Example 1: In a spring-mass system k = 16 N/m and m = 1 kg . If the mass is displaced by .05 m and released from rest, find its subsequent motion.

Using the initial conditions I get

x(0)= D = 0 .05m

So the solution is with the maximum speed of 0.2m/s . The solution x(t) is plotted in figure 4. Also plotted there is the velocity v(t) of the mass as it performs its motion. Notice that from the x(t) curve, the velocity can be easily plotted by taking its slope.

Let me now show you how the solution changes when the initial conditions are different. Suppose instead of pulling the mass and releasing it, I give it an initial velocity of .1m/s toward the right from the equilibrium. In that case

So . Obviously the maximum speed in this case is 0.1m/s, that given in the beginning. The solution looks like shown in figure 5.

Third possibility of initial conditions is when I take the mass to a displacement of .05m and push it towards the equilibrium point with a speed of .1m/sec. Then

Thus the solution is . If we wish to express this as then

and

This gives and . The maximum speed in this case is v_max = 4 x 0.056 = 0.224m/s . So the graph of the motion looks like that shown in figure 6.

From the graph it is very clear that initially the speed of the particle increases in the negative direction and then the particle starts slowing down, stopping at the full compression of the spring, as is clear from the plot of its displacement.

If in the case studied just now, the mass was thrown out instead of being pushed in, it would have a positive velocity to start with but the speed would be decreasing at that moment. Then the mass will travel out to its maximum displacement and would then turn back. The general plot of displacement and velocity versus time would then look as in figure 7. I will leave it for you to work out the numbers for amplitude and initial phase.

Example 2: In the second example I show that about any stable equilibrium point, the motion to a good degree is simple harmonic. let us take two changes of 10 µC each at a distance of half a meter so that is a positive charge of 5 µC is kept at the centre, its experiences no force (see figure 8). The 5 µC charge is confined to move along the line joining the two changes. If displaced by a small distance from its equilibrium position, what kind of motion does it perform?

When the 5 µC is displaced to the right by x, the force on it is

In obtaining the force above, we have used the binomial theorem to expand . Since the force is proportional to the displacement and in direction opposite to it, the charge will perform simple harmonic motion.

Let me now look at some other examples, going beyond the spring-mass system.

Example 3: A disc of mass M and radius R is hanging on a will about a point on its periphery (see figure 9). If it is displaced from its initial position by small angle and released, find its subsequent motion.

This is a case where a rigid body is moving under distributed forces so we use angular momentum to describe its motion. The equation of its motion therefore is

By transformation theorem,

So the equation of motion becomes

This means that in general the motion of the disc would be simple harmonic and will be given as

The initial conditions in this case give C = 0 and D = θ₀ . Therefore the solution in the present case is .

Example 4: As the final example here, let me take a particle moving in a potential .The potential has a minimum at x₀ given by

You can yourself check that the second derivative at this point is positive and its value is 8B. For very small displacements x about this point we have the change in the potential energy given as

which by binomial theorem or the Taylor series expansion leads to

This gives an equivalent spring constant of k=8B and frequency of oscillation .

Having solved these examples I now wish to discuss a very important topic of phase and phase difference in a simple-harmonic motion. I will spend some time discussion phasor diagrams give a feel for the phase.