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# 8.3.2 SHM Energy in time-domain

If we go with $\displaystyle v=\omega {{x}_{0}}\cos \omega t$, then we can write $\displaystyle KE=\frac{1}{2}m{{v}^{2}}$ as $\displaystyle KE=\frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}{{\cos }^{2}}\omega t$

Two energy cycles per oscillation

Notice that there are two complete KE cycles in one velocity cycle. Physically, this is because KE is a scalar whereas velocity is a vector. So the KE is the same regardless whether v is in the positive or negative direction. Mathematically, this is because $\displaystyle {{\cos }^{2}}\omega t=\frac{{1+\cos 2\omega t}}{2}$, which is a raised sinusoid with frequency 2f.

The PE can be derived from \displaystyle \begin{aligned}PE&=TE-KE\\&=\frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}-\frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}{{\cos }^{2}}\omega t\\&=\frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}(1-{{\cos }^{2}}\omega t)\\&=\frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}{{\sin }^{2}}\omega t\end{aligned}

Similarly, there are two PE cycles in one oscillation cycle because PE is maximum at both the extreme positions, and zero whenever the oscillation crosses the equilibrium position.

For completeness’s sake, we present TE, PE and KE in the same graph below.

Exam Tip

During examinations, students lose marks when they sketch the PE and KE graphs to have the shape of the humpy McDonald’s sign. Do realize that the PE and KE graphs are raised sinusoidal functions. They should thus have the shape of smooth sinusoids.

Video Explanation

Energy-Time Graphs of SHM

Concept Test

1435

# 8.3.1 Derivation of TE Formula

To produce an oscillation with amplitude x0, the system must first be displaced from its equilibrium position to $\displaystyle x={{x}_{0}}$ . This requires an external force to pull or push against the restoring force. The work done by this external force corresponds to the (total) energy of oscillation, TE.

Recall that $\displaystyle a=-{{\omega }^{2}}x$  and the restoring force is $\displaystyle \displaystyle {{F}_{{net}}}=-m{{\omega }^{2}}x$ . Since the external force must match the restoring force at every position, $\displaystyle \displaystyle {{F}_{{ext}}}=m{{\omega }^{2}}x$ at every position.

Since the area under the Fx graph is work done, we can derive the formula for TE to be $\displaystyle TE=\frac{1}{2}({{x}_{0}})(m{{\omega }^{2}}{{x}_{0}})=\frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}$

If the restoring force is the only force acting on the system (i.e. no other dissipative forces), then the system will oscillate forever (after it has been set in motion). During the oscillation, the restoring force causes energy to be converted between the kinetic energy KE and potential energy PE of the system. From the principle of conservation of energy, we know that, $\displaystyle TE=KE+PE$

Maximum PE occurs at the extreme positions, where KE is zero.

Conversely, maximum KE occurs at the equilibrium position, where PE is zero. So

So $\displaystyle \frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}$ corresponds to not just TE, but also PEmax and KEmax. $\displaystyle TE=KE+PE=K{{E}_{{\max }}}=P{{E}_{{\max }}}=\frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}$

# 8.2.2 Natural Frequency

One of the amazing things about a SHM is that its period is independent of the amplitude of oscillation. Instead, it is dependent on the system’s restoring force and inertia.

Take for example the vertical spring-mass system.

If we denote the extension of the spring at the equilibrium position by e, then we can encapsulate the dynamics in one N2L equation \displaystyle \begin{aligned}({{F}_{{net}}}&=ma)\\mg-(ke+kx)&=ma\end{aligned}

But $\displaystyle mg=ke$, so \displaystyle \begin{aligned}-kx&=ma\\a&=-\frac{k}{m}x\end{aligned}

Comparing this with the SHM equation $\displaystyle a=-{{\omega }^{2}}x$, we can deduce that $\displaystyle {{\omega }^{2}}=\frac{k}{m}$

Hence, the natural frequency is $\displaystyle {{f}_{n}}=\frac{1}{{2\pi }}\sqrt{{\frac{k}{m}}}$  and the natural period is $\displaystyle {{T}_{n}}=2\pi \sqrt{{\frac{m}{k}}}$ .

