Simple Harmonic Motion

Here we discuss a simple kind of oscillatory motion also called periodic motion. One familiar example is a simple pendulum bob which undergoes periodic motion if displaced from its equilibrium position. We will discuss a motion called simple harmonic motion abbreviated as SHM and verify that the oscillation of a pendulum is approximately simple harmonic under certain condition. Another familiar example of the periodic motion is the oscillation of a body of a particular mass attached to one end of a spring whose other end is fixed about its equilibrium position.

Simple Harmonic Motion (SHM)

A frictionless spring-mass system is shown in Figure 1 in which one end of a massless spring is fixed while the other end is attached to a box of mass $m$ and the box can oscillate about the equilibrium position when displaced.

Figure 1 A box oscillates about its mean position O after being displaced.
Figure 2 When the particle in the circle moves in uniform circular motion its projection on the diameter along x-axis oscillates back and forth.

The massless spring is an ideal spring; you'll never get a real massless spring. We neglect any non-conservative or dissipative forces which may act during the oscillation such as friction. When you displace the box from its equilibrium position, a restoring force is exerted on the box by the spring which tends to keep the box in its equilibrium position. Let $x$ be the displacement of the box from its equilibrium position, then the restoring force $F$ exerted by the spring on the box is

\[F = - kx \tag{1} \label{1}\]

where $k$ is the proportionality constant called force constant. The negative sign indicates that the force exerted by the spring on the box is in opposite direction of the displacement. The Eq. \eqref{1} shows that the restoring force (the force exerted by the spring) is directly proportional to the displacement from the equilibrium position.

The oscillation in which the restoring force is directly proportional to the displacement from the equilibrium position is called simple harmonic motion (SHM). The motion of the box is horizontal and we have chosen our x-coordinate along this motion. And $a_x$ is the acceleration of the box at any instant during its motion. We can express Eq. \eqref{1} as,

\[{a_x} = \frac{{{d^2}x}}{{d{t^2}}} = - \frac{k}{m}x \tag{2} \label{2}\]

It means the acceleration $a_x$ of the box is also directly proportional to the displacement from the equilibrium position. In Figure 1 the absolute value of the maximum displacement of the box from the equilibrium position is called amplitude $A$ of the oscillation which is always positive.

Consider that a particle is moving in uniform circular motion. Now we check and see whether the simple harmonic motion is the same as the motion of the projection of the particle on a diameter of the circle or not. In Figure 2 a particle at point $p$ moves in a circle of radius $R$ whose one diameter is along the x-axis of our coordinate system.

The arrow which joins the origin and the particle on the circle pointing the particle is called phasor. The phasor shows that the angular displacement of the particle at the point $p$ is $\theta$. If you draw a line $pp'$ perpendicular to the diameter of the circle, you'll find that the point ${p'}$ is the projection of the point $p$ on the diameter. As the particle moves in uniform circular motion its projection also moves (oscillates) along the diameter.

For our convenience when the point $p$ moves in uniform circular motion, its projection ${p'}$ moves along the diameter of the circle (the particle moves in the circle but we can say the point moves for our convenience). The maximum displacement of $p'$ from the origin (equilibrium position) is equal to the radius of the circle and therefore the amplitude of oscillation of the point $p'$ is $A = R$. At initial time $t = 0$ the angular displacement of the particle at the point $p$ is $\theta$ so the initial position of its projection at the point ${p'}$ is $x_i = R\cos \theta = A\cos \theta $.

The particle in the circle moves with angular velocity $\omega$ and at any later time $t$ the angular displacement of the particle is $\theta + \omega t$. Note that the initial angular displacement $\theta$ at time $t = 0$ is called the phase angle of the particle. So, the position of $p'$ on the diameter at time $t$ is,

\[x = A\cos (\theta + \omega t) \tag{3} \label{3}\]

The Eq. \eqref{3} gives the position of the point $p'$ at any instant of time $t$. Notice that the value $\theta + \omega t$ is an angle and $\theta $ is the phase angle. The velocity $v_x$ and the acceleration $a_x$ of the point $p'$ (which is the projection of the particle at the point $p$) at time $t$ are

\[{v_x} = \frac{{dx}}{{dt}} = - A\omega \sin (\theta + \omega t) \tag{4} \label{4}\]

\[{a_x} = \frac{{{d^2}x}}{{d{t^2}}} - A{\omega ^2}\cos (\theta + \omega t) \tag{5} \label{5}\]

You know from Eq. \eqref{3} that $x = A\cos (\theta + \omega t)$ and so we can write Eq. \eqref{5} as,

\[{a_x} = - {\omega ^2}x \tag{6} \label{6}\]

The Eq. \eqref{6} shows that the acceleration of the point $p'$ is directly proportional to the displacement from its equilibrium position. Hence, the motion of the projection of the particle moving in uniform circular motion on the diameter of the circle is a simple harmonic motion. Therefore, you can compare Eq. \eqref{2} and Eq. \eqref{6} and obtain

\[\omega = \sqrt {\frac{k}{m}} \tag{7} \label{7}\]

The total time taken to complete one rotation or oscillation is called time period of the rotation or oscillation denoted by $T$ and it i always a positive quantity. In one complete rotation the particle in the circle undergoes an angular displacement of $2\pi$ and if $T$ is the time period, the angular velocity is $\omega = \frac{{2\pi }}{T}$.

