Question

A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in 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 s interval 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/s, what is the length of the rod?

Answer 1

Answer:

44.1 m

Explanation:

Given:

• $V_a$ = speed of sound in air = 343 m/s
• $V_r$ = speed of sound in the rod = $15V_a$
• $\Delta t$ = times interval between the hearing the sound twice = 0.12 s

Assumptions:

• $l$ = length of the rod
• $t$ = time taken by the sound to travel through the rod
• $T$ = time taken by the sound to travel to through air to the same point = $t+\Delta t = t+0.12\ s$

We know that the distance traveled by the sound in a particular medium is equal to the product of the speed of sound in that medium and the time taken.

For traveling sound through the rod, we have

$l=V_r t\\\Rightarrow t = \dfrac{l}{V_r}$..........eqn(1)

For traveling sound through the air to the women ear for traveling the same distance, we have

$l=V_aT\\\Rightarrow l=V_a(t+0.12)\\\Rightarrow l=V_a(\dfrac{l}{V_r}+0.12)\,\,\,\,\,\,(\textrm{From eqn (1)})\\\Rightarrow l=V_a(\dfrac{l}{15V_a}+0.12)\\\Rightarrow l=\dfrac{l}{15}+0.12V_a\\\Rightarrow l-\dfrac{l}{15}=0.12V_a\\\Rightarrow \dfrac{14l}{15}=0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = 44.1\ m$

Hence, the length of the rod is 44.1 m.