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A particle, of unit mass moves along the x-axis under the influence of a potential, V (x) = x (x - 2 )2 . The particle is found to be in stable equilibrium at the point x = 2. Find out the time period of oscillation of the particle.Write answer upto two decimal places

Correct answer is '2.22'. Can you explain this answer?
Verified Answer
A particle, of unit mass moves along the x-axis under the influence of...
The Potential is

V = x[x2 - 2x + 4]

V = x3 - 2x2 + 4x

For equilibrium  



Given x0 = 2 then = 2√2



2.22
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A particle, of unit mass moves along the x-axis under the influence of...
Given information:
- The potential function of the particle is V(x) = x(x - 2)^2.
- The particle is in stable equilibrium at x = 2.

Understanding the problem:
- The particle is in stable equilibrium at x = 2, which means it is in a position where it will not move unless an external force is applied.
- The potential function V(x) represents the energy of the particle at different positions along the x-axis.

Deriving the equation of motion:
- The equation of motion for a particle under the influence of a potential energy function is given by Newton's second law: F = -dV/dx, where F is the force acting on the particle.
- In this case, the force acting on the particle is F = -dV/dx = -d/dx(x(x - 2)^2).
- Taking the derivative, we get F = -2(x - 2)(x - 1).

Understanding the force:
- The force F = -2(x - 2)(x - 1) is a restoring force.
- At x = 2, the force is zero, indicating that the particle is in stable equilibrium.
- When the particle is displaced from x = 2, the force acts to restore it back to the equilibrium position.

Calculating the time period of oscillation:
- The time period of oscillation is the time taken for the particle to complete one full oscillation between two extreme positions.
- The time period can be determined by finding the time it takes for the particle to travel from one extreme position to another and back.

Finding the extreme positions:
- The extreme positions occur where the force F is maximum or minimum.
- The force F = -2(x - 2)(x - 1) is zero at x = 1 and x = 2.
- Therefore, the extreme positions are x = 1 and x = 2.

Calculating the time period:
- The time period T can be calculated using the formula T = 2π/ω, where ω is the angular frequency.
- The angular frequency ω is given by ω = √(k/m), where k is the force constant and m is the mass of the particle.
- In this case, the force constant k can be determined by evaluating the force F at x = 1 or x = 2.
- At x = 1, F = -2(1 - 2)(1 - 1) = 0, indicating that the particle is at rest.
- Therefore, we can conclude that the force constant k is positive.
- Since the mass of the particle is given as unit mass, we can assume m = 1.

Calculating the force constant:
- Let's calculate the force constant by evaluating the force F at x = 2.
- At x = 2, F = -2(2 - 2)(2 - 1) = 2.
- Therefore, the force constant k = 2.

Calculating the angular frequency:
- The angular frequency ω = √(k/m) = √(2/1) = √2.

Calculating the time period:
- The time period T
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Community Answer
A particle, of unit mass moves along the x-axis under the influence of...
I think the answer will be 3.14 (pi). This question was asked on gate 2012. There was also the answer 3.14.
Solution:
V(x)= x(x-2)²
= x³ -4x²+4x
Now, Force F(x) = -dV/dx
= -3x² +8x-4
As the particle is in stable equilibrium at x=2, taking y=x-2,
F(y) = -3(y+2)² + 8(y+2) - 4
=-3y² -12y -12 +8y+16 -4
= -3y² -4y
Here the force component f(y) = -4y will cause simple harmonic motion. So, f(y) = -4y²

Comparing with f(y) = -ky², k= 4

So, T= 2π/w = 2π√(m/k) = 2π/√4 (as given, m=1)
or, T = π
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A particle, of unit mass moves along the x-axis under the influence of a potential, V (x) = x (x - 2 )2. The particle is found to be in stable equilibrium at the point x = 2. Find out the time period of oscillation of the particle.Write answer upto two decimal placesCorrect answer is '2.22'. Can you explain this answer?
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