A particle of mass m is restricted to move without friction on the x-axis. This particle is connected to a fixed point P a distance d from the x-axis by a spring with constant k and length at rest l_0. Working in the two-dimensional plane (x, y), we will denote the coordinates of the particle by (x, 0) while the point P has coordinates (0, d). Find the equilibrium positions of the particle and discuss their stability. In the case where the equilibrium positions are stable, find the period of the small oscillations near equilibrium.
Hint: You will want to distinguish the cases d > l_0, and d < l_0 (ignore the special case d = l_0).
Sorry, I do not know what l_0 means.
To Anonymous: l_0 is the length at rest.
I will call the unstrained length Lo
If d > Lo
then the spring is always under tension
spring length = ( d*2+x^2)^0.5
spring tension = [( d*2+x^2)^0.5 - Lo ]k
equilibrium at x = 0 and stable
x component of restoring spring force = [( d*2+x^2)^0.5 - Lo ]k * [- x / ( d*2+x^2)^0.5]
so
m d^2x/dt^2 = [( d*2+x^2)^0.5 - Lo ]k * [- x / ( d*2+x^2)^0.5]
for small motion x<<d
m d^2x/dt^2 = [d - Lo ]k * [- x / d] = - k x * (1-Lo/d)
oh my, looks like a plain old spring with a new K = original spring k [ 1 -Lo/d]
Use your old spring equations with that new K
T = 2 pi sqrt { m/k [ 1 -Lo/d] }
For the second case you have an unstable mess. There are equilibrium points at a positive x and at a negative x.
To find the equilibrium positions of the particle, we need to consider the forces acting on it.
Let's assume the particle is displaced from its equilibrium position by a distance x in the positive x-direction. The spring will exert a restorative force on the particle towards the equilibrium position, which can be modeled by Hooke's Law. The magnitude of this force is given by F = -kx, where k is the spring constant.
Additionally, there is the force of gravity acting on the particle, which is proportional to the mass of the particle and can be written as F_gravity = -mg, where g is the acceleration due to gravity.
Since there is no friction, the net force acting on the particle should be zero at equilibrium. Therefore, we can write the equation for equilibrium as:
F_net = -kx - mg = 0
Solving this equation for x gives us the equilibrium position(s) of the particle.
Case 1: d > l_0
In this case, the spring will be stretched when the particle is at equilibrium, meaning the equilibrium position will have a positive displacement. Solving the equation, we have:
-kx - mg = 0
-kx = mg
x = -mg/k
So, the equilibrium position is x = -mg/k.
Case 2: d < l_0
In this case, the spring will be compressed when the particle is at equilibrium, meaning the equilibrium position will have a negative displacement. Solving the equation, we have:
-kx - mg = 0
-kx = mg
x = -mg/k
So, the equilibrium position is x = -mg/k.
Now, let's discuss the stability of the equilibrium positions. To determine stability, we can consider the second derivative of the potential energy function. If the second derivative is positive, the equilibrium position is stable; if it is negative, the equilibrium position is unstable.
The potential energy function for a spring is given by U(x) = (1/2)kx^2. Taking the second derivative of U(x) with respect to x, we have:
U''(x) = k
Since k is positive, the second derivative U''(x) is positive, indicating that the equilibrium positions are stable in both cases.
Finally, let's calculate the period of small oscillations near the equilibrium positions.
The period of oscillation can be found using the formula:
T = 2π√(m/k)
Substituting the given values, we have:
T = 2π√(m/k)
Therefore, the period of small oscillations near the equilibrium positions is 2π√(m/k).