Simple Harmonic Motion of a particle of mass

[Himanshu]

Homework Statement

A particle of mass m is constrained to move along the x-axis. The potential energy is given by, V (x) = a + bx + cx2, where a, b, c are positive constants. If the particle is disturbed slightly from its equilibrium position, then it follows that

(a) it performs simple harmonic motion with period 2pi*Sqrt(m/2c)
(b) it performs simple harmonic motion with period 2pi*Sqrt(ma/2b^2).
(c) it moves with constant velocity
(d) it moves with constant acceleration

Homework Equations

Force=-Gradient of V(x)
F=-(b+2cx)
a=-(b+2cx)/m
T=2pi*Sqrt(Displacement/Acceleration)

The Attempt at a Solution

I cannot proceed further. For an object to perform SHM 'a' should be proportional to -x. But how do I find this proportionality over the sum.

 

[rl.bhat]

Shm

Dimensionally b/m must be an acceleraton. So you can write a + b/m = a' = -2c/m*x. So k becomes 2c/m.

 

[andrevdh]

The potential energy of a SHM oscillator is given by

$$U = 1/2 kx^2$$

it therefore seems that for the given equation the origin is not situated at the equilibrium point of the oscillator. Maybe you should try and shift the reference system and see what you get.

 

[Himanshu]

The potential energy of a SHM oscillator is given by

$$U = 1/2 kx^2$$

it therefore seems that for the given equation the origin is not situated at the equilibrium point of the oscillator. Maybe you should try and shift the reference system and see what you get.

On Shifting the origin to the vertex of the parabola I get the origin as (-b/2c, (b^2-4ac)/2c^2) with reference to the previous coordinate system. Therefore the abscissa x' in the new coordinate system is (x+b/2c).

On putting this value in the PE equation I get U=1/2 kx'^2=1/2 k(x+b/2c)^2. How do I proceed from here onwards?

 

[rl.bhat]

Take c common from the equation, and write
V = c(a/c + b/c*x + x^2).You can wright this as c(x + alpha)^2 where alpha = b/2c and (alpha)^2 = a^2/c^2.
Now proceed using relevant euqations.

 

[andrevdh]

My maths are a bit rusty in this respect so I did it the long and hard way. I rewrote the original equation in terms of the shifted reference system (where the origin coincides with the position of minimum potenetial energy). That is

$$x = x_1 - \frac{b}{2c}$$

that got me to

$$V(x_1) = a - \frac{3b^2}{4c} + cx_1^2$$