A musician finds that a simple pipe resonates with a fundamental standing wave o

Question

A musician finds that a simple pipe resonates with a fundamental standing wave of of 440Hz in a room at temperature 21 C. The musician then walks into another room where the air temperature seems uncomfortably high, and she decides to measure the temperature. She does not have a conventional thermometer with her but she does have a 440 Hz tuning fork. Playing the tuning fork and pipe together produces beats at 5.0 Hz. Speed of sound is given by: Vs = 331.4 + 0.6T, where T is the temperature in degrees Celsius.

i) Assuming the pipe itself does not expand significantly, calculate the sound velocity in the hot room.

ii) Find the temperature in the room.

Explanation / Answer

Beat frequency = 5 Hz

Speed of sound increases with increase in temperature.

f_pipe - f_fork = 5Hz ............(i) (for hot room)

f_fork remains constant at 440 Hz in the hot room

For the pipe in 21*C, v_21/4l = 440 (v_21 = velocity at 21*C) (As fundamental frequency for an open pipe is f = v/4l)

l = length of pipe

v_21 = 1760l ........... (ii)

Now, v_21 as per the given velocity-temperature releation

= v(T =21) =331.4 + (0.6*21) = 344 m/s

v_21 = 1760l = 344

l = 0.1954 m

Now, frequency of pipe in hot room = 445 hz (as f_beat = 5Hz)

v_hot = 445(4l) = 1780l (v_hot = velocity in hot room)

= 445(4*0.1954)

v_hot = 379.91 m/s

331.4 + 0.6T = 379.91

T = 80.85 *C

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