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 at and k with their usual meanings)
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a
y = 4sin[ω(t+2)+k(x-2)+π6]
b
y = 4 sin[ω(t+2)+k(x)+π6]
c
y = 4 sin[ω(t-2)-k(x-4)+5π6]
d
y = 4 sin[ω(t-2)-k(x-4)+π6]
answer is D.
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Detailed Solution
SHM equation of the particle at x = 4 is y = 4 sin[ω(t+2)+π6]∴ Wave equation, replacing by [t-(x-4v)], isy = 4 sin(ω(t-(x-4)v-2)+π6) = 4 sin[ω(t-2)-k(x-4)+π6]