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A moving particle has an acceleration a ...

A moving particle has an acceleration a = -bx, where b is a positive constant and x is the distance of the particle from the equilibrium position. What is the time period of oscillation of the particle?

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When a particle executing SHM is at a distance of 0.03 m from the equilibrium position, its acceleration is 0.296m*s^(-2) . What is the time period of oscillation of the particle?

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Knowledge Check

  • A particle moves such that its acceleration 'a' is give a= -bx where x is the displacement from equilibrium position and b is constant. The period of oscillation is

    A
    `(2pi)/b`
    B
    `(2pi)/sqrtb`
    C
    `sqrt((2pi)/b)`
    D
    `2sqrt((pi)/b`

  • A particle starts oscillating simple harmonically from the equilibrium position with time period T. The ratio of K.E and P.E of the particle at time t = T/12

    A
    1 : 4
    B
    2 : 1
    C
    3 : 1
    D
    4 : 1

  • Acceleration of a particle with SHM at 8 cm away from equilibrium 128am//s^2 . What is the time period?

    A
    1.57s
    B
    2.57s
    C
    3s
    D
    8s

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    The maximum displacement of a vibrating particle from the equilibrium position is called _____

    The ratio of kinetic energy and potential energy of a particle executing SHM at a distance of 2 cm from its equilibrium position is 3 : 2. What is the amplitude of vibration of the particle?

    Acceleration of a particle having SHM is 0.12 ml S^2 . When the particle is 0.03 mt away from its equilibrium position what will be its time period?

    m is the mass of a particle and omega is the angular frequency of the particle executing S.H.M. The time period of the oscillation will be

    A particle vibrating simple harmonically has an acceleration of 16cm*s^(-2) when it is at a distance of 4 cm from the mean position. Its time period is

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