The frequency of rotation or oscillation is the number of rotations or cycles or oscillations completed per unit time. In one complete rotation of a particle in a circle, the angular displacement of the particle in time period $T$ is $2\pi$. So the angular frequency of the rotation is $\frac{{2\pi }}{T}$ which is the same as the angular velocity $\omega$. Therefore, the angular frequency of oscillation of the point $p'$ is the same as the angular velocity of the point $p$ and we denote both with the same letter $\omega$. Substituting the value of $\omega = \frac{{2\pi }}{T}$ in \eqref{7} and we get

\[T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{m}{k}} \tag{8} \label{8}\]

Since, one rotation or one oscillation occurs in time $T$, the frequency of oscillation is,

\[f = \frac{1}{T} = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} \tag{9} \label{9}\]

The frequency is always positive and its SI unit is hertz denoted by Hz that is, $1{\rm{Hz}} = 1{\rm{cycle/s}}$. One interesting thing in Eq. \eqref{8} is that the time period $T$ is independent of the amplitude $A$. A body which undergoes simple harmonic motion is called harmonic oscillator. For a given harmonic oscillator, the time period of oscillation is independent of the amplitude of the oscillation.

Again consider the spring-mass system as in Figure 1 where a box oscillates about its equilibrium position. If no dissipative forces act on the system, the total mechanical energy is conserved. The potential energy of the spring force at elongation $x$ from the equilibrium position is $\frac{1}{2}k{x^2}$. And the kinetic energy at that instant is $\frac{1}{2}mv_x^2$. So the total mechanical energy of the system is

\[E = \frac{1}{2}k{x^2} + \frac{1}{2}mv_x^2 \tag{10} \label{10}\]

Now substituting the values of $x$ from Eq. \eqref{3} and $v_x$ from Eq. \eqref{4} in Eq. \eqref{10}, and solving you'll get:

\[\begin{align*} E &= \frac{1}{2}k{[A\cos (\theta + \omega t)]^2} + \frac{1}{2}m{[ - A\omega \sin (\theta + \omega t)]^2}\\ &= \frac{1}{2}k{A^2} \tag{11} \label{11} \end{align*}\]

Vertical Simple Harmonic Motion

The harmonic oscillator in Figure 1 oscillates horizontally along x-axis of the coordinate system shown. Now we discuss about what happens when the box is hanged by the spring vertically. Again we neglect any other non-conservative forces that may act during the oscillation and suppose the mass of the spring is zero. You should know that the zero mass of the spring is an idealized concept. But we can neglect the mass of the spring if the spring force is large enough in comparison to the mass of the spring. In Figure 3 the box is attached to the free end of the vertically hanged spring.

Figure 3 A spring is hanged vertically downwards with a box of mass m which oscillates up and down about its mean position when displaced.

The weight of the box $\vec w = m\vec g $ stretches the spring by a distance of $l$. In Figure 3 we have chosen the positive y-axis to be vertically downwards in our coordinate system. The weight of the box is exactly balanced by the spring force at elongation $l$ that is, $w = mg = kl$. And, therefore, the equilibrium position of the box is the position at which the weight of the box is balanced by the spring force. At an instant, suppose the box is at a distance $y$ below the equilibrium position and the total upward force exerted by the spring on the box is $ - k(l + y)$. Note that the negative y-axis is in upward direction and the upward force is negative. If $a_y$ is the vertical acceleration of the box, the total force acting on the box below the equilibrium position is:

\[\begin{align*} - k(l + y) + mg &= m{a_y}\\ {\rm{or,}}\quad m{a_y} &= - ky \tag{12} \label{12} \end{align*}\]

It means the force exerted by the spring on the harmonic oscillator is $F = m{a_y} = - ky$. If you make the positive x-axis vertically downwards and replace $y$ by $x$ in equation Eq. \eqref{12}, you'll get $F = - kx$ which is the same as Eq. \eqref{1}. So the box oscillating vertically in Figure 3 is a simple harmonic motion.

Angular Simple Harmonic Motion

We have a disk suspended at the centre by a string as shown in Figure 4. When the disk is slightly rotated by applying a tangential force at the rim, the disk oscillates about its equilibrium position. The rotating effect is caused by the torque $\tau = I\alpha $ where $I$ is the moment of inertia of the disk about the rotation axis.

Figure 4 The restoring torque in the string tends to keep the disk in its equilibrium position which in turn results angular simple harmonic motion.

When the disk is displaced from its equilibrium position, a restoring torque is developed and it tends to restore the disk into the equilibrium position. And it is found that the restoring torque $\tau$ is directly proportional to the angular displacement $\theta$. Hence, we can write

\[\tau = - \kappa \theta \tag{13} \label{13}\]

Where $\kappa$ is called the torsion constant. The negative sign indicates that the restoring torque acts in opposite direction of the angular displacement. The Eq. \eqref{13} shows the angular form of simple harmonic motion called angular simple harmonic motion. Again we have;

\[\begin{align*} I\alpha &= - \kappa \theta \\ {\rm{or,}}{\kern 1pt} \quad \alpha &= \frac{{{d^2}\theta }}{{d{t^2}}} = - \frac{k}{I}\theta \quad \tag{14} \label{14} \end{align*}\]

Comparing Eq. \eqref{14} with Eq. \eqref{2}, the role of $k/m$ in Eq. \eqref{2} is played by $\kappa/I$ in Eq. \eqref{14}. The moment of inertia $I$ is the rotational analogue of mass $m$. And, therefore, you can immediately find the equation of angular frequency $\omega$ by replacing $k/m$ in Eq. \eqref{7} by $\kappa/I$:

\[\omega = \sqrt {\frac{\kappa }{I}} \tag{15} \label{15}\]

Now you can easily find the time period and frequency of angular simple harmonic motion after knowing the angular frequency in Eq. \eqref{15}.

Mechanics

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