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maks197457 [2]
3 years ago
8

A 40-cmcm-long tube has a 40-cmcm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. A

s the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length LL is 42.5 cmcm , 58.5 cmcm
What is the frequency of the tuning fork? Assume velocity of sound = 343 m/s
Physics
1 answer:
Licemer1 [7]3 years ago
7 0

Answer:

1070 Hz

Explanation:

First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:

The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength \frac{\lambda}{2}.

Hence, we have

\dfrac{\lambda}{2}=58.5-42.5=16 \text{ cm}

\lambda=32\text{ cm}=0.32 \text{ m}.

The speed of a wave is the product of its wavelength and its frequency.

v=f\lambda

f=\dfrac{v}{\lambda}

f=\dfrac{343}{0.32}=1070 \text{ Hz}

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