Question

The period of oscillation of a particle in SHM is 4s and its amplitude of vibration is 4cm. The distance of the particle 1/3 s after passing the mean position is1)1.33cm 2)2cm 3)3cm 4)2.33cm

Answer 1

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Answer 2

answer : option (2)2cm is correct choice

given,

The period of oscillation of a particle in SHM, T = 4s.

amplitude of vibration, A = 4cm

so the equation of particle executing SHM, y = Asin(2πt/T)

= (4cm)sin(2πt/4)

= (4cm)sin(πt/2)

distance of particle from the mean position after t = 1/3s will be , y = (4cm)sin(π × 1/3/2)

= (4cm)sin(π/6)

= 4cm × 1/2 = 2cm

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