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The displacement (in m) of a particle of mass 100 gm from its mean position is given by the equation $y = 0.05 \sin 3 π (5 t + 0.4)$ m

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The time period of motion is

1 /30 \sec

The maximum acceleration of the particle is

11.2 π^{2} {m / s}^{2}

The force acting on the particle is zero when the displacement is 0.05 m

The time period of motion is

1 /7 .5 \sec

answer is C.

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Detailed Solution

The time period does not depend on mass of the sphere

A = 0.05

$ω = 15 π$

$\frac{2 π}{T} = 15 π$

$T = \frac{2}{15} s$

$a_{n a x} = A ω^{2}$

$= 0.05 {(15 π)}^{2}$

= 0.5 × 225

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