A tuning fork is sounded above a resonating tube (one end closed),which resonates at a length of 0.80 m and again at 1.40 m.What is the frequency of the fork when the speed of sound is taken to be 340 m/s?
A)243 Hz
B)155 Hz
C)425 Hz
D)567 Hz
E)748 Hz

Question

Quizplus · Accepted Answer

The resonating length at the first resonance is equal to ¼ of the wavelength (since one end is closed).
Thus, λ/4 = 0.80 m 
⇒ λ = 3.20 m 
The resonating length at the second resonance is equal to ¾ of the wavelength. 
Thus, 3λ/4 = 1.40 m 
⇒ λ = 1.87 m
From these two equations, we get:
3.20 m = 4 × (1.87 m)/3 
Thus, the frequency of the tuning fork is:
f = v/λ = 340/1.87 ≈ 181.28 Hz
But, we need to multiply this frequency by the ratio of the lengths of the tube at the two resonances to get the correct frequency of the tuning fork. 
Thus, f′ = f × (0.8/1.4) = 181.28 × 0.8/1.4 ≈ 104 Hz
Now, we need to find the nearest option to this frequency from the given choices.
We can see that option B, 155 Hz, is the closest choice to our calculated frequency of 104 Hz.
Thus, the correct answer is:
Answer: B

A Tuning Fork Is Sounded Above a Resonating Tube (One