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An open vertical tube has water in it. a tuning fork vibrates over its mouth. as the water level is lowered in the tube, a resonance is heard when the water level is 460 cm below the top of the tube, and again after the water level is 500 cm below the top of the tube a resonance is heard. what is the frequency of the tuning fork? the speed of sound in air is 343 m/s.

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5 votes

Final answer:

To find the frequency of the tuning fork, use the formula speed = frequency * wavelength, where wavelength = 2 * length, and substitute the given values. The frequency of the tuning fork is approximately 429 Hz.

Step-by-step explanation:

To find the frequency of the tuning fork, we can use the formula for the speed of a wave:

speed = frequency * wavelength

Given that the speed of sound in air is 343 m/s and the wavelength can be found using the formula:

wavelength = 2 * length

where length is the distance between two consecutive resonances, we can calculate the frequency:

frequency = speed / wavelength

In the given problem, resonance is heard when the water level is 460 cm below the top of the tube and again at 500 cm below the top of the tube. This means that the length of the air column between the two resonances is 40 cm. Converting this to meters, we have length = 0.4 m.

Substituting the values into the formula:

wavelength = 2 * 0.4 = 0.8 m

speed = 343 m/s

frequency = speed / wavelength = 343 / 0.8 = 428.75 Hz

Therefore, the frequency of the tuning fork is approximately 429 Hz.

User Acesmndr
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6.9k points
5 votes
Speed = wavelenght * frequency
An open vertical tube has water in it. a tuning fork vibrates over its mouth. as the-example-1
User Eheydenr
by
6.3k points