9. Oscillations and Waves

— P R O B L E M S —

Any motion that repeats itself in regular time intervals is called an oscillation, periodic motion or harmonic motion.

If there is no friction, the time to complete one oscillation is constant and it is called the period ($T$) of the oscillation. Frequency ($f$) is the opposite of period and defined as the number of oscillations completed per unit time. Unit of period is s and unit of frequency is $s^{-1}$ or Hz (Hertz).

$$ f = \frac{1}{T}.$$

One of the most common oscillations is simple harmonic motion (SHM). A system is undergoing simple harmonic motion only if its acceleration ($a$) is proportional to the negative of the displacement ($x$). Because the next external force $F$ is proportional to acceleration according to Newton, the condition for SHM can be written as

$$ F \propto a \propto -x.$$

A block attached to a spring undergoing SHM in lieu with the compression and expansion of the spring is a good example of SHM because, according to Hooke’s law, $F=-kx$ where $F$ is the restoring force and $k$ the spring constant.

When $x$ is positive, $F$ is negative. When you pull the block to the right, the spring exerts a restoring force toward the right and vice versa. The maximum displacement in the positive or negative direction is called the amplitude ($A$) of the SHM.

The period and frequency of a simple harmonic oscillator (SHO, a system undergoing SHM) does not depend on the amplitude. It does not matter how far you pull the block from its equilibrium position ($x=0$), once you let it go it will go through a SHM with the same $T$ and $f$. Period is completely determined by the properties of the block and the spring.

We can use the example of this spring-block system to derive a general equation for SHM. The differential equation of SHM can be derived using Hooke’s law and Newton’s second law.

\begin{equation} \label{shmd} ma = F \Rightarrow m\frac{d^2x}{dt^2} = -kx \Rightarrow \frac{d^2x}{dt^2} = -\frac{k}{m}x \end{equation}

where $m$ is the mass of the block and $t$ is time. A solution of this differential equation (describing $x$ as a function of $t$) can be intuitively derived using the following diagram.

Seven snapshots of the spring-block system are shown above from $t=0$ to $t=3T/2$ where $T$ is the period. The next external force on the block is shown for each step. It is clear that the motion of the block from $+A$ to $-A$ and back can be described by a cosine function of the form

$$ x(t) = A \cos \left(\frac{2\pi}{T} t\right) $$

which is drawn in the figure above. A cosine function repeats every multiple of $2\pi$ and the motion of the block repeats every period T. Therefore, the function $\cos(2\pi t/T)$ will repeat every integer multiple of the period because $2\pi 1T/T=2\pi$ and so on.

But we know angular frequency (similar to angular velocity) $\omega=d\theta/dt=2\pi/T$ because the period is constant. So the equation becomes

\begin{equation} \label{shm} x(t) = A \cos \left(\omega t\right) \end{equation}

which is drawn as a x-t curve in the following figure.

However, this is valid only if $x=A$ at $t=0$ which might not be the case. In order to make the equation more general, we introduce a phase angle $\phi$ and write

$$ x(t) = A \cos \left(\omega t + \phi \right).$$

Adding a phase angle $+\phi$ is equivalent to shifting the curve to the left by an angle $\phi$ as shown below.

Figure b shows a shift to the left. A negative phase angle corresponds to a shift to the right.

Now, if Equation \eqref{shm} is the equation of motion (EOM) of a harmonic oscillator, we can find its velocity and acceleration via differentiation:

\begin{align} v(t) = \frac{dx}{dt} = -\omega A \sin (\omega t), \label{v} \\ a(t) = \frac{dv}{dt} = -\omega^2 A \cos (\omega t) = -\omega^2 \label{a} x. \end{align}

Velocity is a negative sine function and acceleration a negative cosine function as shown below.

Note that the figure is shown for a displacement along the $y$ axis. We see that when position is maximum ($A$), velocity is zero and acceleration is maximum in the negative direction. Maximum velocity $v_{max}=\omega A$ and maximum acceleration $a_{max}=\omega^2 A$.

