The time period of oscillation of a particle in S-H-M- is - sec- If its acceleration at extreme position 1 m s-2- its velocity at mean position is

The correct option is A 0.5 ms^ 1 T=π sec ω =2ππ=2radsec a=1ms^2 a= ω ^2A #160;; A=14m v_max=ω A=2×14=0.5ms ...

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Standard XII

Physics

SHM expression

The time peri...

Question

The time period of oscillation of a particle in S.H.M. is $π$ sec. If its acceleration at extreme position $1 m$ $s^{- 2}$ , its velocity at mean position is

0.5 m s^{- 1}

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2 m s^{- 1}

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π / 2 m s^{- 1}

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π m s^{- 1}

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Solution

The correct option is A $0.5 m s^{- 1}$
$T = π s e c$
$ω = \frac{2 π}{π} = 2 \frac{r a d}{s e c}$
$a = 1 \frac{m}{s^{2}}$
$a = - ω^{2} A; A = \frac{1}{4} m$
$v_{m a x} = ω A = 2 \times \frac{1}{4} = 0.5 \frac{m}{s}$

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The time period of oscillation of a particle in S.H.M. is π sec. If its acceleration at extreme position 1ms−2, its velocity at mean position is

The correct option is A 0.5 ms−1T=πsecω=2ππ=2radseca=1ms2a=−ω2A;A=14mvmax=ωA=2×14=0.5ms

The time period of oscillation of a particle in S.H.M. is $π$ sec. If its acceleration at extreme position $1 m$ $s^{- 2}$ , its velocity at mean position is

The correct option is A $0.5 m s^{- 1}$
$T = π s e c$
$ω = \frac{2 π}{π} = 2 \frac{r a d}{s e c}$
$a = 1 \frac{m}{s^{2}}$
$a = - ω^{2} A; A = \frac{1}{4} m$
$v_{m a x} = ω A = 2 \times \frac{1}{4} = 0.5 \frac{m}{s}$