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In a stationary wave of frequency 200 Hz...

In a stationary wave of frequency 200 Hz, the distance between a node and a neighbouring antinode is 0.4m . What the wavelength and velocity of the wave ?

A

1.6 m, 320 m/s

B

2.0 m, 360 m/s

C

1.8m, 340 m/s

D

2.2 m, 380 m/s

Text Solution

Verified by Experts

The correct Answer is:
A
`lamda/4 = 0.4 therefore lamda=0.4xx4=1.6m`
`v=nlamda`
`=200xx1.6=320`
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