Resonance & Tubes: Frequency of Tuning Fork

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An open vertical tube filled with water produces resonance when a tuning fork is placed over its mouth, with resonance occurring at water levels of 17 cm and 49 cm. The difference in these measurements is 32 cm, which is crucial for calculating the wavelength. The wavelength is determined using the formula λ = 2l, leading to λ = 0.64 m. The frequency of the tuning fork is then calculated using the speed of sound (343 m/s) divided by the wavelength, resulting in a frequency of approximately 536 Hz. The calculations highlight the importance of accurately measuring the distances for precise frequency determination.
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Homework Statement


An open vertical tube is filled with water and a tuning fork vibrates over its mouth. As the water level is lowered in th etube, resonance is heard when the water level has dropped 17 cm, and again after 49 cm of distance exists from the water to the top of the tube. What is the frequency of the tuning fork?


Homework Equations


lamda=2xlength, f=v/ lamda, speed of sound is 343m/s


The Attempt at a Solution


49-17=35

lamda=2l (or is it lamda=4l cause its a closed system?)
=2x.35
=0.70m

f=v/lamda
=343m/s/0.70m
= 4.9x10 exponent 2 Hz?
 
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m.l. said:
1. Homework Statement
An open vertical tube is filled with water and a tuning fork vibrates over its mouth. As the water level is lowered in th etube, resonance is heard when the water level has dropped 17 cm, and again after 49 cm of distance exists from the water to the top of the tube. What is the frequency of the tuning fork?

lamda=2xlength, f=v/ lamda, speed of sound is 343m/s

3. The Attempt at a Solution
49-17=35

Isn't 49 - 17 = 32 ?
 
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