Accoustics Problem 1 - Resonance and Speed of Sound

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A tuning fork at 256 Hz creates a stationary wave in a partially submerged open-end tube by compressing air, with water reflecting the air to maintain the wave. Elevating the tube alters sound intensity due to changes in resonance, as resonance occurs at specific lengths related to the tube's configuration. The tube acts as a closed pipe resonator, and the speed of sound can be calculated using the formula for resonance in closed pipes. To determine resonance positions, one must consider the harmonic frequencies associated with the tube's length. Understanding the relationship between tube elevation and sound intensity is crucial for grasping resonance concepts.
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1. A tuning fork (frequency is 256 Hz) is placed above an open end tube that is partially in water.

a. Explain how a stationnary wave is created.

b. The tube is elevated a little. Explain why the sound intensity changes in therms of resonance.

c. The tube is elevate progressively from a postion of maximum intensity to the next position with the same property. If the tube is now 65 cm above water, what is the speed of sound in it?

My anwsers:

a: The fork compresses the air in the tube and the water reflects this aire, thus creating a continuous stationnary wave.

b: Need help... what is the link between power and the length of a open-close end tupe?

c: Same with this one...
 
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what kind of resonator is this? open or closed pipe resonator? What is the formula to find where resonance will occur in the (open or closed pipe resonator, depending on which one it is) that will do for c. Now b, you do not really need to relate power with resonance. It seems to me that they just want to know what resonace means.
 

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