A particle of mass m moves in the potential energy U shoen above. The period of the motion when the particle has total energy E is

Options

A. `2pisqrt((m)/(k))+4sqrt((2E)/(mg^2))`
B. `2pisqrt((m)/(k))`
C. `pisqrt((m)/(k))+2sqrt((2E)/(mg^2))`
D. `2sqrt((2E)/(mg^2))`

Explanation

To find the period of motion for a particle of mass \( m \) moving in a potential energy curve with total energy \( E \), we can break down the solution into several steps.

### Step 1: Analyze the Potential Energy Curve
The potential energy \( U \) is given as:
- For the left side: \( U = \frac{1}{2} k x^2 \) (which represents a harmonic oscillator)
- For the right side: \( U = mgx \) (which represents a linear potential)

### Step 2: Determine the Motion on the Left Side
For the left side of the curve:
- The force \( F \) can be derived from the potential energy:
\[ F = -\frac{dU}{dx} = -\frac{d}{dx}\left(\frac{1}{2} k x^2\right) = -kx \]
- This indicates that the motion is simple harmonic motion (SHM) with:
\[ a = -\frac{k}{m}x \]
- The angular frequency \( \omega \) is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]
- The time period \( T \) for a complete oscillation in SHM is:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

### Step 3: Calculate the Time for Left Side Motion
Since the particle moves from the mean position to the extreme position and back to the mean position, the time taken for this half cycle (left side) is:
\[ T_{\text{left}} = \frac{T}{2} = \pi \sqrt{\frac{m}{k}} \]

### Step 4: Determine the Motion on the Right Side
For the right side of the curve:
- The force \( F \) is:
\[ F = -mg \]
- The particle will accelerate downwards under gravity until it reaches the maximum displacement where its kinetic energy is zero.

### Step 5: Calculate the Maximum Velocity
At the lowest point (where potential energy is minimum), the kinetic energy is equal to the total energy \( E \):
\[ \frac{1}{2} mv^2 = E \implies v = \sqrt{\frac{2E}{m}} \]

### Step 6: Calculate Time to Stop
Using the equation of motion:
\[ v = u - gt \]
Setting \( v = 0 \) (when the particle stops), we have:
\[ 0 = \sqrt{\frac{2E}{m}} - gt \implies t = \frac{\sqrt{\frac{2E}{m}}}{g} = \frac{1}{g} \sqrt{\frac{2E}{m}} \]
This is the time taken to stop, denoted as \( t \).

### Step 7: Calculate Total Time for Right Side Motion
The time to go down and come back to the origin is:
\[ T_{\text{right}} = 2t = 2 \left(\frac{1}{g} \sqrt{\frac{2E}{m}}\right) = \frac{2}{g} \sqrt{\frac{2E}{m}} \]

### Step 8: Calculate Total Period of Motion
The total period \( T_{\text{total}} \) is the sum of the left and right side times:
\[ T_{\text{total}} = T_{\text{left}} + T_{\text{right}} = \pi \sqrt{\frac{m}{k}} + \frac{2}{g} \sqrt{\frac{2E}{m}} \]

### Final Expression
Thus, the period of the motion when the particle has total energy \( E \) is:
\[ T = \pi \sqrt{\frac{m}{k}} + \frac{2}{g} \sqrt{\frac{2E}{m}} \]