A particle of mass 200g executes simple harmonic motion. A restoring force is provided by a spring of spring constant 80N/m. Find the time period.

0.31 s

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0.52 s

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0.1 s

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0.9 s

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Solution

The correct option is A 0.31 s
The time period is $T = 2 π \sqrt{\frac{m}{k}}$
$= 2 π \sqrt{\frac{200 \times 10^{- 3} k g}{80 N m^{- 1}}}$
$= 2 π \times 0.05 s = 0.31 s .$

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A particle of mass 200 gm executes S.H.M. The restoring force is provided by a spring of force constant 80 N / m. The time period of oscillations is.

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A particle of mass 200g executes simple harmonic motion. A restoring force is provided by a spring of spring constant 80N/m. Find the time period.

The correct option is A 0.31 sThe time period is T=2π√mk =2π√200×10−3kg80Nm−1 =2π×0.05s=0.31s.

The correct option is A 0.31 s
The time period is $T = 2 π \sqrt{\frac{m}{k}}$
$= 2 π \sqrt{\frac{200 \times 10^{- 3} k g}{80 N m^{- 1}}}$
$= 2 π \times 0.05 s = 0.31 s .$