Starting from the origin a body osillates simple harmonicall with a period of 2 s. A fter what time will its kinetic energy be 75% of the total energy?

Options

A. `1/6s`
B. `1/4s`
C. `1/3s`
D. `1/12s`

Explanation

To solve the problem step by step, we will analyze the kinetic energy in simple harmonic motion (SHM) and find the time when the kinetic energy is 75% of the total energy.

### Step 1: Understand the relationship between kinetic energy and total energy in SHM
In SHM, the total mechanical energy (E) is constant and is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where:
- \( m \) is the mass of the body,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude of the motion.

The kinetic energy (KE) at any point in time is given by:
\[ KE = \frac{1}{2} m v^2 \]
where \( v \) is the velocity of the body.

### Step 2: Express kinetic energy in terms of displacement
The velocity \( v \) in SHM can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( x \) is the displacement from the mean position. Therefore, the kinetic energy can be rewritten as:
\[ KE = \frac{1}{2} m (\omega^2 (A^2 - x^2)) \]

### Step 3: Set up the equation for kinetic energy being 75% of total energy
We want to find when the kinetic energy is 75% of the total energy:
\[ KE = \frac{3}{4} E \]
Substituting the expressions for KE and E, we get:
\[ \frac{1}{2} m (\omega^2 (A^2 - x^2)) = \frac{3}{4} \left(\frac{1}{2} m \omega^2 A^2\right) \]

### Step 4: Simplify the equation
Cancel out the common terms:
\[ A^2 - x^2 = \frac{3}{4} A^2 \]
This simplifies to:
\[ x^2 = A^2 - \frac{3}{4} A^2 = \frac{1}{4} A^2 \]
Thus, we find:
\[ x = \frac{A}{2} \]

### Step 5: Find the time corresponding to this displacement
The displacement in SHM is given by:
\[ x = A \sin(\omega t) \]
Substituting \( x = \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t) \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t) \]

### Step 6: Solve for \( t \)
Taking the inverse sine:
\[ \omega t = \sin^{-1}\left(\frac{1}{2}\right) \]
We know that:
\[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \]
Now, we need to find \( \omega \):
\[ \omega = \frac{2\pi}{T} \]
Given that the period \( T = 2 \) seconds, we have:
\[ \omega = \frac{2\pi}{2} = \pi \]

Substituting back:
\[ \pi t = \frac{\pi}{6} \]
Thus:
\[ t = \frac{1}{6} \text{ seconds} \]

### Final Answer
The time after which the kinetic energy will be 75% of the total energy is:
\[ t = \frac{1}{6} \text{ seconds} \]