A particle vibrating simple harmonically has an acceleration of $16 {c m s}^{- 2}$ when it is at a distance of $4 c m$ from the mean position. Its time period is

1 s

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2.572 s

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3.142 s

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6.028 s

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Solution

The correct option is C $3.142 s$
The equation of motion for an SHM is given as
$x = A s i n (ω t + ϕ)$
$⟹ a = \frac{d^{2} x}{d t^{2}}$
$= - A ω^{2} s i n (ω t + ϕ)$
$= - ω^{2} x$
Hence for the given position,
$16 = ω^{2} (4)$
$⟹ ω = 2$
Hence the time period of the motion is $T = \frac{2 π}{ω} = \frac{2 π}{2} = π \approx 3.142 s$

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A particle vibrating simple harmonically has an acceleration of 16cms−2 when it is at a distance of 4cm from the mean position. Its time period is

The correct option is C 3.142sThe equation of motion for an SHM is given asx=Asin(ωt+ϕ)⟹a=d2xdt2=−Aω2sin(ωt+ϕ)=−ω2xHence for the given position,16=ω2(4)⟹ω=2Hence the time period of the motion is T=2πω=2π2=π≈3.142s

A particle vibrating simple harmonically has an acceleration of $16 {c m s}^{- 2}$ when it is at a distance of $4 c m$ from the mean position. Its time period is