A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Question:

A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is [

Options :

(a) T/4
(b) T/8
(c) T/12
(d) T/2

Answer :

Answer Steps :

step 1

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The kinetic energy of a particle executing a linear SHM is 16 J, when it is in the mean position. If the amplitude is 25 cm, periodic time of the particle in second is (Mass of the particle = 51.2 g)

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Related Questions

The kinetic energy of a particle executing a linear SHM is 16 J, when it is in the mean position. If the amplitude is 25 cm, periodic time of the particle in second is (Mass of the particle = 51.2 g)

A body of mass ‘M’ executes SHM Along x-axis with frequency ‘n’. At a distance x from the mean position, the body has kinetic energy ‘K’ and potential energy ‘P’. The amplitude of oscillation of the SHM is

The bob of a simple pendulum of ‘length ‘l’ is pulled through an angle ‘θ’ from its equilibrium position and them released. When it passes through its equilibrium position, its speed is given by: (g = acceleration due to gravity)

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