Simple Harmonic Motion

Simple Harmonic Motion (Condition $F \propto -x$)

Simple Harmonic Motion (SHM) is the simplest type of oscillation and is a fundamental concept in physics. It is a special case of periodic motion where the restoring force acting on the oscillating object is directly proportional to its displacement from the equilibrium position and acts in the opposite direction to the displacement.

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Condition for Simple Harmonic Motion

The defining characteristic of SHM is the nature of the force acting on the oscillating body. For linear SHM, if the displacement from the equilibrium position is $x$, the net restoring force $\vec{F}$ acting on the object must be:

$ \vec{F} = -k\vec{x} $

where $\vec{x}$ is the displacement vector from the equilibrium position, $\vec{F}$ is the restoring force vector, and $k$ is a positive constant called the force constant or spring constant (analogous to the spring constant in Hooke's Law, as SHM is often exemplified by a mass on a spring). The negative sign indicates that the force is always directed towards the equilibrium position (opposite to the displacement).

In one dimension, taking the equilibrium position at $x=0$, the force is $F = -kx$. If $x$ is positive, $F$ is negative (towards the origin). If $x$ is negative, $F$ is positive (towards the origin). This type of force is called a linear restoring force.

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Differential Equation of SHM

Applying Newton's Second Law ($F_{net} = ma$) to an object of mass $m$ undergoing linear SHM:

$ F = m a $

$ -kx = m \frac{d^2x}{dt^2} $

Rearranging, we get the second-order linear differential equation for SHM:

$ \frac{d^2x}{dt^2} + \frac{k}{m} x = 0 $

This equation is the mathematical representation of simple harmonic motion. Any system whose motion is described by this equation undergoes SHM.

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Solution to the SHM Differential Equation

The general solution to this differential equation is a sinusoidal function:

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

where:

• $x(t)$ is the displacement from equilibrium as a function of time $t$.
• $A$ is the amplitude, the maximum displacement from equilibrium.
• $\omega$ is the angular frequency of the oscillation.
• $(\omega t + \phi)$ is the phase of the motion at time $t$.
• $\phi$ is the initial phase constant (or phase angle), determined by the initial conditions (position and velocity at $t=0$).

By taking the second derivative of $x(t)$ and substituting it into the differential equation, we find the relationship between $\omega$, $k$, and $m$:

$ \frac{dx}{dt} = -A\omega \sin(\omega t + \phi) $

$ \frac{d^2x}{dt^2} = -A\omega^2 \cos(\omega t + \phi) = -\omega^2 x(t) $

Substituting into $\frac{d^2x}{dt^2} + \frac{k}{m} x = 0$:

$ -\omega^2 x + \frac{k}{m} x = 0 $

$ \left(\frac{k}{m} - \omega^2\right) x = 0 $

This must hold for all values of $x$ (as $x$ varies during oscillation), so the term in the parenthesis must be zero:

$ \frac{k}{m} - \omega^2 = 0 \implies \omega^2 = \frac{k}{m} $

$ \omega = \sqrt{\frac{k}{m}} $

The angular frequency of SHM is determined by the force constant and the mass. From the angular frequency, we can find the frequency ($\nu$) and period ($T$) of the SHM:

$ \nu = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $

$ T = \frac{1}{\nu} = 2\pi \sqrt{\frac{m}{k}} $

These formulas show that the period and frequency of SHM depend only on the properties of the oscillating system (mass and force constant) and not on the amplitude of the oscillation (as long as the oscillation is within the limit where the force is linearly proportional to displacement). This is a key characteristic of SHM.

Examples of systems that undergo SHM (at least approximately for small displacements): a mass attached to an ideal spring, a simple pendulum swinging with a small amplitude.

Simple Harmonic Motion And Uniform Circular Motion

There is a deep mathematical and physical connection between Simple Harmonic Motion and Uniform Circular Motion (UCM). SHM can be viewed as the projection of UCM onto a diameter of the circle.

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The Connection

Consider a particle moving with constant angular speed $\omega$ in a circle of radius $A$ centred at the origin in the xy-plane. Let the particle start at an angle $\phi$ with the positive x-axis at $t=0$ and move counterclockwise.

(Image Placeholder: A circle centered at the origin. A point P moves on the circle with angular velocity omega. Draw the projection of P onto the x-axis (point Q). As P moves in a circle, Q moves back and forth along the x-axis between -A and +A. Indicate the radius A of the circle and the angle (omega t + phi) of point P with the x-axis.)

The coordinates of the particle P at time $t$ are $x(t) = A \cos(\omega t + \phi)$ and $y(t) = A \sin(\omega t + \phi)$.

Consider the projection of the position of particle P onto the x-axis. This projected point, let's call it Q, moves back and forth along the x-axis. The position of Q at time $t$ is simply its x-coordinate, $x(t) = A \cos(\omega t + \phi)$.

