Question

When the air temperature is −40.0°C, 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 6.42×103 m/s. (a) Determine the length of the rail.

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

Answer 1

(a)

Speed of sound in air vs temperature

s = (331 + 0.606T) m/s

T is temperature in ºC

s = 331 + 0.606(–40) = 307 m/s

time for sound to travel through air is t = L/307 where L is length of rail

time for sound to travel through Al is t = L/6420

difference is 0.02 sec

L/307 – L/6420 = 0.02

0.310157L = 2

L = 6.45 m

b) repeat above for diff temp.

Speed of sound in air vs temperature

s = (331 + 0.606T) m/s

T is temperature in ºC

s = 331 + 0.606(45) = 358.27 m/s

time for sound to travel through air is t = 6.45/358.27 where L is length of rail

time for sound to travel through Al is t = 6.45/6420

6.45/358.27 – 6.45/6420

= 0.017s

That is, 17ms