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The kinetic energy of a particle executing S.H.M. is 16J when it is at its mean position if the amplitude of oscillation is 25cm , and the mass of the particle is 5.12kg then time period of the oscillation is:

(A) 20πsec
(B) 2πsec
(C) π5sec
(D) 5πsec

User Aderesh
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Final answer:

The time period of oscillation for a particle with a given mass, amplitude, and kinetic energy at its mean position is calculated using the formulas for kinetic energy and period of a simple harmonic oscillator, resulting in a time period of 5π seconds.

Step-by-step explanation:

The kinetic energy (KE) of a particle at its mean position during simple harmonic motion (SHM) is given by KE = ½mv², where m is the mass of the particle, and v is its velocity at that point. When the particle is at the mean position, its potential energy is zero, and its kinetic energy is at the maximum, which is equal to the total energy of the system. To find the time period (T) of oscillation, we use the formula T = 2π√(m/k), where k is the spring constant which can be derived from the total energy and amplitude (A).

Given that the kinetic energy at the mean position is 16J, the mass is 5.12kg, and the amplitude of oscillation is 25cm (0.25m), we can find the spring constant using the relation KE = ½kA². After determining k, we substitute it into the time period formula to find T. The calculation yields the time period of the oscillation as 5π seconds.

User Michael Fromin
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