This equation is the general equation for the displacement of an object undergoing simple harmonic motion. Thus, the motion of the projection of a particle in uniform circular motion onto a diameter is Simple Harmonic Motion.

Similarly, the projection onto the y-axis, $y(t) = A \sin(\omega t + \phi)$, also represents SHM. The difference between the x and y projections is just a phase shift of $90^\circ$ ($ \sin(\theta) = \cos(\theta - \pi/2) $ or $ \sin(\theta) = \cos(\pi/2 - \theta) $).

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Advantages of this Analogy

This analogy is useful because:

• It provides a visual way to understand SHM.
• Properties of SHM (like velocity and acceleration) can be easily derived from the corresponding properties in UCM.
• Complex oscillatory phenomena that are combinations of SHM can sometimes be visualized using combinations of circular motions.

The angular frequency $\omega$ in the SHM equation is the same as the constant angular speed of the reference particle in UCM. The amplitude $A$ of the SHM is equal to the radius of the reference circle in UCM.

Velocity And Acceleration In Simple Harmonic Motion

Since Simple Harmonic Motion is a specific type of motion, the velocity and acceleration of an object undergoing SHM vary continuously over time. We can find expressions for these quantities by taking the time derivatives of the displacement equation.

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Displacement

Let the displacement from the equilibrium position be given by $x(t) = A \cos(\omega t + \phi)$.

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Velocity

The velocity $v(t)$ of the object is the rate of change of its displacement with respect to time:

$ v(t) = \frac{dx}{dt} = \frac{d}{dt} [A \cos(\omega t + \phi)] $

$ v(t) = A [-\sin(\omega t + \phi) \cdot \frac{d}{dt}(\omega t + \phi)] $

$ v(t) = A [-\sin(\omega t + \phi) \cdot \omega] $

$ v(t) = -A\omega \sin(\omega t + \phi) $

The velocity in SHM is also a sinusoidal function, but it is $90^\circ$ out of phase with the displacement (a sine function is a cosine function shifted by $90^\circ$). The maximum speed occurs when $\sin(\omega t + \phi) = \pm 1$, and its magnitude is $v_{max} = A\omega$. This maximum speed occurs when the object passes through the equilibrium position ($x=0$). The velocity is zero at the extreme positions ($x = \pm A$), where $\cos(\omega t + \phi) = \pm 1$ and $\sin(\omega t + \phi) = 0$.

We can also express velocity in terms of displacement using the identity $\sin^2\theta + \cos^2\theta = 1$. From $x = A\cos(\omega t + \phi)$, $\cos(\omega t + \phi) = x/A$. From $v = -A\omega\sin(\omega t + \phi)$, $\sin(\omega t + \phi) = -v/(A\omega)$.

$ \left(\frac{x}{A}\right)^2 + \left(\frac{-v}{A\omega}\right)^2 = 1 $

$ \frac{x^2}{A^2} + \frac{v^2}{A^2\omega^2} = 1 $

$ \frac{v^2}{A^2\omega^2} = 1 - \frac{x^2}{A^2} = \frac{A^2 - x^2}{A^2} $

$ v^2 = \omega^2 (A^2 - x^2) $

$ v = \pm \omega \sqrt{A^2 - x^2} $

This formula gives the speed of the object at any displacement $x$. The sign depends on the direction of motion (positive for increasing x, negative for decreasing x). It clearly shows that $v=0$ when $x = \pm A$ and $v_{max} = \omega A$ when $x=0$.

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Acceleration

The acceleration $a(t)$ of the object is the rate of change of its velocity with respect to time:

$ a(t) = \frac{dv}{dt} = \frac{d}{dt} [-A\omega \sin(\omega t + \phi)] $

$ a(t) = -A\omega [\cos(\omega t + \phi) \cdot \frac{d}{dt}(\omega t + \phi)] $

$ a(t) = -A\omega [\cos(\omega t + \phi) \cdot \omega] $

$ a(t) = -A\omega^2 \cos(\omega t + \phi) $

Notice that $A \cos(\omega t + \phi) = x(t)$. So, the acceleration can be expressed directly in terms of the displacement:

$ a(t) = -\omega^2 x(t) $

This confirms the defining condition of SHM from Newton's Second Law, $a = - (k/m)x$, since $\omega^2 = k/m$. The acceleration is also a sinusoidal function, $180^\circ$ out of phase with the displacement (a negative cosine function). The maximum acceleration occurs at the extreme positions ($x = \pm A$), where its magnitude is $a_{max} = A\omega^2$. The acceleration is zero at the equilibrium position ($x=0$). The acceleration is always directed towards the equilibrium position, just like the restoring force.

(Image Placeholder: Three aligned graphs with time (t) on the x-axis. The first graph is Displacement (x) vs t, showing a cosine curve. The second graph is Velocity (v) vs t, showing a negative sine curve (cosine shifted by +90 degrees). The third graph is Acceleration (a) vs t, showing a negative cosine curve (cosine shifted by +180 degrees). Show the phase relationships.)

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