One way to interpret the $\displaystyle {{\omega }^{2}}=\frac{k}{m}$ relationship is that a higher k value results in a stronger restoring force (per unit displacement), and a smaller m value results in a larger acceleration (per unit displacement), and hence fn. Some people call it the “stiffness-to-inertia ratio” ratio.

Demonstrations

Spring-Mass and Pendulum

Video Explanation

Derivation of Natural Frequency Formula for Spring-Mass System

Interesting

Lissajous Figures in the Sand

Lissajous Figures on CRO

# 8.2.1 Restoring Force

if we multiply the mass of the oscillating body m to both sides of the a-x equation, we get $\displaystyle \displaystyle ma=-m{{\omega }^{2}}x$

But $\displaystyle {{F}_{{net}}}=ma$, so $\displaystyle \displaystyle {{F}_{{net}}}=-m{{\omega }^{2}}x$

This tells us that a SHM experiences a net force that is (1) proportional to its displacement and (2) opposite in direction to its displacement. Since this net force is always directed towards the equilibrium position, it is called the restoring force.

Take for example the spring-mass system.

(a) $\displaystyle x=0$

At the equilibrium position, the downward weight mg is balanced by the upward tension force T of the spring. The restoring force is zero at the equilibrium position.

(b) $\displaystyle x>0$(We are using the “downward is positive” sign convention)

When the mass is below the equilibrium position, so the spring is further extended by x (compared to when it is at the equilibrium position). This results in the upward tension increasing by kx. But the downward weight is fixed as mg. So the net force is simply kx and upward.

(c) $\displaystyle x<0$(We are using the “downward is positive” sign convention)

When the mass is below the equilibrium position, so the spring is shorter by x (compared to when it is at the equilibrium position). This results in the upward tension decreasing by kx. But the downward weight is fixed as mg. So the net force is simply kx and downward.

In summary, the magnitude of the net force is kx, and its direction is always towards the equilibrium position. So the spring-mass system does indeed have a restoring force whose magnitude is proportional to the displacement, directed towards the equilibrium position. This is why when the mass is displaced from the equilibrium position and released, it goes into simple harmonic motion.

Concept Test

1422

# 8.1.3 v-x and a-x Equations

In this section, we will obtain the ax and vx equations which we can use to calculate the acceleration and velocity of a SHM at any displacement.

a-x equation

We begin by writing down the three time-domain equations. \displaystyle \begin{aligned}x&={{x}_{0}}\sin \omega t\\v&=\omega {{x}_{0}}\cos \omega t\\a&=-{{\omega }^{2}}{{x}_{0}}\sin \omega t\end{aligned}

Replacing $\displaystyle {{x}_{0}}\sin \omega t$ in the third equation by x, we obtain $\displaystyle a=-{{\omega }^{2}}x$

This is a very important result because it shows that the acceleration of a SHM is directly proportional in magnitude but opposite in sign to the displacement.

The a-x graph is thus a straight line passing through the origin, with the negative gradient corresponding to $\displaystyle -{{\omega }^{2}}$.

v-x equation

To obtain the v-x equation, we rewrite the x-t and v-t equations as $\displaystyle \sin \omega t=\frac{x}{{{{x}_{0}}}}$ $\displaystyle \cos \omega t=\frac{v}{{\omega {{x}_{0}}}}$

Using the trigonometric identity $\displaystyle {{\sin }^{2}}x+{{\cos }^{2}}x=1$, we can get rid of the t terms, \displaystyle \begin{aligned}{{\sin }^{2}}\omega t+{{\cos }^{2}}\omega t&=1\\{{(\frac{x}{{{{x}_{0}}}})}^{2}}+{{(\frac{v}{{\omega {{x}_{0}}}})}^{2}}&=1\\{{\omega }^{2}}{{x}^{2}}+{{v}^{2}}&={{\omega }^{2}}{{x}_{0}}^{2}\\v&=\pm \omega \sqrt{{{{x}_{0}}^{2}-{{x}^{2}}}}\end{aligned}

The v-x graph is an ellipse, with intercepts at $\displaystyle x=\pm {{x}_{0}}$  and $\displaystyle v=\pm \omega {{x}_{0}}$.

