A simple pendulum of mass m executes SHM with total energy E. if at an instant it is at one of extreme positions, then its linear momentum after a phase shift of `(pi)/(3)` rad will be

Options

A. `sqrt(2mE)`
B. `sqrt((3mE)/(2))`
C. `2 sqrt(mE)`
D. `sqrt((2mE)/(3))`

Answer

To solve the problem step by step, we will follow the reasoning and calculations outlined in the video transcript.

### Step 1: Understand the Energy of the Simple Harmonic Motion (SHM)
The total energy \( E \) of a simple pendulum executing SHM is given by the formula:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where:
- \( m \) is the mass of the pendulum,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude of the motion.

### Step 2: Determine the Displacement at Phase Shift
At one of the extreme positions, the displacement \( x \) is equal to the amplitude \( A \). When the phase shifts by \( \frac{\pi}{3} \) radians, we can express the new displacement as:
\[ x = A \cos\left(\frac{\pi}{3}\right) \]
Calculating \( \cos\left(\frac{\pi}{3}\right) \):
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
Thus, the displacement becomes:
\[ x = A \cdot \frac{1}{2} = \frac{A}{2} \]

### Step 3: Calculate the Velocity
The velocity \( v \) of the pendulum can be calculated using the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
Substituting \( x = \frac{A}{2} \):
\[ v = \omega \sqrt{A^2 - \left(\frac{A}{2}\right)^2} \]
Calculating \( \left(\frac{A}{2}\right)^2 \):
\[ \left(\frac{A}{2}\right)^2 = \frac{A^2}{4} \]
Thus, we have:
\[ v = \omega \sqrt{A^2 - \frac{A^2}{4}} = \omega \sqrt{\frac{3A^2}{4}} = \omega \cdot \frac{\sqrt{3}}{2} A \]

### Step 4: Calculate the Linear Momentum
The linear momentum \( p \) is given by:
\[ p = m \cdot v \]
Substituting the expression for \( v \):
\[ p = m \cdot \left(\frac{\sqrt{3}}{2} A \omega\right) = \frac{\sqrt{3}}{2} m A \omega \]

### Step 5: Relate \( A \omega \) to Total Energy
From the energy equation, we know:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
We can express \( A \omega \) in terms of \( E \):
\[ A \omega = \sqrt{\frac{2E}{m}} \]
Substituting this back into the momentum equation:
\[ p = \frac{\sqrt{3}}{2} m \cdot \sqrt{\frac{2E}{m}} = \frac{\sqrt{3}}{2} \sqrt{2Em} \]

### Final Step: Simplify the Expression
Thus, the final expression for the linear momentum is:
\[ p = \sqrt{3E} \]

### Conclusion
The linear momentum after a phase shift of \( \frac{\pi}{3} \) radians is:
\[ p = \sqrt{3E} \]