Equation \eqref{shmd} and \eqref{a} can be used to derive an expression for the angular frequency as

$$ \frac{d^2x}{dt^2} = a(t) = -\omega^2 x = -\frac{k}{m}x \Rightarrow \omega=\sqrt{\frac{k}{m}}.$$

Because $\omega=2\pi/T=2\pi f$, we can also write

$$ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}, \ T = 2\pi \sqrt{\frac{m}{k}}. $$

We use the spring-block system to derive an expression for total energy of an oscillator in general. Total energy $E$ of s spring-block system is the summation of kinetic energy $E_k=mv^2/2$ and potential energy $E_p=kx^2/2$.

$$ E = E_k + E_p = \frac{1}{2} mv^2 + \frac{1}{2} kx^2 = \frac{1}{2}m A^2\omega^2 \sin^2(\omega t) + \frac{1}{2}k A^2 \cos^2(\omega t) $$

where we have replaced $v$ using Eq. \eqref{v} and $x$ using Eq. \eqref{shm}. Remembering $\omega=\sqrt{k/m}$ we can easily find that total energy

\begin{equation} \label{E} E = \frac{1}{2} kA^2 \end{equation}

is a constant that depends only on the spring constant and the amplitude ($E\propto A^2$).

Kinetic and potential energies change with the changing velocity and position, respectively, but the total energy always remains the same as shown the figure above. A relation between the speed of the block and its position can also be found as shown below.

$$ \frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \Rightarrow v = \sqrt{\frac{k}{m}(A^2-x^2)} = \omega\sqrt{A^2-x^2}. $$

The mysterious relation between SHM and uniform circular motion (UCM) reminds us of Plato’s allegory of the cave. The relation can be established by the following experiment.

A peg is rotating at constant angular velocity (or frequency) with respect to the center of an elevated vertical disk. A lights are shone from above, we see a shadow of the peg on wall below. As the peg goes through a two-dimensional UCM on the disk, its shadow goes through a one-dimensional SHM on the wall.

If $\mathbf{r}(t)$ is the radius or position vector of the peg as a function of time, $x$-component of $\mathbf{r}$ will be the position of the shadow. If $\theta(t)$ is the angle of $\mathbf{r}(t)$ with $x$-axis and $|\mathbf{r}|=A$, remembering that $\omega=\theta/t$, we can easily recreate the aforementioned equation of SHM:

$$ x(t) = A \cos(\omega t). $$

The variation of $x$ with changing $\theta=\omega t$ is shown in the figure above.

We will consider a simple pendulum defined as a pendulum with a point mass known as the pendulum bob. The bob is suspended from a string of length $L$. Mass $m$ of the bob and the mass of the string are negligible. The bob is subjected to earth’s gravity and the string’s tension as shown below.

We can derive an equation of motion of such a pendulum in analogy with Eq. \eqref{shmd}. If the pendulum has moment of inertia $I$ and the bob has the angular acceleration $\alpha$ because of the torque $\tau$, then

$$ I\alpha = \tau \Rightarrow mL^2 \frac{d^2\theta}{dt^2} = -L mg\sin\theta \Rightarrow \frac{d^2\theta}{dt^2} = -\frac{g}{L} \sin\theta $$

where we have used the relation $\tau = -Lmg\sin\theta$ and $I=mL^2$. Here, $g$ is the gravitational acceleration (or field) near the surface of earth and $\theta$ the angle of the string with the vertical axis. If $\theta< 15^\circ$, we can use the small angle approximation to write $\sin\theta \approx \theta$ and the equation becomes

\begin{equation} \label{shmr} \frac{d^2\theta}{dt^2} = -\frac{g}{L} \theta \end{equation}

which looks very similar to the last part of Eq. \eqref{shmd} if you remember the analogy between linear (translational) and angular (rotational) variables. Using the analogy between equations \eqref{shmd} and \eqref{shmr}, we find that the angular frequency of the pendulum

$$ \omega = \sqrt{\frac{g}{L}}, \ f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}, \ T = 2\pi\sqrt{\frac{L}{g}}. $$

In case of a spring-block system, frequency depends on the properties of both the spring ($k$) and the block ($m$). Similarly in case of a earth-pendulum system, frequency depends on the properties of both the earth ($g$) and the pendulum ($L$).

