A tuning fork with a frequency of f = 536 hz is placed near the top of the tube shown below. the water level is lowered so that the length l slowly increases from an initial value of 20.0 cm. determine the next two values of l that correspond to resonant modes. (assume that the speed of sound in air is 343 m/s.)

Question

Clare

Physics

23 March, 11:54

A tuning fork with a frequency of f = 536 hz is placed near the top of the tube shown below. the water level is lowered so that the length l slowly increases from an initial value of 20.0 cm. determine the next two values of l that correspond to resonant modes. (assume that the speed of sound in air is 343 m/s.)

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Keshawn Rush · Answer 1

At a speed 343 m/s and frequency of 536 Hz;

the wavelength is computed by velocity / frequency = 343 / 536 = 0.640 m or 64 cm

If the tube is open at the top there is a concentrated at the top and a node at the bottom at resonance.

i. e: at L = lambda / 4

The next resonance happens at 3 lambda/4 then 5 lambda/4 etc.

these resemble to 64 / 4 = 16 cm.

therefore,

the lengths from the open end to the water level corresponding to the first 2:

L1 = v / 4f = 343 / 4*536 = 16cm

L2 = 3v / 4f = 3*343 / 4*536 = 48 cm