Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 25% of the total energy?

Options

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

Answer

To solve the problem of finding the time after which the kinetic energy of a body oscillating simple harmonically is 25% of the total energy, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential energy in SHM
In simple harmonic motion (SHM), the total mechanical energy (E) is constant and is the sum of kinetic energy (KE) and potential energy (PE). The total energy can be expressed as:
\[ E = KE + PE \]

### Step 2: Define the total energy
The total energy in SHM is given by the maximum potential energy when the body is at the maximum displacement (amplitude A):
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant.

### Step 3: Set up the condition for kinetic energy
We are given that the kinetic energy is 25% of the total energy:
\[ KE = 0.25 E \]
This implies that the potential energy must be 75% of the total energy:
\[ PE = 0.75 E \]

### Step 4: Express potential energy in terms of displacement
The potential energy at a displacement \( x \) from the mean position is given by:
\[ PE = \frac{1}{2} k x^2 \]

### Step 5: Relate potential energy to total energy
From the previous steps, we can write:
\[ 0.75 E = \frac{1}{2} k x^2 \]
Substituting \( E \) from Step 2:
\[ 0.75 \left(\frac{1}{2} k A^2\right) = \frac{1}{2} k x^2 \]

### Step 6: Simplify the equation
Canceling \( \frac{1}{2} k \) from both sides (assuming \( k \neq 0 \)):
\[ 0.75 A^2 = x^2 \]

### Step 7: Solve for displacement \( x \)
Taking the square root:
\[ x = A \sqrt{0.75} = A \frac{\sqrt{3}}{2} \]

### Step 8: Use the SHM equation to find time
The displacement in SHM can be expressed as:
\[ x = A \sin(\omega t) \]
where \( \omega = \frac{2\pi}{T} \) and \( T = 2 \, \text{s} \) (the period given).

Thus:
\[ A \frac{\sqrt{3}}{2} = A \sin(\omega t) \]
Cancelling \( A \) (assuming \( A \neq 0 \)):
\[ \frac{\sqrt{3}}{2} = \sin(\omega t) \]

### Step 9: Find \( \omega \)
Using the period \( T = 2 \, \text{s} \):
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \, \text{rad/s} \]

### Step 10: Substitute \( \omega \) into the equation
Now substituting \( \omega \) into the equation:
\[ \frac{\sqrt{3}}{2} = \sin(\pi t) \]

### Step 11: Solve for time \( t \)
The value of \( \sin^{-1}(\frac{\sqrt{3}}{2}) \) corresponds to \( \frac{\pi}{3} \):
\[ \pi t = \frac{\pi}{3} \]
Dividing both sides by \( \pi \):
\[ t = \frac{1}{3} \, \text{s} \]

### Final Answer
The time after which the kinetic energy is 25% of the total energy is:
\[ t = \frac{1}{3} \, \text{s} \]

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