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marysya [2.9K]
3 years ago
7

A vibrating tuning fork of frequency 512 Hz is held over a water column with one end closed and the other open. As the water lev

el is allowed to fall, a loud sound (resonance) is heard at specific water levels. Assume you start with the tube full of water, and begin steadily lowering the water level. What is the water level (as measured from the top of the tube) for the third such resonance?
Physics
1 answer:
aleksklad [387]3 years ago
8 0

Answer:

Explanation:

This is an example of resonance of air column with one end closed.

For third resonance , length of water column

= 5  x λ / 4

λ ( wave length ) = velocity / frequency

343 / 512

= .67 m

= 67 cm

length of air column

= 5 x λ / 4

= 5/4 x 67 cm

= 83.75 cm

water level will be 83.75 cm below.

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This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

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(b) t=0.984 s

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Explanation:

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g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

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t=t1+t2

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(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

7 0
3 years ago
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