A particle starts oscillating simple harmoniclaly from its equilibrium postion. Then the ratio fo kinetic and potential energy of the principle at time `T/12` is (`U_(mean)`=0, T= Time period)

Options

A. `2:1`
B. `3:1`
C. `4:1`
D. `1:4`

Answer

To solve the problem step by step, we need to find the ratio of kinetic energy (K) to potential energy (U) of a particle oscillating in simple harmonic motion (SHM) at time \( T/12 \).

### Step 1: Understand the Motion
The particle starts oscillating from its equilibrium position, which means its initial displacement \( x(0) = 0 \). The displacement of a particle in SHM can be described by the equation:
\[ x(t) = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega = \frac{2\pi}{T} \) is the angular frequency.

### Step 2: Calculate Displacement at \( T/12 \)
At time \( t = \frac{T}{12} \):
\[ x\left(\frac{T}{12}\right) = A \sin\left(\omega \frac{T}{12}\right) = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{12}\right) = A \sin\left(\frac{\pi}{6}\right) \]
Since \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \):
\[ x\left(\frac{T}{12}\right) = A \cdot \frac{1}{2} = \frac{A}{2} \]

### Step 3: Calculate Velocity at \( T/12 \)
The velocity \( v \) in SHM can be calculated using:
\[ v = \omega \sqrt{A^2 - x^2} \]
Substituting \( x = \frac{A}{2} \):
\[ v = \omega \sqrt{A^2 - \left(\frac{A}{2}\right)^2} = \omega \sqrt{A^2 - \frac{A^2}{4}} = \omega \sqrt{\frac{3A^2}{4}} = \frac{\sqrt{3}}{2} \omega A \]

### Step 4: Calculate Kinetic Energy (K)
The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} m v^2 \]
Substituting the expression for \( v \):
\[ K = \frac{1}{2} m \left(\frac{\sqrt{3}}{2} \omega A\right)^2 = \frac{1}{2} m \cdot \frac{3}{4} \omega^2 A^2 = \frac{3}{8} m \omega^2 A^2 \]

### Step 5: Calculate Potential Energy (U)
The potential energy \( U \) in SHM is given by:
\[ U = \frac{1}{2} k x^2 \]
where \( k = m \omega^2 \). Substituting \( x = \frac{A}{2} \):
\[ U = \frac{1}{2} k \left(\frac{A}{2}\right)^2 = \frac{1}{2} k \cdot \frac{A^2}{4} = \frac{1}{8} k A^2 \]
Substituting \( k = m \omega^2 \):
\[ U = \frac{1}{8} m \omega^2 A^2 \]

### Step 6: Calculate the Ratio \( \frac{K}{U} \)
Now we can find the ratio of kinetic energy to potential energy:
\[ \frac{K}{U} = \frac{\frac{3}{8} m \omega^2 A^2}{\frac{1}{8} m \omega^2 A^2} = \frac{3}{1} = 3 \]

### Final Answer
The ratio of kinetic energy to potential energy at time \( T/12 \) is:
\[ \frac{K}{U} = 3 \]