A tuning fork with a frequency of 440Hz is held just above the top of a uniform

Question

A tuning fork with a frequency of 440Hz is held just above the top of a uniform glass tube containing water. The glass tube can help excite the column of air above the water to form standing waves. The level of the column of air above water can be changed by a spigot at the bottom of the glass tube. As the water is dreained out, the sound intensity of the fork is enhanced when the air column has a length of 0.6 m and again when the air column has a length of 1m. What is the speed of sound in air.?

Explanation / Answer

frequesncy = 440 Hz

frequency = v / { 4L / (2n + 1 ) } = (2n + 1) * v / 4L

so 440 = ( 2n +1 ) * v / (4*0.6 )

also 440 = ( 2(n+1) + 1 ) * v / (4*1)

equating both equations

so.... ( 2n + 1 ) / 0.6 = 2n + 3

2n + 1 = 1.2n + 1.8

0.8n = 0.8

n = 1

so v = 440 *4*0.6 / 3 = 352 m/sec

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