Potential energy of a particle with mass m is U=kx3 , where k is a positive constant. The particle is oscillating about the origin on x-axis. If the amplitude of oscillation is a, then its time period, T isa)Proportional to a2b)Independent of ac)Proportional to ad)Proportional to 1/aCorrect answer is option 'D'. Can you explain this answer?

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a)
Proportional to a2
b)
Independent of a
c)
Proportional to √a
d)
Proportional to 1/√a

Correct answer is option 'D'. Can you explain this answer?

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Potential energy of a particle with mass m is U=kx3 , where k is a pos...

P.E.=kx^2

F=-du/dx=-3kx^2

Maximum force=-3ka^2=-mw^2a\

w^2=3ka/m

and w^2 also equals to 4π^2/T^2

Comparing both, we get T is directly proportional to 1/√a.

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Potential energy of a particle with mass m is U=kx3 , where k is a pos...

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Potential energy of a particle with mass m is U=kx3 , where k is a pos...

According to the given question, the potential energy of a particle with mass m is given by U = kx^3, where k is a positive constant.

To analyze the time period of oscillation, we need to consider the motion of the particle.

The motion of the particle can be described by the equation of motion, which is derived from the potential energy function. The equation of motion for this system is given by:

m(d^2x/dt^2) = -dU/dx

Differentiating the potential energy function with respect to x, we get:

dU/dx = 3kx^2

Substituting this value into the equation of motion, we have:

m(d^2x/dt^2) = -3kx^2

Rearranging the equation, we get:

(d^2x/dt^2) = -(3k/m)x^2

This is a non-linear differential equation, and its solution gives the equation of motion for the particle.

To find the time period of oscillation, we can use the small angle approximation, where the amplitude of oscillation (a) is much smaller than the distance between the two extreme points of oscillation.

In this case, we can assume that x < a,="" which="" simplifies="" the="" equation="" of="" motion="" />

(d^2x/dt^2) = -(3k/m)a^2

This is a linear differential equation with a simple harmonic motion solution. The general solution to this equation is given by:

x(t) = A*cos(ωt + φ)

Where A is the amplitude of oscillation, ω is the angular frequency, t is time, and φ is the phase constant.

The time period of oscillation (T) is given by the reciprocal of the angular frequency:

T = 2π/ω

From the equation of motion, we can see that the angular frequency ω is proportional to the square root of the constant (3k/m)a^2.

Therefore, the time period of oscillation (T) is inversely proportional to the amplitude of oscillation (a).

Hence, the correct answer is option D: Proportional to 1/a.

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Potential energy of a particle with mass m is U=kx3 , where k is a positive constant. The particle is oscillating about the origin on x-axis. If the amplitude of oscillation is a, then its time period, T isa)Proportional to a2b)Independent of ac)Proportional to √ad)Proportional to 1/√aCorrect answer is option 'D'. Can you explain this answer?

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