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A transverse wave is described by equati...

A transverse wave is described by equation ` y =y_0 sin 2pi (ft -(x)/( lambda) ) `.The maximum particle velocity is equal to 4 times wave velocity if

A

`lambda=(piY_(0))/(4)`

B

`lambda=(piY_(0))/(2)`

C

`lambda=piY_(0)`

D

`lambda=2piY_(0)`

Text Solution

Verified by Experts

The correct Answer is:
B

`Y=Y_(0)sin2pi(ft-(x)/(lambda))`
Maximum particle velocity `(v_(max))_(p)`
`=Aomega=A2pif=2Y_(0)pif`
wave velocity `(v)=(omega)/(k)=(2pif)/(2pi//lambda)=flambda`
given, `(v_(max))_(P)=4v`
`implies2Y_(0)pif=4flambdaimplieslambda=(Y_(0)pi)/(2)`
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