The speed of sound in air is
[tex]c_a=343 m/s[/tex]
The pulse moving in air covers a distance of [tex]L=4.9 m[/tex] with this velocity, so the time it takes is
[tex]t_a= \frac{L}{c_a}= \frac{4.9 m}{343 m/s}=0.0143 s [/tex]
The problem says that the second pulse (moving in the metal bar) arrives with a delay of [tex]\delta t = 10.4 ms = 0.0104 s[/tex]. Therefore, the time [tex]t_m[/tex] at which the pulse in the metal arrives at the microphone is
[tex]t_m = t_a + \Delta t=0.0143 s+0.0104 s=0.0247 s[/tex]
The distance covered by this pulse is still [tex]L=4.9 m[/tex], therefore we can find the speed of the pusle in metal (so, the speed of sound in this metal):
[tex]c_m= \frac{L}{t_m}= \frac{4.9 m}{0.0247 s} =198 m/s[/tex]
