Question

2. When the air temperature is -40.0 degreeC, a hammer strikes one end of an aluminum rail. A microphone located at the opposite end of the rail detects two pulses of sound, one that travels through the rail and the other travels through the air. The two pulses are detected at 20.0 ms apart. The speed of sound in aluminum is 6.42×10^3 m/s.

(a) Determine the length of the rail.

(b) If the air temperature increases to +45.0 degree C, determine the separation in time between the arrivals of the two pulses

Answer 1

speed of sound in air at 0 C = 331 m/s

at T = -40 C

speed in air at - 40C = 331((1 + 0.606*T) = 306.76 m/s

speed in aluminium = 6420 m/s

let L be the length of the rod and ta and tr the times it takes for the sound to travel through air and the rod for the same distance then

l= 306.76*ta = 6420tr

ta-tr = 20 ms = 0.020s diff. in times

l = 307(0.020+tr) = 6420tr

l = 6.45 m -length of the rod

at 45 C speed of sound = 331 + 0.606*45 = 358 m/s

time to travel in air = 6.45/358 = 18 ms

time of travel in aluminimum tr = 6.45/6420 = 1 ms

diff. in times = 18 -1 = 17 ms, at 45 C