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, we will follow these steps:

### Step 1: Understand the Initial Conditions
At the extreme position of the pendulum, the displacement \( x \) is equal to the amplitude \( A \). The total energy \( E \) of the pendulum in simple harmonic motion (SHM) is given by:
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant equivalent for the pendulum.

### Step 2: Determine the Phase Shift
We are given that the pendulum undergoes a phase shift of \( \frac{\pi}{3} \) radians from the extreme position. The extreme position corresponds to a phase of \( \frac{\pi}{2} \) radians. Therefore, after the phase shift, the new phase \( \phi \) is:
\[ \phi = \frac{\pi}{2} - \frac{\pi}{3} = \frac{3\pi}{6} - \frac{2\pi}{6} = \frac{\pi}{6} \]

### Step 3: Calculate the New Displacement
Using the equation of motion for SHM:
\[ x(t) = A \sin(\omega t + \phi) \]
At the new phase \( \phi = \frac{\pi}{6} \):
\[ x = A \sin\left(\frac{\pi}{6}\right) = A \cdot \frac{1}{2} = \frac{A}{2} \]

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

### Step 5: Calculate the Linear Momentum
The linear momentum \( p \) is given by:
\[ p = mv \]
Substituting the expression for \( v \):
\[ p = m \cdot \frac{\omega A \sqrt{3}}{2} = \frac{m A \omega \sqrt{3}}{2} \]

### Step 6: Relate Momentum to Total Energy
We know that the total energy \( E \) is also given by:
\[ E = \frac{1}{2} m v^2 \]
Substituting \( v = \frac{\omega A \sqrt{3}}{2} \):
\[ E = \frac{1}{2} m \left(\frac{\omega A \sqrt{3}}{2}\right)^2 = \frac{1}{2} m \cdot \frac{3 \omega^2 A^2}{4} = \frac{3}{8} m \omega^2 A^2 \]

### Step 7: Express Momentum in Terms of Total Energy
From the expression for momentum:
\[ p = \frac{m A \omega \sqrt{3}}{2} \]
We can express \( A \omega \) in terms of \( E \):
\[ E = \frac{1}{2} m \omega^2 A^2 \implies A \omega = \sqrt{\frac{2E}{m}} \]
Substituting this into the momentum expression:
\[ p = \frac{m \sqrt{\frac{2E}{m}} \sqrt{3}}{2} = \frac{\sqrt{3m}}{2} \cdot \sqrt{2E} = \frac{\sqrt{6mE}}{2} \]

### Final Answer
Thus, the linear momentum after a phase shift of \( \frac{\pi}{3} \) radians is:
\[ p = \frac{\sqrt{6}}{2} \sqrt{mE} \]