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?

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

Assumptions:

[tex]l[/tex] = length of the rod
[tex]t[/tex] = time taken by the sound to travel through the rod
[tex]T[/tex] = time taken by the sound to travel to through air to the same point = [tex]t+\Delta t = t+0.12\ s[/tex]

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

[tex]l=V_r t\\\Rightarrow t = \dfrac{l}{V_r}[/tex]..........eqn(1)

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

[tex]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[/tex]

Hence, the length of the rod is 44.1 m.