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A transverse wave is propagating along +...

A transverse wave is propagating along `+x` direction. At `t=2` sec the particle at `x=4`m is at `y=2` mm. With the passage of time its `y` coordinate increases and reaches to a maximum of 4 mm. The wave equation is (using `omega` and `k` with their usual meanings)

A

`y=4sin[omega(t+2)+k(x-2)+(pi)/(6)]`

B

`y=4sin[omega(t+2)+k(x)+(pi)/(6)]`

C

`y=4sin[omega(t-2)-k(x-4)+(5pi)/(6)]`

D

`y=4sin[omega(t-2)-k(x-4)+(pi)/(6)]`

Text Solution

Verified by Experts

The correct Answer is:
D
SHM equation of the particle at `x=4` is
`y=4sin[omega(t+2)+(pi)/(6)]`
Wave equation, replacing t by `[t-((x-4)/(v))]` is
`y=4sin(omega(t-((x-4))/(v)-2)+(pi)/(6))`
`=4sin[omega(t-2)-k(x-4)+(pi)/(6)]`
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Knowledge Check

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