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

Air temperature is -40 ⁰ C .  

To calculate the sound in air at any given temperature,

C_air = 33.5 + (0.6 * T_c) m/s , where Tc = specific temperature

Therefore, speed of air at -40⁰C = 331.5 + ( 0.6* (-40 ) m/s

= 307.5 m/s

Speed of aluminium = 6420 m/s (i'm assuming it to be this you have given as the actual speed of aluminium is 6320 m/s)

pulses detection time difference = 20 ms = .02s= T

therefore, x/S1 -x/S2 = T ( S1 = speed of sound, S2= speed of aluminium )

=> x/307.5 - x / 6420 = .02

x = 6.459 m

(b) speed of sound at air temperature +45 ⁰C :- 331.5 + (06 * 45)

= 601.5 m/s

T1 = x/v1 = 6.45 / 601.5 = 0.0107 s

T2 = x/v2 = 6.45/ 6420 = 0.001

=> 0.0170-0.001 = 0.0097 s = 9.7 * 10 ^-3 s

(Difference between the two is the separation between the arriva; of the two pulses.)