You should be able to deduce that the oscillation progresses in the clockwise direction along the elliptical graph. This is because for the top half of the graph, the velocity is positive, so the displacement should progress towards the right (to become more positive). Likewise, for the bottom half of the graph, the velocity is negative, so the displacement should progress towards the left (to become more negative).

Explanation Video

a-x and v-x Graph of SHM

Concept Test

1405

# 8.1.2 v-t and a-t Equations

Remember that $\displaystyle v=\frac{{dx}}{{dt}}$ and $\displaystyle a=\frac{{dv}}{{dt}}$?

If we begin with $\displaystyle x={{x}_{0}}\sin \omega t$

and differentiate both sides with respect to time, we obtain $\displaystyle \displaystyle v=\omega {{x}_{0}}\cos \omega t$

and differentiate both sides with respect to time, again we obtain $\displaystyle a=-{{\omega }^{2}}{{x}_{0}}\sin \omega t$

With just two steps of differentiation, we learn that

1. In a SHM, displacement, velocity and acceleration all vary sinusoidally with time.
2. Acceleration leads velocity by a quarter cycle, and velocity in turn leads displacement by a quarter cycle.
3. Maximum velocity $\displaystyle {{v}_{{\max }}}=\omega {{x}_{0}}$
4. Maximum acceleration $\displaystyle {{a}_{{\max }}}={{\omega }^{2}}{{x}_{0}}$.

Video Explanation

x-t, v-t and a-t Graphs of SHM

Concept Test

1401

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# 8.1.1 x-t Equation

A mass hung on a spring (often called a spring-mass system) is the simplest example of a simple harmonic motion (SHM). When displaced from the equilibrium position and released, the mass will oscillate periodically about the equilibrium position.

The variation with time t of the displacement x of the mass can be described by the equation $\displaystyle x={{x}_{0}}\sin \omega t$

where xo is the amplitude,

T is the period, and

ω is the angular frequency

Angular Frequency

The angular frequency ω is a rather abstract concept. If you are new to SHM, just think of it as the frequency multiplied by $\displaystyle 2\pi$ for the time being. The appendix contains a fuller discussion when you are ready. $\displaystyle \omega =2\pi f$

Since $\displaystyle f=\frac{1}{T}$, ω can also be written as $\displaystyle \omega =\frac{{2\pi }}{T}$

You should also realize that mathematically speaking, we need the $\displaystyle \omega =\frac{{2\pi }}{T}$ term in $\displaystyle x={{x}_{0}}\sin \omega t$ to “scale the graph horizontally” so that one sinusoidal cycle corresponds to one period T.

Must it be the “sine” function?

No, it does not have to be $\displaystyle \sin \omega t$. It is often written as $\displaystyle x={{x}_{0}}\cos \omega t$ too. It all depends on when we choose to start time t from. In fact, the general equation for SHM is $\displaystyle x={{x}_{0}}\sin (\omega t+\phi )$. When we say “sinusoidal”, we can mean any time-shifted version of $\displaystyle x={{x}_{0}}\sin \omega t$.

Why is SHM called SHM?

SHM is “simple harmonic” because its motion is described by a single sinusoidal function. A square wave, for example, requires the summation of an infinite number of sinusoidal functions $\displaystyle \text{square wave}=\sin (x)+\frac{1}{3}\sin (3x)+\frac{1}{5}\sin (5x)+...$.

