First slide

Simple hormonic motion

Question

Two particles A and B execute simple harmonic motion according to the equations $y_{1} = 3 \sin ωt and y_{2} = 4 \sin (ωt + \frac{π}{2}) + 3 \sin ωt$ . The phase difference between them is

Moderate

A

$\frac{π}{2}$
B

$\tan^{- 1} (\frac{4}{3})$
C

$\tan^{- 1} (\frac{3}{4})$
D

None of these

Solution

Representing a quantity with phase ( $ωt$ ) along X-axis any quantity with a phase ( $ωt + ϕ$ ) can be represented by a vector making an angle $ϕ$ with X-axis in the anticlockwise sense and any quantity with a phase ( $ωt - ϕ$ ) can be represented by a vector making an angle $ϕ$ with X-axis in the clockwise sense.
Phase difference $ϕ = \tan^{- 1} (\frac{4}{3})$
Alternatively, given $y_{1} = 3 \sin ωt - - - - - (i)$
And $y_{2} = 4 \sin (ωt + \frac{π}{2}) + 3 \sin ωt$
= 4 cos $ωt + 3 \sin ωt = 5 [\frac{4}{5} \cos ωt + \frac{3}{5} \sin ωt]$
=5[sin $ϕ \cos ωt + cosϕ sinωt]$
= 5 sin( $ωt + ϕ$ )-----------------(ii)
$\cos ϕ = \frac{3}{5}; \sin ϕ = \frac{4}{5} and \tan ϕ = \frac{4}{3}$
From (i) and (ii), the result follows.

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