A tuning fork is sounded above a resonating tube (one end closed),which resonates at a length of 0.200 m and again at 0.600 m.What is the frequency of the fork when the speed of sound is taken to be 340 m/s?
A)567 Hz
B)425 Hz
C)1 700 Hz
D)950 Hz

Question

Quizplus · Accepted Answer

The resonating tube has one end closed, so it will have resonant frequencies given by

f = nf0/2L,

where n is an integer, f0 is the fundamental frequency (corresponding to n=1), and L is the length of the tube.

For the first resonant length of 0.200 m, we have

f1 = f0/2L = f0/0.4.

For the second resonant length of 0.600 m, we have

f2 = 3f0/2L = 3f0/1.2.

Setting these two expressions equal to each other and solving for f0 gives

f0 = (2L/3) * (340/0.400 - 340/1.200) = 425 Hz.

Therefore, the correct choice is B.

A Tuning Fork Is Sounded Above a Resonating Tube (One