The velocity of a particle that executes S.H.M. is at its mean position 0.866 m/s.Velocity at a displacement half of its amplitude from mean position is

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Answer 1

$v_{m a x} = A ω = 0.866 \frac{m}{s}$
At $x = \frac{A}{2}; v = ω A^{2} - x^{2}$
$v = \frac{3}{2} ω A = \frac{3}{2} v_{m a x}$
$v = \frac{3}{2} \times 0.866 = 0.75 \frac{m}{s}$

Answer 2

$v_{m a x} = A ω = 0.866 \frac{m}{s}$
At $x = \frac{A}{2}; v = ω A^{2} - x^{2}$
$v = \frac{3}{2} ω A = \frac{3}{2} v_{m a x}$
$v = \frac{3}{2} \times 0.866 = 0.75 \frac{m}{s}$

The velocity of a particle that executes S.H.M. is at its mean position 0.866 m/s.Velocity at a displacement half of its amplitude from mean position is

1 m $s^{- 1}$

1.414m $s^{- 1}$

0.5m $s^{- 1}$

0.75m $s^{- 1}$

$v_{m a x} = A ω = 0.866 \frac{m}{s}$
At $x = \frac{A}{2}; v = ω A^{2} - x^{2}$
$v = \frac{3}{2} ω A = \frac{3}{2} v_{m a x}$
$v = \frac{3}{2} \times 0.866 = 0.75 \frac{m}{s}$

The time period of oscillation of a S.H.O. is $\frac{π}{2} s$ . Its acceleration at a phase angle $\frac{π}{3}$ rad from extreme position is $2 m / s^{2}$ , what is its velocity at a displacement equals to half of its amplitude from mean position?

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

A pendulum is going S.H.M. The velocity of the bob in the mean position is $v$ . If now its amplitude is doubled, keeping the length same, its velocity in the mean position will be :

For a particle executing SHM having amplitude 'a' the speed of the particle is one half of its maximum speed when its displacement from the mean position is

A particle executing SHM has amplitude of 4 cm., and its acceleration at a distance of 1 cm from the mean position is 3 cm $s^{- 2}$ . Its velocity be when it is at a distance of 2 cm from its mean position is

Revise with Concepts

Velocity and Acceleration in SHM

Graphical Representation of Displacement, Velocity and Acceleration

The velocity of a particle that executes S.H.M. is at its mean position 0.866 m/s.Velocity at a displacement half of its amplitude from mean position is

1 ms−1

1.414ms−1

0.5ms−1

0.75ms−1

vmax​=Aω=0.866sm​ At x=2A​;v=ωA2−x2​ v=23​​ωA=23​​vmax​ v=23​​×0.866=0.75sm​

The time period of oscillation of a S.H.O. is 2π​ s. Its acceleration at a phase angle 3π​ rad from extreme position is 2 m/s2, what is its velocity at a displacement equals to half of its amplitude from mean position?

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

A pendulum is going S.H.M. The velocity of the bob in the mean position is v. If now its amplitude is doubled, keeping the length same, its velocity in the mean position will be :

For a particle executing SHM having amplitude 'a' the speed of the particle is one half of its maximum speed when its displacement from the mean position is

A particle executing SHM has amplitude of 4 cm., and its acceleration at a distance of 1 cm from the mean position is 3 cms−2. Its velocity be when it is at a distance of 2 cm from its mean position is

Revise with Concepts

Velocity and Acceleration in SHM

Graphical Representation of Displacement, Velocity and Acceleration

1 m $s^{- 1}$

1.414m $s^{- 1}$

0.5m $s^{- 1}$

0.75m $s^{- 1}$

$v_{m a x} = A ω = 0.866 \frac{m}{s}$
At $x = \frac{A}{2}; v = ω A^{2} - x^{2}$
$v = \frac{3}{2} ω A = \frac{3}{2} v_{m a x}$
$v = \frac{3}{2} \times 0.866 = 0.75 \frac{m}{s}$

The time period of oscillation of a S.H.O. is $\frac{π}{2} s$ . Its acceleration at a phase angle $\frac{π}{3}$ rad from extreme position is $2 m / s^{2}$ , what is its velocity at a displacement equals to half of its amplitude from mean position?

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

A pendulum is going S.H.M. The velocity of the bob in the mean position is $v$ . If now its amplitude is doubled, keeping the length same, its velocity in the mean position will be :

A particle executing SHM has amplitude of 4 cm., and its acceleration at a distance of 1 cm from the mean position is 3 cm $s^{- 2}$ . Its velocity be when it is at a distance of 2 cm from its mean position is