A particle, doing simple harmonic motion, at a distance 3 cm from mean position has acceleration 12 cm / s 2. What is its time period?

Question

A particle, doing simple harmonic motion, at a distance 3 cm from mean position has acceleration 12 cm / s 2. What is its time period? (A) 0.5 s (B) 1 s (C) 2 s (D) 3.14 s

Tardigrade Admin · Accepted Answer

The time period of a particle executing SHM is given as T=2 π √(x/a)=2 π √( text Displacement / text Acceleration ) =2 π √(3 × 10-2/12 × 10-2) =2 π × (1/2) =π=3.14 s

Q. A particle, doing simple harmonic motion, at a distance $3 c m$ from mean position has acceleration $12 c m / s^{2}$ . What is its time period?

0.5 s

1 s

2 s

3.14 s

The time period of a particle executing SHM is given as
$T = 2 π \frac{x}{a} = 2 π \frac{Displacement}{Acceleration}$ $= 2 π \frac{3 \times 1 0 ^{- 2}}{12 \times 1 0 ^{- 2}}$
$= 2 π \times \frac{1}{2}$
$= π = 3.14 s$

Q. A particle, doing simple harmonic motion, at a distance 3cm from mean position has acceleration 12cm/s2. What is its time period?

0.5 s

1 s

2 s

3.14 s

The time period of a particle executing SHM is given as T=2πax​​=2π Acceleration Displacement ​​ =2π12×10−23×10−2​​ =2π×21​ =π=3.14s

Q. A particle, doing simple harmonic motion, at a distance $3 c m$ from mean position has acceleration $12 c m / s^{2}$ . What is its time period?

The time period of a particle executing SHM is given as
$T = 2 π \frac{x}{a} = 2 π \frac{Displacement}{Acceleration}$ $= 2 π \frac{3 \times 1 0 ^{- 2}}{12 \times 1 0 ^{- 2}}$
$= 2 π \times \frac{1}{2}$
$= π = 3.14 s$