7. A sound wave begins traveling through a thin metal rod at one end with a speed that is 15 times the speed of sound in air. If an observer at the other end of the rod hears the sound twice, one from the sound traveling through the rod and one from the sound traveling through the air, with a time delay of 0.12 s, how long is the rod? The speed of sound in air is 343 m/s.

   L = 44,096 m

Explanation:

The speed of the sound wave is constant therefore we can use the relations of uniform kinematics

             v = x / t

the speed of the wave in the bar is

            v = 15 v or

            v = 15 343

             v = 5145 m / s

The sound at the bar goes the distance

             L = v t

Sound in the air travels the same distance

             L = v_air (t + 0.12)

as the two recognize the same dissonance,

             v t = v_air (t +0.12)

             t (v- v_air) = 0.12 v_air

              t = 0.12 v_air / (v -v_air)

l

et's calculate

             t = 0.12 343 / (5145 - 343)

             t = 8.57 10-3 s

The length of the bar is

              L = 5145 8.57 10-3

              L = 44,096 m

e) none of the above answers are correct

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation