First slide

Simple hormonic motion

Question

Two simple harmonic motions are represented by the equations $y_{1} = 0.1 \sin (100 πt + \frac{π}{3}) and y_{2} = 0.1 \cos πt .$ The phase difference of the velocity of particle l, with respect to the velocity of particle 2 is :

Moderate

A

$\frac{- π}{6}$
B

$\frac{π}{3}$
C

$\frac{- π}{3}$
D

$\frac{π}{6}$

Solution

$v_{1} = \frac{{dy}_{1}}{dt} = 0.1 \times 100 π \cos (100 πt + \frac{π}{3}) = 10 π \cos (100 πt + \frac{π}{3}) v_{2} = \frac{{dy}_{2}}{dt} = 0.1 \times 100 π \sin (100 πt) = 10 πcos (\frac{π}{2} + 100 πt) ∴ Phase difference = \frac{π}{3} - \frac{π}{2} = - \frac{π}{6}$

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