First slide

Simple hormonic motion

Question

Two simple harmonic motions given by $y_{1} = 10 \sin 100 πt$ and $y_{2} = 5 \cos (100 πt - π / 6)$ are super posed on a particle. The amplitude of the resulting motion of the particle is

Moderate

A

15
B

$5 \sqrt{6}$
C

$5 \sqrt{7}$
D

$10 \sqrt{3}$

Solution

$y_{1} = 10 \sin 100 πt = 10 \cos (10 πt - \frac{π}{2})$
Phase difference $ϕ = (100 πt - \frac{π}{6}) - (100 πt - \frac{π}{2}) = \frac{π}{3}$
$∴$ Resultant amplitude $= \sqrt{10^{2} + 5^{2} + 2 .10.5 \cos \frac{π}{3}} = 5 \sqrt{7}$

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