Time to Reach 75% Kinetic Energy in Simple Harmonic Oscillation

Given:
- Period of oscillation, T = 2 seconds
- Total energy = Potential energy + Kinetic energy
- We need to find the time at which kinetic energy is 75% of the total energy

Approach:
- In a simple harmonic oscillator, the total energy remains constant.
- At any given time, the total energy is the sum of potential and kinetic energy.
- The kinetic energy is maximum when the displacement is zero and potential energy is zero when the displacement is maximum.
- At the mean position (origin), kinetic energy is maximum and potential energy is zero.

Steps to Find the Time:
1. Let's assume the total energy at the mean position is E.
2. As kinetic energy is 75% of the total energy, kinetic energy at that time = 0.75E.
3. We know that kinetic energy, KE = (1/2) * m * v^2, where m is the mass and v is the velocity.
4. At the mean position, velocity is maximum and displacement is zero.
5. Therefore, kinetic energy at the mean position = (1/2) * m * vmax^2 = E.
6. As kinetic energy is proportional to the square of velocity, the ratio of kinetic energy to total energy will be the square of the ratio of velocity to maximum velocity.
7. So, (v/vmax)^2 = 0.75
8. Solving for v/vmax, we get v/vmax = sqrt(0.75) = 0.866
9. The time taken to reach this velocity is one-fourth of the time period, which is 0.5 seconds.
Therefore, after 0.5 seconds from the origin, the kinetic energy will be 75% of the total energy in a simple harmonic oscillator with a period of 2 seconds.