First slide

Simple harmonic motion

Question

Two particles A and B execute simple harmonic motion according to the equations ) $y_{1} = 3 \sin ω t$ and $y_{2} = 4 \sin (ω t + \frac{π}{2}) + 3 \sin ω t$ 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 $(ω t)$ along x-axis any quantity with a phase $(ω t + ϕ)$ can be represented by a vector making an angle $ϕ$ with X-axis in the anticlockwise sense and any quantity with a phase $(ω t - ϕ)$ can be represented by a vector making an angle $ϕ$ with X-axis in the clockwise sense.
$Phase difference ϕ = \tan^{- 1} (\frac{4}{3})$
$Alternatively, given y_{1} = 3 \sin ω t . ......(i)$
$and y_{2} = 4 \sin (ω t + \frac{π}{2}) + 3 \sin ω t$
$\begin{array}{l} = 4 \cos ω t + 3 \sin ω t \\ = 5 [\frac{4}{5} \cos ω t + \frac{3}{5} \sin ω t] \\ = 5 [\sin ϕ \cos ω t + \cos ϕ \sin ω t] \\ = 5 \sin (ω t + ϕ) . . . . . . . . . (ii) \end{array}$
From (i) and (ii), the result follows.
$\cos ϕ = \frac{3}{5}; \sin ϕ = \frac{4}{5} and \tan ϕ = \frac{4}{3}$

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