 6420m{s^{ - 1}}$)
(A) $221.3$
(B) $18.55$
(C) $22.13$
(D) $185.5$

Answer 1

We need to find the time taken by the sound wave to cover the total length of the rod through the metal and then find the time taken to cover the same length through the air. Then we need to find the ratio of the time in both cases.

Formula Used In this solution, we will be using the following formula,
$t = \dfrac{D}{S}$
where $t$ is the time, $D$ is the distance and $S$ is the speed.

Complete step by step solution
We are given that one child strikes one end of the rod. Let us consider the length of the rod to be $L$.
So the sound waves in the air as well as in the metal has to travel a distance of $L$. Let ${V_1}$ be the speed of the sound waves in the metal and ${V_2}$ be the speed of the waves in air.
Let the time taken for the sound waves to reach from one point to another through the metal be ${t_1}$. And similarly the time taken to travel in air be ${t_2}$
Now the time taken when the speed and distance are given, can be calculated by the formula,
$t = \dfrac{D}{S}$
So now by substituting the values in the case of the metal we get,
${t_1} = \dfrac{L}{{{V_1}}}$
And similarly in the case of air, we get
${t_2} = \dfrac{L}{{{V_2}}}$
In the question we are asked to find the ratio of the time taken by the sound waves in air to that in the metal. So we have the ratio as, $\dfrac{{{t_2}}}{{{t_1}}}$.
By substituting the values we have,
$\dfrac{{{t_2}}}{{{t_1}}} = \dfrac{{\dfrac{L}{{{V_2}}}}}{{\dfrac{L}{{{V_1}}}}}$
On cancelling the $L$ we get,
$\dfrac{{{t_2}}}{{{t_1}}} = \dfrac{{{V_1}}}{{{V_2}}}$
Now the speed of sound waves in the metal is given in the question as, ${V_1} = 6420m{s^{ - 1}}$ and the speed of sound waves in air is ${V_2} = 346m{s^{ - 1}}$
So substituting the values we have,
$\dfrac{{{t_2}}}{{{t_1}}} = \dfrac{{6420}}{{346}}$
On calculating this we have the ratio as,
$\dfrac{{{t_2}}}{{{t_1}}} = 18.55$

So the correct answer is option B.

Note: The speed of sound in a medium depends on the medium that it is travelling through. The speed of the sound waves also changes with the temperature and the pressure of the medium. As the temperature of the medium increases, the speed of the sound in the medium also increases.