A particle is undergoing simple harmonic motion with a period of $2$ seconds and an amplitude of $2$ metres. Its maximum speed in m $s^{- 2}$ is?

4

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2

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\frac{π}{2}

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2 π

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Solution

The correct option is D $2 π$
Maximum speed $v = r ω$
$ω = 2 π / t$
$t = 2 s, r = 5 m$
$v = 2 \times \frac{2 π}{2} = 2 π m / s$

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[EAMCET (Med.) 2000]

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A particle is undergoing simple harmonic motion with a period of 2 seconds and an amplitude of 2 metres. Its maximum speed in m s−2 is?

The correct option is D 2πMaximum speedv=rωω=2π/tt=2s,r=5mv=2×2π2=2πm/s

A particle is undergoing simple harmonic motion with a period of $2$ seconds and an amplitude of $2$ metres. Its maximum speed in m $s^{- 2}$ is?

The correct option is D $2 π$
Maximum speed $v = r ω$
$ω = 2 π / t$
$t = 2 s, r = 5 m$
$v = 2 \times \frac{2 π}{2} = 2 π m / s$