Question

A particle of mass m moves under a conservative force with potential energy V ( x )= cx/(x2+a2),where c and a are positive constants. Find the position of stable equilibrium and the period of small oscillations about it. (b) If the particle starts from this point with velocity v, find the r

Answer 1

The position of stable equilibrium is x = ±a. The period of small oscillations about the stable equilibrium is 2πa²√(4.0 kg / c).

Step-by-step explanation:

To find the position of stable equilibrium, we need to find the critical points of the potential energy function V(x). To do this, we find where the derivative of V(x) equals zero. Taking the derivative of V(x) with respect to x gives us:

V'(x) = c(a² - x²) / (x² + a²)²

Setting V'(x) equal to zero, we get:

a² - x² = 0

This implies x² = a², so x = ±a. Therefore, the position of stable equilibrium is x = ±a.

To find the period of small oscillations about the stable equilibrium, we use the formula:

T = 2π√(m/k)

Where m is the mass of the particle and k is the force constant. In this case, the force constant k is given by the second derivative of V(x) with respect to x:

k = V''(x) = 2c(a⁴ - 3x²a² + x⁴) / (x² + a²)³

When evaluating the force constant at the stable equilibrium x = ±a, we obtain:

k = V''(±a) = 2c(a⁴ - 4a⁴) / (a² + a²)³ = -4ca² / 4a⁶ = -c/a⁴

Substituting m = 4.0 kg and k = -c/a⁴ into the period formula, we get:

T = 2π√(4.0 kg / (-c/a⁴)) = 2πa²√(4.0 kg / c)

Therefore, the period of small oscillations about the stable equilibrium is 2πa²√(4.0 kg / c).

Answer 2

The position of stable equilibrium is -a

And the period of small oscillations must be: c/(ma^3)

Step-by-step explanation:

Since the potential is:

`V(x) = (c x)/(a^2+x^2)`

We first look for a position of stable equilibrium. This posiiton must satisfy two considtions, that the first derivative of the potential must vanish at this point and the second derivative must be positive.

`V'(x) = (c)/(a^2+x^2)-(2 c x^2)/(\left(a^2+x^2\right)^2)`

Which vanish for

x = a ; x =-a

The second derivative of V(x) is:

`V''(x) = (8 c x^3)/(\left(a^2+x^2\right)^3)-(6 c x)/(\left(a^2+x^2\right)^2)`

And:

`V''(a) = -(c)/(2 a^3)\\V''(-a) = (c)/(2 a^3)\\`

Therefore:

a)

The position of stable equilibrium is -a

And the period of small oscillations must be:

`\omega = √(2 V''(-a)/m) = \sqrt{(c)/(a^3 m)}`

(c/(ma^3))^1/2

b)

Let's find the maximum amplitude if the particle starts at this point with velocity v

If this is the case, the total energy will be:

(mv^2)/2

And the maximum amplitude will be

x = a^3/c mv^2 = (m v^2 a^3)/ c