The SHM of a particle is given by $x (t) = 5 \cos (2 πt + \frac{π}{4})$ in MKS units. Calculate the displacement and the magnitude of acceleration of the particle at $t = 1.5 s$

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$- 3.54 m, 140 m / s^{2}$

$- 3.0 m, 100 m / s^{2}$

$+ 2.54 m, 200 m / s^{2}$

$+ 3.55 m, 120 m / s^{2}$

answer is C.

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Detailed Solution

Displacement $x (t) = 5 \cos (2 π \times \frac{3}{2} \times \frac{π}{4}) = 5 \cos (\frac{13 π}{4}) = - 3.54 m$ $∴ Velocity (v) = \frac{d x}{dt} = - 10 πsin (2 π + \frac{π}{4})$

$∴ Acceleration, a = \frac{dv}{dt} = - 20 π^{2} \cos (2 πt + \frac{π}{4})$

$= - 20 π^{2} \cos (2 πt + \frac{π}{4}) = 20 π^{2} \cos \frac{13 π}{4} = 140 m / s^{2}$

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