Starting from origin a body oscillates simple harmonically with a period of 2seconds after what time will it's kinetic energy be 75% of the total energy

To find the time at which the kinetic energy of the body is 75% of the total energy, we can use the equation for the total energy of a simple harmonic oscillator:

Total Energy = 1/2 * m * ω^2 * A^2

where m is the mass of the body, ω is the angular frequency, and A is the amplitude of the oscillation.

In a simple harmonic motion, the kinetic energy is given by:

Kinetic Energy = 1/2 * m * ω^2 * A^2 * sin^2(ωt)

To find the time at which the kinetic energy is 75% of the total energy, we need to solve the equation:

0.75 * Total Energy = Kinetic Energy

Substituting the expressions for total energy and kinetic energy, we get:

0.75 * (1/2 * m * ω^2 * A^2) = 1/2 * m * ω^2 * A^2 * sin^2(ωt)

Simplifying the equation, we have:

0.75 = sin^2(ωt)

Taking the square root of both sides, we get:

√0.75 = sin(ωt)

Now, we need to find the value of ωt that satisfies this equation. Taking the inverse sine of both sides, we have:

ωt = arcsin(√0.75)

Using a calculator, we can find the value of arcsin(√0.75) to be approximately 0.968 radians.

Finally, we can find the time at which the kinetic energy is 75% of the total energy by dividing ωt by the angular frequency ω:

t = (ωt) / ω

Since the period T is given as 2 seconds, we know that ω = 2π / T. Substituting this value, we have:

t = (0.968) / (2π / 2)

Simplifying the expression, we get:

t = 0.968 / π

Using a calculator, we can find the value of t to be approximately 0.308 seconds.

Therefore, after approximately 0.308 seconds, the kinetic energy of the body will be 75% of the total energy.