Question

An air column in a glass tube is open at one end and closed at the other by a movable piston, and a 384Hz tuning fork is held at the open end. Resonance is heard when the piston is at a distance d1 = 22.8cm from the open end and again when it is at a distance d2 = 68.3cm from the open end. 

What speed of sound is implied by these data?

Determine the end correction

How far from the open end will the piston be when the next resonance is heard?

Answer 1

For the tubes that open at one end the resonant frequency can be found as follows:

$$f_n=\dfrac{nv}{4L}.$$

(a) The first resonance appears at the first harmonic frequency:

$$v=\dfrac{4f_1L}{1}=\dfrac{4\cdot384\ Hz\cdot0.228\ m}{1}=350\ \dfrac{m}{s}.$$

The second resonance appears at the third harmonic frequency:

$$v=\dfrac{4f_3L}{3}=\dfrac{4\cdot384\ Hz\cdot0.683\ m}{3}=350\ \dfrac{m}{s}.$$

Therefore, the speed of sound equals $350\ \dfrac{m}{s}.$

(b) The next resonance appears at the fifth harmonic frequency, therefore, we can find the length from the same formula:

$$L=\dfrac{nv}{4f_n}=\dfrac{5\cdot350\ \dfrac{m}{s}}{4\cdot384\ Hz}=1.14\ m.$$