First slide

Simple hormonic motion

Question

Two simple harmonic motions given by $x_{1} = Acosωt$ and $x_{2} = Asinωt$ are superposed on a single particle. Then the particle starts its motion from a point whose distance from equilibrium position is

Difficult

A

A
B

$A / \sqrt{2}$
C

$A / 2$
D

$A / 2 \sqrt{2}$

Solution

Resultant distance from equilibrium position is $x = x_{1} + x_{2} = Asinωt + Acosωt$
$\Rightarrow x = \sqrt{2} A . \cos (ωt - \frac{π}{4})$
The particle starts its motion at t = 0.
$∴$ Initially $x = \sqrt{2} Acos (ω \times 0 - \frac{π}{4}) = A$

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