A particle performing S.H.M is found at its equilibrium t = 1s and it is found to have a speed of 0.25m/s at t = 2s.If the period of oscillation is 6s.Calculate amplitude of oscillation

Engineering

Physics

SHM of Simple Pendulum

Basics of Simple Harmonic Motion

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Question

A particle performing S.H.M is found at its equilibrium t = 1s and it is found to have a speed of 0.25m/s at t = 2s.If the period of oscillation is 6s.Calculate amplitude of oscillation

$\frac{3}{2 π} m$

$\frac{3}{4 π} m$

$\frac{6}{π} m$

$\frac{3}{8 π} m$

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Given,

ω = \frac{2 π}{6} = \frac{π}{3}

Let the displacement of particle be,

y = Asin(ωt + ϕ)

At t = 1, y = 0

\frac{πt}{3} + ϕ = π

ϕ = \frac{2 π}{3}

Velocity at t = 10.25m/s

v = Aωcos (\frac{2 π}{3} + \frac{2 π}{3}) = \frac{Aω}{2}

0.25 = \frac{πA}{6}

A = \frac{3}{2 π}

Option A is the correct answer