The Half-Amplitude Position

SHM is not constant speed motion. It is slower when it is near the extreme positions and faster when it is near the equilibrium position.

It is very useful to “memorize” that $\displaystyle \cos 60{}^\circ =0.5$ or $\displaystyle \sin 30{}^\circ =0.5$. Because then you can “see” without any calculations that

The time taken to travel between $\displaystyle x={{x}_{0}}$ and $\displaystyle x=0.5{{x}_{0}}$ is exactly $\displaystyle \frac{{60{}^\circ }}{{360{}^\circ }}T=\frac{1}{6}T$

but

The time taken to travel between $\displaystyle x=0$ and $\displaystyle x=0.5{{x}_{0}}$  is exactly $\displaystyle \frac{{30{}^\circ }}{{360{}^\circ }}T=\frac{1}{{12}}T$

Animation

Displacement-Time Graph

Video Explanation

Solving Half-Amplitude Problems by Look-Look-See-See

Concept Test

1419

Beyond Syllabus

How is SHM Simple Harmonic?

# 9.4.3 The Three Polarizer Problem

Consider the set up illustrated below. Do you understand why there is no light after polarizer Y? That’s right. Since polarizers X and Y are aligned perpendicularly to each other, all the light is cut out.

Now. What If we insert a third polarizer Z between X and Y? Oh my god, there is now some light after polarizer Y (as long as Z’s polarization direction is neither vertical nor horizontal). Adding another layer of polarization cuts out LESS light than before! How is that possible?

To understand this surprising outcome, you must realize that a polarizer is more than just a filter. It does not just attenuate the light passing through. It also re-aligns the polarization direction of the light!

By choosing the polarization direction of the 3rd polarizer (Z) to be neither vertical nor horizontal, we avoid having two consecutive polarizers which are perpendicular to each other. This prevents the light from being totally cut off.

Let’s confirm this with some math. Since X and Z are misaligned by angle θ, the intensity of light after passing through Z is $\displaystyle {{\mathrm{I}}_{1}}={{\mathrm{I}}_{0}}{{\cos }^{2}}\theta$

Since Z and Y are misaligned by angle $\displaystyle 90^\circ -\theta$, the intensity of light after passing through Y is \displaystyle \begin{aligned}{{\mathrm{I}}_{2}}&={{\mathrm{I}}_{1}}{{\cos }^{2}}(90{}^\circ -\theta )\mathrm{I}\\&={{\mathrm{I}}_{0}}{{\cos }^{2}}\theta {{\cos }^{2}}(90{}^\circ -\theta )\\&={{\mathrm{I}}_{0}}{{\cos }^{2}}\theta {{\sin }^{2}}(\theta )\\&=\frac{{{{\mathrm{I}}_{0}}}}{4}{{\sin }^{2}}2\theta \end{aligned}

In fact, the maximum intensity of $\displaystyle \displaystyle \frac{{{{\mathrm{I}}_{0}}}}{4}$ is achieved by arranging Z to have polarization direction midway between that of X and Y, i.e. 45°, 135°, 225°, 315°, etc.

Demonstration

Teddy Comes Back

Three Polarizers

Video Explanation

The Third Polarizer

Concept Test

1831

# 9.4.2 Polarization

Imagine an attempt to transmit a rope wave through a vertical slit. If the rope is oscillating vertically, then the wave would travel unimpeded through the slit. On the other hand, if the rope is vibrating horizontally, then the wave would be totally absorbed by the slit. Welcome to the magical world of polarization.

It is useful to note that polarization is a phenomenon unique to transverse waves. It is not applicable to longitudinal waves (since they can only vibrate in one direction). In fact, the fact that light can be polarized is evidence that light is a transverse wave.