Oscillations of particles in a medium or field can generate a wave in that medium or field. Particles do not travel, only oscillate with respect to their equilibrium positions and, as a result, a wave travels along the medium or field carrying energy from one place to another. Waves traveling on a medium are called mechanical waves, for example, sound waves in air and water waves on water. Waves traveling on a field are called field waves, for example, electromagnetic waves. And the waves giving rise to the fundamental particles in nature (electrons, protons, neutrons, etc.) are called matter waves.

Consider the duck in this figure to be like a particle on the surface of water. As the wave travels in the $+x$ direction, the duck only oscillates up and down in a SHM. The length of a wave, the distance between two consecutive crests or troughs is called the wavelength $\lambda$. The wave travels a distance $\lambda$ in time $T$, the period of the wave, the time for one oscillation. Therefore, the wave speed or velocity

$$ v = \frac{\lambda}{T} = f\lambda. $$

Waves can be transverse (অনুপ্রস্থ) or longitudinal (অনুদৈর্ঘ্য). If you shake the left-end of a spring vertically like in figure (a), a wave will propagate along the spring horizontally toward the right. The wave is transverse because the disturbance of the medium (the segments of the spring) is perpendicular to the propagation of the wave. If you shake the left-end abruptly forward and backward like in figure (b), the direction of the wave and the disturbance will both be in the horizontal direction. This is a longitudinal wave. The following figure shows seven snapshots (a–g) of this longitudinal wave at seven different instances.

As the wave propagates along $+x$ direction, the blue dot oscillates horizontally like a spring. The period and frequency of the SHM and the wave are exactly the same. The wavelength for longitudinal waves is calculated from the distance between two consecutive compressions or expansions of the medium.

A wave can be described using a function $y$ that describes the vertical position of a particle at any point along the $x$-axis as a function of time, i. e. using $y(x,t)$.

In this figure $y(x=2)$ represents the particle of the medium at $x=2$ cm from the origin and $y(x=2,t=0)$ gives the initial position of the particle at $t=0$ shown by the intersection of the dashed curve with the $x$-axis. So $y(x=2,t=3)$ represents the position of the same particle after 3 s given by the intersection of solid curve with the horizontal axis. In 3 s the wave has traveled 6 cm (shown by the red arrowed line) to the right and the particle at $x=2$ cm has traveled 6 cm downward.

We have to compare this wave with a sine function in order to construct $y(x,t)$. A wave completes one wavelength in time $T$ and a sine function completes $2\pi$ angle in time $T$. Hence, the ratio of the angular and linear positions

$$ \frac{\theta}{x} = \frac{2\pi}{\lambda} \Rightarrow \theta = \frac{2\pi}{\lambda}x. $$

The dashed curve (at $t=0$) in the figure above can be modeled using the sine function

$$ y(x,t=0) = A\sin\theta_0 = A \sin\left(\frac{2\pi}{\lambda}x_0\right). $$

In order to generalize this equation for any time, we can use the relation $x=x_0+vt\Rightarrow x_0=x-vt$ and write

$$ y(x,t) = A \sin\left[\frac{2\pi}{\lambda}(x-vt)\right] = A \sin\left(\frac{2\pi}{\lambda}x - \frac{2\pi}{\lambda}vt\right). $$

