The time period of oscillation of a particle that executes S.H.M. is $1.2 s e c$ . The time starting from extreme position, its velocity will be half of its velocity at mean position is

Question

The time period of oscillation of a particle in S-H-M- is  pi  sec- If its acceleration extreme position 1 m s-2- - its velocity mean position is0-5 ms-1- pi ms-1- 2 ms-1-  pi-2 ms-1-

Toppr Admin · Accepted Answer

T-x3C0-sec-x3C9-2-x3C0-x3C0-2radseca-1ms2a-x2212-x3C9-2A-A-14mvmax-x3C9-A-2-xD7-14-0-5ms

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
$0.5 m s^{- 1}$
$π m s^{- 1}$
$2 m s^{- 1}$
$π / 2 m s^{- 1}$

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

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 is0.5 ms−1π ms−12 ms−1π/2 ms−1

T=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
$0.5 m s^{- 1}$
$π m s^{- 1}$
$2 m s^{- 1}$
$π / 2 m s^{- 1}$

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