First slide

Stationary waves

Question

A resonance occurs with a tuning fork and an air column of size 12 cm. The next higher resonance occurs with an air column of 38 cm. What is the frequency of the tuning fork? Assume that the speed of sound is 312 m/s.

Moderate

A

500 Hz
B

550 Hz
C

600 Hz
D

650 Hz

Solution

Since the standing wave mode has a displacement antinode at the opening, there is a displacement node at the water-air interface. By increasing the height of the air column, to go from one harmonic to the next, an additional length equal ½ of wavelength is required. Hence
$\frac{λ}{2} = (0 .38 - 0 .12) m \Rightarrow λ = 0 .52 m$
Finally. from $v = fλ$ , we find that $f = v / λ = 312 / 0 .52 = 600 Hz$

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