Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy?

Let's solve this step by step.

Step 1: Understanding the system
The body is undergoing simple harmonic motion (SHM) with a period of 2 seconds. The total mechanical energy of a body in SHM is constant and is the sum of its kinetic energy (KE) and potential energy (PE).

The total energy (E) in SHM is given by: E = (1/2) * m * ω² * A² where:

m is the mass of the body,
ω is the angular frequency,
A is the amplitude of oscillation.
The angular frequency (ω) is related to the period (T) by the formula: ω = 2π / T.

Given that the period T is 2 seconds, we can calculate ω: ω = 2π / 2 = π radians per second.

Step 2: Kinetic and potential energy in SHM
In SHM, the kinetic energy (KE) and potential energy (PE) vary during the motion. At any point in the oscillation, the total energy is divided into KE and PE.

KE = (1/2) * m * v²
PE = (1/2) * m * ω² * (A² - x²) where:
v is the velocity,
x is the displacement from the equilibrium position.
At any position, the total energy E is the sum of KE and PE: E = KE + PE.

Step 3: Condition for 75% of total energy as KE
We are asked to find the time at which the kinetic energy is 75% of the total energy. This means: KE = 0.75 * E.

We know the relationship between KE and PE in SHM: KE = (1/2) * m * ω² * (A² - x²), PE = (1/2) * m * ω² * x².

Since the total energy E is constant, we can write: E = KE + PE = (1/2) * m * ω² * A².

Now, we know that at any point in SHM: KE = 0.75 * E.

This gives: (1/2) * m * ω² * (A² - x²) = 0.75 * (1/2) * m * ω² * A².

Simplifying: A² - x² = 0.75 * A², x² = 0.25 * A², x = 0.5 * A.

Step 4: Finding the time
Now, we know that the displacement x = 0.5 * A. In SHM, the displacement x is related to the amplitude A by the equation: x = A * cos(ωt).

Thus, we have: 0.5 * A = A * cos(ωt).

Canceling A from both sides: 0.5 = cos(ωt).

Taking the inverse cosine: ωt = cos⁻¹(0.5).

We know that cos⁻¹(0.5) = π/3, so: ωt = π/3.

Now, solving for t: t = (π/3) / ω.

Since ω = π, we get: t = (π/3) / π = 1/3 seconds.

Final Answer:
The time after which the kinetic energy will be 75% of the total energy is 1/3 seconds.