A tuning fork of frequency 500 is sounded on a resonance tube
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Answered by
1
Answer:
what is the que in this ?
Answered by
0
Explanation:
350 m/s
Step by step explanation:
The first resonance occurs at \lambda/4, while the second resonance occurs at 3\lambda/4.
Therefore,
17+e=\lambda/4 and 52+e=3\lambda/4.
If we subtract the two equations from each other:
52-17=\lambda/2\\\lambda=70\:cm=0.7\:m.
Hence,
v=f*\lambda\\v=500*0.7=350\:m/s
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