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Q.

Two SHMs are given by $Y_{1} = A \sin (\frac{π}{2} t + ϕ) and Y_{2} = B \sin (\frac{2 πt}{3} + ϕ) .$ The phase difference between these two after '1' sec is

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a

$π$

b

$\frac{π}{2}$

c

$\frac{π}{4}$

d

$\frac{π}{6}$

answer is D.

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Detailed Solution

$Phase difference = ϕ_{2} - ϕ_{1}$
$= \frac{2 πt}{3} + ϕ - \frac{π}{2} t - ϕ = \frac{π}{6} t$

$t = 1, \frac{π}{6}$

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Two SHMs are given by Y1=A sinπ2t+ϕ and Y2=B sin2πt3+ϕ. The phase difference between these two after '1' sec is