Physics Lab » Oscillations

Jul 28 2008

Oscillations

An oscillation is the periodic motion of a particle moving from a maximum position to a minimum position relative to some fixed origin. An oscillatory motion occurs when the particle is displaced from its equilibrium point and a certain “restoring” force acts to restore its position back to the equilibrium point. The equilibrum point is the location at which the particle’s potential energy is minimum and all forces acting on the particle is zero, we designate this point as the origin.

Equation of Motion

We are only concerned with restoring forces that are time-independant, that is; the force is dependant only on the position of the particle and not on time and there is no frictional losses.

$m\frac{d^2r}{dt^2} = F(r)$

(1)

We will consider only one dimensional motion in the x direction. If F is a continious function with continious derivatives of all orders, then F can be expanded using MacLaurin Series (Taylor Series Expansion around the point x=0).

$F(x) = F_0 + x( \frac{dF}{dx} )_0 + \frac{x^2}{2!}(\frac{d^2F}{dx^2})_0 + \frac{x^3}{3!}(\frac{d^3F}{dx^3})_0 + \ldots$

(2)

But we designated the origin to be the point where no forces act on the particle, so F0 = 0. If we only consider small displacements of the particle we can ignore the higher order terms have our approximation of the force as:

$F(x) =(\frac{dF}{dx})_0 x = -kx$

(3)

The Force is negative because F is a restoring force, that is it acts to restore the particle to its original position, it cannot do that unless F points back towards the origin. Substituting equation 3 into equation 1 we have:

$m\frac{d^2r}{dt^2} + kx = 0$

(4)

Equation 4 is a rather simple ordinary differential equation with a known solution. If we divide both sides of the equation by m, and let $\omega_0^2 = \frac{k}{m}$ , the equation then becomes:

$\frac{d^2r}{dt^2} + \omega_0^2x = 0$

(5)

The solution to the above differential equation is a sinusoidal function:

$x(t) = A_0\cos(\omega_0 t - \alpha)$

(6)

To solve for the two unknowns $A_0$ and $\alpha$ we require two initial conditions. Given the initial position and the initial speed of the mass, we can calculate the two unknowns. We shall take the initial position to be some distance from the origin, and the initial speed to be 0 m/s. This makes the unknowns easier to solve, although non-zero starting speeds are not too complicated. With our initial conditions, the equation of motion for a mass on a spring is:

$x(t) = A_0\cos(\omega_0t)$

(7)

Where $A_0$ is the initial position of the mass. In this case the initial position is also the maximum position of the mass, which is called the Amplitude.

Potential Energy

The potential energy of the particle is the work required to pull the particle from its equilibrium point to a point x. The work, and thus the potential energy of the particle in our one dimensional case is defined as:

$U=W=-\int_0^x Fdx$

(8)

By integrating equation 8 we get the expression for the potential energy at position x.

$U(x)&=-\int_0^x Fdx \\&= -\int_0^x -kx \cdot dx$

$U(x)&=\frac{1}{2}kx^2$

(9)

By substituting equation 7 into equation 9, we get the potential energy of the mass as a function of time.

$U(t) = \frac{1}{2}kA_0^2\cos^2(\omega_0 t)$

(10)

Kinetic Energy

The kinetic energy of a particle is given by the following equation:

$T = \frac{1}{2}mv^2 = \frac{1}{2}m (\frac{dx}{dt})^2$

(11)

The derivative of the position with respect to time is the velocity of the particle, we can determine the velocity of the particle by differentiating equation 7 with respect to time.

$v(t) = \frac{dx}{dt} = \frac{d}{dt}A_0\cos(\omega_0 t) = -A_0\omega_0\sin(\omega_0 t)$

(12)

Substituting equation 12 into equation 11 we can determine the kinetic energy of the particle as a function of time:

$T(t) = \frac{1}{2}mA_0^2\omega_0^2 \sin^2(\omega_0 t)$

(13)

Total Energy

The total energy of the particle is the sum of the potential and kinetic energy. By combining equation 10 and equation 13, we get the total energy of the particle and thus the total energy of the system.

$E(t) = T(t) + U(t) &= \frac{1}{2}mA_0^2\omega_0^2 \sin^2(\omega_0 t) + \frac{1}{2}kA_0^2\cos^2(\omega_0 t) \\ E(t) &= \frac{1}{2}mA_0^2(\frac{k}{m}) \sin^2(\omega_0 t) + \frac{1}{2}kA_0^2\cos^2(\omega_0 t) \\ E(t) &= \frac{1}{2}kA_0^2 ( \sin^2(\omega_0 t) + \cos^2(\omega_0 t) )$

(16)

Remember that $\sin^2(A) + \cos^2(A) = 1$ , substituting that into equation 16 we find that the total energy of the system is:

$E(t) = \frac{1}{2}kA_0^2$

(17)

Notice that the total energy of the system is independant of time, that is the total energy of the system does not change and therefore energy is conserved. Also, notice that the value of the total energy is the same as the potential energy when the particle is at its maximum position. This is obvious because when the particle is at its maximum position, the particle has no speed and therefore no kinetic energy, all the energy has been converted into potential energy and stored in the spring.

One response so far

One Response to “Oscillations”

# Yugeneon 10 Feb 2010 at 4:54 pm

This is beautiful!! It is a wonderful resource, and a great teaching aid!! Thank you!!

Oscillations

Oscillations

Equation of Motion

Potential Energy

Kinetic Energy

Total Energy

One Response to “Oscillations”

Leave a Reply