A tuning fork with a frequency of f = 524 Hz is placed near the top of the tube

Question

A tuning fork with a frequency of f = 524 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.) shorter length What is the relationship between the length of the tube and the sound wavelength for a tube that is open at one end and closed at the other, m longer length Note that the wavelength of the sound wave remains constant as the water level is varied, m

Explanation / Answer

for resonance f = 524 Hz

v/(4*L1) = 524

343/(4*L1) = 524

L1 = 0.163 m = 16.3 m

next resonance is

3*v/(4*L) = 524

3*343/(4*L) = 524

L = 0.4909 m = 49.09 cm

next resonance is

5*v/(4*L) = 524

5*343/(4*L) = 524

L = 0.818 m = 81.8 cm

so beyond 20 cm

shorter length is 49.1 cm

longer length is 81.8 cm

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