InterviewSolution
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InterviewSolution
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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)]` |
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Answer» 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)]` |
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