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A transerse sinusodial wave of amplitude...

A transerse sinusodial wave of amplitude `2 mm` is setup in a long uniform string. Snapshot of string from `x =0` to `x = pi` meter is taken at `t = 0`, which is shown. Velocity of point `P` is in `-y` direrction. Magnitude of relative velocity of `P` with respect to `Q` is `2 cm//s`. Choose the correct options : wave equation is

A

Displacement of particle at position P from its mean position as function of time is given by `Y=-(2xx10^(-3))sin5t(m)`

B

wave equation is `Y=(2xx10^(-3))sin(5t+2x+(pi)/(6))(m)`

C

wave equation is `Y=(2xx10^(-3))sin(5t+2x+(5pi)/(6))(m)`

D

wave equation is `Y=(2xx10^(-3))sin(5t-2x+(pi)/(6))(m)`

Text Solution

Verified by Experts

The correct Answer is:
A, B

Wave is heading towards `-ve` x-direction
Equation of R with respect to time `=Asin(omegat+(pi)/(6))`
wave equation `=Asin(omega+kx+(pi)/(6))`
given `|vecV_(P)-vecV_(Q)|=2omegaA=2(cm)/(s)`
`omega(2xx10^(-3))=10^(-2)(m)/(s)`
`omega(2xx10^(-1))=2impliesomega=5(rad)/(s)`
from snapshot `lamda=pimimpliesk=(2pi)/(lamda)=(2pi)/(pi)=2m^(-1)`
wave equation `y=(2xx10^(-3))sin(5t+2x+(pi)/(6))(m)`
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