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The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation is

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$5 πsec$

$\frac{π}{5} \sec$

$2 πsec$

$20 πsec$

answer is A.

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Detailed Solution

At mean position, the kinetic energy is maximum.
Hence

$\frac{1}{2} {ma}^{2} ω^{2} = 16 \frac{1}{2} \times 5.12 \times {(25 \times 10^{- 2})}^{2} ω^{2} = 16$
On putting the values we get $ω = 10 \Rightarrow T = \frac{2 π}{ω} = \frac{π}{5} \sec$

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