Question

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)]`

Answer 1

Correct Answer - 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)]`