Physics Bootcamp

Subsection 13.2.1 Equation of Motion and Solution

To study the block analytically, we choose \(x \) coordinate system that has a positive \(x \) axis pointed in the direction in which the spring stretches and we place the origin at the place where the block is located when spring has length equal to the equilibrium length, i.e., when there is zero force on the block.

Let \(x \) denote the \(x \) coordinate of the block at instant \(t \text{.}\) With the arrangement in Figure 13.2.2, when the block's position has \(x \gt 0 \text{,}\) the force on the block is pointed left, i.e., \(F_x \lt 0\text{,}\) and when the block's position has \(x \lt 0\text{,}\) we have \(F_x \gt 0\text{,}\) where \(F_x\) is the \(x \) component of the force vector \(\vec F\) as usual. Therefore, force by spring has \(x\)-component given by

Therefore, \(x\)-equation of motion of the block is

We would like to get a formula that tells us \(x(t) \) for given initial position and velocity, \(x_0\) and \(v_0\text{.}\) This is called solving the equation of motion. If you solve Eq. (13.2.5) for acceleration, they don't take you anywhere.

You need to think of Eq. (13.2.5) as a differential equation and solve that differential equation. We will not do that here. We will just borrow the solution from the folks who have solved this equation.

The solution is more compactly written if we replace \(k/m\) by another symbol, actually, the square of a symbol \(\omega^2\text{,}\) since we want to make sure this number is always positive.

Then, position of the block that starts at \(x=x_0\) and \(x\) component of the velocity \(v=v_0\) is completely determined by

The oscillator moves about the origin symmetrically. The largest \(x\) away from the equilibrium is called the amplitude, which is obtained by combining the coefficients of the cosine and sine terms as follows. The amplitude is usually denoted by \(A\text{.}\)

The velocity at intstant \(t \) is given by

from which we get the largest speed in the motion to be

which occurs when the block passes through the equilibrum position, i.e., \(x=0\text{.}\)

Contrast these formulas for \(x\) and \(v\) to the constant acceleration motion for acceleration \(a_x \text{.}\)

We don't have this situation here. The acceleration of a simple harmonic oscillator oscialltes in time and is definitely not constant.

Subsection 13.2.2 Multiple Ways of Writing the Solution

The solution in Eq. (13.2.7) is good for some purposes but not convenient for some other purposes. For instance, if we wanted to know the furthest from the equilibrium the block will go, its not \(x_0\text{,}\) and it is not clear right away how we would get the answer. There is another way to write a sum of cosine and sine of the same argument using the trig formulas for sum of angles.

For instance, we can write Eq. (13.2.7), which we now refer to as two-amplitudes form, also in the following form, which we will call amplitude-phase form.

where \(A \) is the amplitude and \(\phi \) the phase constant. Then, the velocity is

Another way to write in amplitude-phase form is as a sine function.

Our convention is to write as positive amplitude times a cosine and use the following to convert anything written in sine to cosine, which makes writing in sine superflous.

If we find an expression with negative amplitude times a cosine, we can convert that to positive amplitude times cosine by

The new constants \(A \) and \(\phi \) are related to \(x_0\) and \(v_0\) in Eq. (13.2.7). To find the relation, we expand the right side in this equation,

and compare the result to Eq. (13.2.7). This gives the following two equations.

Squarring and adding these will give us \(A^2 \) and dividing will give us \(\tan\phi\text{.}\) We take the notation that \(A \) be positive.

The velocity corresponding to the form (13.2.11) is

This says that the velocity oscillates between \(-\omega A \) and \(\omega A\text{.}\) Clearly, oscillator has turns in its motion, where velocity must be zero. Turning point at \(x=+A\) and \(x=-A\) correspond to \(v=0\text{.}\)

Subsection 13.2.3 Meaning of \(\omega \)

The meaning of \(\omega\) can be gleaned from the periodic behavior of the solution given in Eq. (13.2.7). We observe that

That is, the position is periodic in time with a period \(T = \dfrac{2\pi}{\omega}\text{.}\) Furthermore, velocity in Eq. (13.2.9) is also periodic with the same period. That is, the motion at \(t + \dfrac{2\pi}{\omega}\) is identical to the motion at \(t\text{.}\)

Therefore, we can state \(\omega \) in terms of the period of the oscillator.

Now, in one period \(T\text{,}\) an oscillator will complete one cycle. Therefore, \(1/T\) would be the number of cylces in unit time. This is called frequency of the oscillator and denoted by \(f\text{,}\) and sometimes by the Greek letter “nu”, \(\nu \text{.}\)

Now, writing \(\omega\) in terms of the frequency \(f\text{,}\) we have

where \(2\pi \) is an angle in radian. Therefore, \(\omega \) is frequency such that for each cycle we count \(2\pi\) radian instead of one. This is what you would expect of angle variable for something moving in a circle. That is why we call \(\omega\) the angular frequency. Note that the block attached to a spring is not moving in circle - its just an interpretation of \(\omega\) as it relates to the regular frequency, which is the number of cycles of oscillations per unit time.

The unit of period \(T \) is seconds. Therefore, the unit of regular frequency, \(f \) is per second or cycles per second, which is given its own name, Hertz (Hz). The unit for \(\omega\) is radians per second.

Subsection 13.2.4 Amplitude and Phase

The solution in the form (13.2.11) tells us that \(x \) will oscillate between \(+ A \) and \(-A \) since cosine will oscillate between \(+1 \) and \(-1\) as illustrated in Figure 13.2.7. That is, \(A \) is the amplitude of the oscillation.

In the solution in the form Eq. (13.2.11), the argument of the cosine, viz., \((\omega t + \phi) \) is called phase of the oscillator, and \(\phi \) is called the phase constant. We usually think of phase as an angle in a circle as illustrated in Figure 13.2.8.

But what about \(\phi\text{?}\) How to think about it? It has the information about where in the cycle, the oscillator is. Actually, \(\phi\) is best understood by comparing two identical oscillators, oscillator 1 with \(\phi = 0\) and oscillator 2 with \(\phi = -\pi/2\text{,}\) as presented in Figure 13.2.9.

From the pictures of sine and cosine, you can see that cosine at \(\theta=0\) is at the same place in cycle as sine would be at \(\theta=\pi/2\text{.}\) You might say sine is behind by \(\pi/2\) radian. that is oscillator 2 with \(\phi=-\pi/2\) is \(\pi/2\) radian behind oscillator 1.