Aparticle 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

The position of stable equilibrium is -a

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

Explanation:

Since the potential is:

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.

Which vanish for

x = a   ; x =-a

The second derivative of V(x) is:

And:

Therefore:

a)

The position of stable equilibrium is -a

And the period of small oscillations  must be:

(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

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

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