Systems Executing SHM

Some Systems Executing Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a ubiquitous phenomenon in physics and engineering. While the mathematical description ($F = -kx$ or $a = -\omega^2 x$) defines SHM, many real-world systems exhibit this type of motion, at least under certain conditions. Here, we examine two classic examples:

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Oscillations Due To A Spring ($ T = 2\pi\sqrt{m/k} $)

The most common and direct example of a system that ideally exhibits SHM is a mass attached to an ideal spring. An ideal spring is one that obeys Hooke's Law perfectly ($F = -kx$) and is massless.

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The Mass-Spring System

Consider a block of mass $m$ attached to one end of a horizontal spring. The other end of the spring is fixed to a rigid support. Assume the block slides on a frictionless horizontal surface. The equilibrium position is where the spring is neither stretched nor compressed (its natural length). Let's set the equilibrium position at $x=0$.

(Image Placeholder: A block attached to a spring on a horizontal surface. Show the equilibrium position (x=0). Show the block displaced to the right (x>0) with the spring stretched, and a force vector F pointing left. Show the block displaced to the left (x<0) with the spring compressed, and a force vector F pointing right. Indicate the force F = -kx.)

If the block is displaced horizontally by a distance $x$ from its equilibrium position, the spring exerts a restoring force on the block. According to Hooke's Law, this force is:

$ F_{spring} = -kx $

where $k$ is the spring constant. This force is directed towards the equilibrium position (negative sign). Assuming no friction, the net force on the block is just the spring force:

$ F_{net} = F_{spring} = -kx $

This net force is a linear restoring force directly proportional to the displacement and opposite in direction. Thus, the motion of the block is Simple Harmonic Motion.

The equation of motion is $m a = -kx$, or $m \frac{d^2x}{dt^2} = -kx$, which is the defining differential equation for SHM.

The angular frequency of this oscillation is given by $\omega = \sqrt{k/m}$.

The frequency is $\nu = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.

The period of oscillation is $T = \frac{1}{\nu} = 2\pi \sqrt{\frac{m}{k}}$.

This formula $T = 2\pi\sqrt{m/k}$ is a fundamental result for the period of a mass-spring system. It shows that the period increases with increasing mass (more inertia to overcome) and decreases with increasing spring stiffness (larger $k$, stronger restoring force). This is consistent with intuition.

The same analysis applies to a mass hanging vertically from a spring (vertical SHM). The equilibrium position is shifted downwards due to gravity, but the oscillations about this new equilibrium are still SHM with the same period $T=2\pi\sqrt{m/k}$, provided the force due to gravity $mg$ is constant and the spring still obeys Hooke's Law relative to its unstretched length.

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The Simple Pendulum ($ T = 2\pi\sqrt{l/g} $)

A simple pendulum is an idealised system consisting of a point mass (bob) suspended from a fixed support by a massless, inextensible string or rod. When the bob is displaced from its equilibrium position (hanging vertically) and released, it swings back and forth. For small angular displacements, the motion of the simple pendulum is approximately Simple Harmonic Motion.

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Restoring Torque on a Simple Pendulum

Consider a simple pendulum of length $l$ with a bob of mass $m$. Let the displacement from the vertical equilibrium position be an angle $\theta$. The forces acting on the bob are the tension in the string ($\vec{T}$) and the gravitational force ($\vec{F}_g = m\vec{g}$) acting vertically downwards.

(Image Placeholder: A simple pendulum suspended at point O, with string of length l and bob of mass m. The bob is displaced by angle theta from the vertical. Show the vertical line representing equilibrium. Show the weight vector mg downwards. Resolve mg into components: mg cos(theta) along the string, and mg sin(theta) perpendicular to the string (tangential). Indicate the restoring force/torque due to mg sin(theta).)

The tension force acts along the string and passes through the pivot point, so it produces no torque about the pivot. The gravitational force $m\vec{g}$ acts at the position of the bob. The torque ($\vec{\tau}$) due to gravity about the pivot point O is given by $\vec{\tau} = \vec{r} \times \vec{F}_g$, where $\vec{r}$ is the position vector of the bob from the pivot, with magnitude $l$.

The magnitude of the torque is $\tau = |\vec{r}| |\vec{F}_g| \sin\alpha$, where $\alpha$ is the angle between $\vec{r}$ and $\vec{F}_g$. In this case, $\alpha = \theta$ (the angle between the position vector along the string and the downward vertical force). The component of force perpendicular to the string is $mg\sin\theta$. The torque is $\tau = l (mg\sin\theta)$. This torque tends to restore the pendulum to the vertical position, so it is a restoring torque.

If we take the angular displacement $\theta$ as positive when displaced to the right, the restoring torque acts to the left (decreasing $\theta$), so the torque has the opposite sign to the displacement $\theta$.

$ \tau_{restoring} = -lmg\sin\theta $

For small angular displacements ($\theta$ in radians), $\sin\theta \approx \theta$.

$ \tau_{restoring} \approx -lmg\theta $

This restoring torque is approximately linearly proportional to the angular displacement $\theta$ and acts in the opposite direction. This is the condition for angular Simple Harmonic Motion.

The equation of motion for rotational dynamics is $\tau_{net} = I\alpha_{angular}$, where $I$ is the moment of inertia about the pivot and $\alpha_{angular} = \frac{d^2\theta}{dt^2}$ is the angular acceleration. For a point mass bob at distance $l$ from the pivot, $I = ml^2$.

$ ml^2 \frac{d^2\theta}{dt^2} = -lmg\sin\theta $

For small angles ($\sin\theta \approx \theta$):

$ ml^2 \frac{d^2\theta}{dt^2} = -lmg\theta $

$ \frac{d^2\theta}{dt^2} + \frac{lmg}{ml^2} \theta = 0 $

$ \frac{d^2\theta}{dt^2} + \frac{g}{l} \theta = 0 $

This is the differential equation for angular SHM, analogous to $\frac{d^2x}{dt^2} + \omega^2 x = 0$. Comparing the two equations, the angular frequency squared for the simple pendulum is $\omega^2 = \frac{g}{l}$.

The angular frequency is $\omega = \sqrt{\frac{g}{l}}$.

The frequency is $\nu = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.

The period of oscillation is $T = \frac{1}{\nu} = 2\pi \sqrt{\frac{l}{g}}$.

This formula $T = 2\pi\sqrt{l/g}$ gives the period of a simple pendulum for small amplitudes. It shows that the period depends only on the length of the pendulum ($l$) and the acceleration due to gravity ($g$) at that location. It is independent of the mass of the bob and the amplitude of oscillation (for small angles). This property makes simple pendulums useful for timekeeping.

The approximation $\sin\theta \approx \theta$ is valid for angles up to about 10-15 degrees. For larger angles, the motion is still periodic but not strictly SHM, and the period depends on the amplitude.

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