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Two simple harmonic motions are given by $y_{1} = a sin [(\frac{π}{2}) t + ϕ]$ and $y_{2} = b sin [(\frac{2 π}{3}) t + ϕ]$ . The phase difference between these after 1 s is:

A

zero

B

$π / 2$

C

$π / 4$

D

$π / 6$

Medium

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Solution

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Correct option is D)

At t = 1 s
$ϕ_{1} = \frac{π}{2} + ϕ$
and $ϕ_{2} = \frac{2 π}{3} + ϕ$
$∴$ $Δ ϕ = ϕ_{2} - ϕ_{1} = \frac{π}{6}$

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