Passing Unpolarized Light through a Polarizer

Light is a transverse wave. If a light wave is propagating along the z-axis, then its E-field could be oscillating in the x-axis, or y-axis, or any direction in the xy plane. It turns out that light waves emitted by most light sources are unpolarized light, which is to say the light beam contains a mixture of waves with E-field oscillating in all possible directions in the xy plane. (Mathematically, we can model natural light as two arbitrary, incoherent, perpendicularly polarized waves of equal amplitude.)

There are materials such as Polaroids which, due to their chemical structure, absorb only electric fields oscillating in the direction of their polymer chains. Electric fields oscillating in the direction perpendicular to the polymer chains are not absorbed. The Polaroid is an example of a polarizer, and the direction of the polymer chains is called the polarization direction.

When a beam of unpolarized light passes through a polarizer, it becomes polarized. After polarization, the oscillation of electric field will only be in the polarization direction of the polarizer (because the electric field oscillating in the perpendicular direction has been absorbed).

Passing Polarized Light through a Polarizer

The fun part begins now. Suppose we pass a beam of polarized light with amplitude A0 and intensity I0 through a second polarizer.

If the 2nd polarizer’s polarization direction is parallel to the 1st polarizer’s, then we get 100% transmission. On the other hand, if the two polarizers’ polarization directions are perpendicular to each other, then we get 0% transmission.

Now, what happens if they are neither parallel nor perpendicular, but misaligned by some angle θ? Easy. We can always resolve the amplitude of the electric field A0 (of the polarized light) into the components parallel and perpendicular to the polarization direction (of the polarizer), A0cosθ and A0sinθ.

The parallel component will be completely passed through, while the perpendicular component will be completely absorbed. In other words, $\displaystyle \displaystyle A={{A}_{0}}\cos \theta$

Since $\displaystyle \displaystyle \mathrm{I}\propto {{A}^{2}}$, we arrive at Malu’s Law $\displaystyle \displaystyle \mathrm{I}={{\mathrm{I}}_{0}}{{\cos }^{2}}\theta$

Video Explanation

What is Polarization

Demonstration

Polarizers

Polarized Light Around Us

LCD Projectors

Newton or Einstein?

Polarization by Reflection

Concept Test

1833

Animation

Cabrillo Applet

# 9.4.1 The Electromagnetic Spectrum

Having gone through your O-Level education, you are probably comfortable with the knowledge that light is a wave. But light is not an ordinary wave. Has it occurred to you that there is something truly unique about light?

Firstly, light is not a mechanical wave. It does not have any oscillating particles. Instead, it has oscillating fields. Two fields in fact, an electric field and a magnetic field (hence the name electromagnetic waves). And the two fields oscillate at right angle to each other.

And since there is no oscillating particles, electromagnetic waves do not require any medium to propagate. The next time you marvel at the beauty of a sunrise, remember that the light managed to reach you, despite the vast vacuum between the Sun and the Earth.

Furthermore, electromagnetic waves propagate at an incredible speed. In vacuum the speed of light is $\displaystyle \displaystyle c=3.00\times {{10}^{8}}\text{ m }{{\text{s}}^{{\text{-1}}}}$. In glass, it is about one-third slower. But still incredibly fast.

Unlike mechanical waves, the intensity I of an EM wave is only dependent on the amplitude A of its oscillating electric or magnetic field ( $\displaystyle \displaystyle \mathrm{I}\propto {{A}^{2}}$ to be specific) and independent of its frequency f ( $\displaystyle \displaystyle \mathrm{I}\propto {{A}^{2}}{{f}^{2}}$ for mechanical waves).

The range of wavelengths for electromagnetic waves is truly mind boggling, from picometers (gamma rays) to kilometers (radio waves). In comparison, the human eye is only sensitive to light of wavelength between 400 nm (violet) to 700 nm (red).

The syllabus DOES require you to “memorize” the wavelengths of the different type of EM waves. You should find the following table useful.

Animation

E and M Oscillations

Song

The Electromagnetic Spectrum Song