Now $2\pi/\lambda\equiv k$, the wave number and $(2\pi t/\lambda)v = (2\pi t/\lambda)(\lambda/T) = (2\pi/T) t = \omega t $ and, hence, the final form of the wave function

\begin{equation} y(x,t) = A \sin(kx-\omega t) \end{equation}

which describes a wave traveling in the $+x$ direction. For a wave traveling in the opposite direction the function is $y(x,t) = A \sin(kx+\omega t)$. Therefore a more general function including the phase angle $\phi$ is

$$ y(x,t) = A \sin(kx\mp \omega t + \phi).$$

The wave number $k$ and angular frequency $\omega$ are related through the wave speed $v$:

$$ v = \frac{\lambda}{T} = \frac{2\pi}{T} \frac{\lambda}{2\pi} = \frac{\omega}{k}. $$

Note that $v$ is the speed of the wave and not the speed of the particles or disturbance of the medium. For a transverse wave, $v$ is typically along $x$-axis and the velocity of the particles ($v_y$) along the $y$-axis. The velocities $v_y$ can be calculated by differentiating $y$ with respect to time.

Equation \eqref{shmd} described a SHM and Eq. \eqref{shm} was a solution of that equation. Similarly, we can derive the wave equation by partially differentiating the wave function with respect to both time $t$ and space $x$. The partial derivatives will give rise to velocity $v_y$ and acceleration $a_y$ of the particles, and slope $\mathcal{S}$ and curvature $\mathcal{C}$ of the wave.

$$ v_y = \frac{\partial}{\partial t} y(x,t) = - \omega A \cos(kx-\omega t). $$

$$ a_y = \frac{\partial}{\partial t} v_y = \frac{\partial^2y}{\partial t^2} = - \omega^2 A \sin(kx-\omega t). $$

$$ \mathcal{S} = \frac{\partial}{\partial x} y(x,t) = kA \cos(kx-\omega t). $$

$$ \mathcal{C} = \frac{\partial}{\partial x} \mathcal{S} = \frac{\partial^2y}{\partial x^2} = -k^2A \sin(kx-\omega t). $$

Dividing the acceleration by the curvature ($a_y/\mathcal{C}$) and remembering that $v=\omega/k$ we find that

\begin{equation} \frac{\partial^2y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2y}{\partial t^2} \end{equation}

which is the infamous linear wave equation. It describes the relation between the curvature of wave (literally) and the acceleration of the particles in the medium.

The total energy carried by a wave can be calculated using the example of a wave on a stretched string. Consider a string with linear density $\mu=dm/dx$.

Kinetic energy of a small mass element $dE_k = dm v_y^2/2 = \mu dx v_y^2 / 2$ where $v_y$ is the velocity of the particles. So total kinetic energy for one wavelength in the string

$$ E_k = \int_0^{E_{k,\lambda}} dE_k = \frac{1}{2} \mu \omega^2 A^2 \int_0^\lambda \cos^2(kx-\omega t) dx = \frac{1}{4} \mu\omega^2 A^2 \lambda. $$

Potential energy of a small mass element $dE_p = kx^2/2$ where $k=\omega^2 dm=\omega^2 \mu dx$ is the spring constant (not wave number). So $dE_p = \mu\omega^2 x^2 dx/2$ and, hence, the total potential energy over one wavelength

$$ E_p = \int_0^{E_{p,\lambda}} dE_p = \frac{1}{2} \mu \omega^2 A^2 \int_0^\lambda \sin^2(kx-\omega t) dx = \frac{1}{4} \mu\omega^2 A^2 \lambda. $$

So total energy carried by the wave

\begin{equation} \label{Ew} E_w = E_k + E_p = \frac{1}{2} \mu\omega^2 A^2 \lambda. \end{equation}

The energy is exactly proportional to $A^2$ similar to Eq. \eqref{E}. Power $P$ can be calculated by dividing the energy within one wavelength by the period, so

$$ P = \frac{E_w}{T} = \frac{1}{2} \mu\omega^2 A^2 \frac{\lambda}{T} = \frac{1}{2} \mu\omega^2 A^2 v. $$

---

Image credits: Samuel J. Ling, Jeff Sanny, and Bill Moebs; University Physics, OpenStax.