SHMMedium

Question

Aparticle execute simple harmonic motion with an angular velocity of 3.5 rad/sec andmaximum acceleration 7.5m/ s2 .The amplitude of oscillation will be:[

Options

A.0.53 cm
B.0.28 m
C.0.61 m
D.0.36 m

Solution

ω = 3.5 radian/sec maximum acceleration of a particle under SHM = ω2 a where a is amplitude of oscillation.
ω2a = 7.5 ⇒(3.5)2a = 7.5
⇒ a = = 0.6 lm

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