Question

One end of a steel bridge is struck by a hammer. The sound produced is transmitted through the bridge and through the air to the other side of the bridge. There is a 1.4 s delay between the arrivals of the two sounds. What is the length of the bridge if the velocity of sound in air is 340 m/s and in steel is 5100 m/s. m

Answer 1

The sound in steel covers the distance L (length of the bridge) in a time $t_S$, with a velocity $v_S=5100 m/s$:
$v_S = \frac{L}{t_S}$
which we can rewrite as
$L=v_S t_S$ (1)

Similarly, the sound in air covers the length of the bridge L in a time $t_A$, with a velocity $v_A=340 m/s$:
$v_A = \frac{L}{t_A}$
which we can rewrite as
$L=v_A t_A$ (2)
We also know that the sound in air arrives with a delay of 1.4 s. This means we can rewrite tA as
$t_A = t_S+1.4 s$ (3)

The length of the bridge is always the same, so we can write (1)=(2) and using the information found in (3):
$v_A (t_S+1.4)=v_S t_S$
Re-arranging, we find
$t_S = \frac{1.4 v_A }{v_S -v_A}=0.1 s$

And at this point we can find the length of the bridge:
$L=v_s t_S = (5100 m/s)(0.1 s)=510 m$