Physics: Class 11 (Book 2) — Chapter 13: Oscillations

• Periodic Motion: Any motion that repeats itself at regular, equal intervals of time (e.g., the orbital revolution of planets).
• Oscillatory Motion: A periodic motion in which the system moves back and forth about a central mean position (e.g., a swinging pendulum, a vibrating violin string).
• Key Distinction: Every oscillatory motion is periodic, but not every periodic motion is oscillatory (e.g., uniform circular motion is periodic but not to-and-fro).

While there is no strict physical boundary between the two terms, they are generally distinguished by their frequency scales:

• Oscillation: Typically refers to low-frequency macroscopic movements, such as the swaying of a tree branch or the motion of a swing.
• Vibration: Typically refers to high-frequency microscopic movements, such as the rapid motion of atoms in a crystal lattice or a musical instrument string.

The temporal characteristics of periodic motion are defined by the time interval of repetition and its rate:

• Period ($T$): The smallest interval of time after which the motion repeats itself. The SI unit of period is the second ( s ).
• Frequency ($\nu$): The number of repetitions or cycles that occur per unit time. It is mathematically defined as the reciprocal of the period:$$\nu = \frac{1}{T}$$The SI unit of frequency is the hertz ( Hz ), where $1 \text{ Hz} = 1 \text{ s}^{-1}$.

Mathematical Problem Statement: On average, a human heart is found to beat 75 \times in a minute. Calculate its frequency and period.

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Step-by-Step Algorithmic Solution Trace:

1. Identify the total number of cycles ($N = 75$) and the total time interval ($t = 60 \text{ s}$).
2. Calculate the frequency ($\nu$) as the number of cycles per second:$$\nu = \frac{75}{60 \text{ s}} = 1.25 \text{ Hz}$$
3. Calculate the time period ($T$) as the reciprocal of the frequency:$$T = \frac{1}{\nu} = \frac{1}{1.25 \text{ s}^{-1}} = 0.8 \text{ s}$$

Final Evaluation Output: Frequency $\nu = 1.25 \text{ Hz}$, Time Period $T = 0.8 \text{ s}$

In the context of oscillations, displacement is generalized beyond spatial coordinates to represent the change in any physical property over time:

• Mechanical Systems: Position coordinates ($x(t)$) or angular position ($\theta(t)$) for a simple pendulum.
• Electrical Systems: The time-varying voltage across a capacitor in an AC circuit.
• Electromagnetic Systems: The oscillating electric ($E$) and magnetic ($B$) field vectors in a propagating light wave.
• Acoustic Systems: Pressure variations in a medium during sound propagation.

Mathematically, the simplest periodic displacement function is represented by a sinusoidal function of time:$$f(t) = A \cos \omega t$$By the Fourier theorem , any complex periodic function can be expressed as a linear superposition of sine and cosine functions of different periods with appropriate coefficients:$$f(t) = A \sin \omega t + B \cos \omega t = D \sin(\omega t + \phi)$$where $D = \sqrt{A^2 + B^2}$ and $\phi = \tan^{-1}(B/A)$.

The instantaneous displacement $x(t)$ of a particle executing SHM along the x-axis is given by:$$x(t) = A \cos(\omega t + \phi)$$Where the parameters are defined as follows:

• Amplitude ($A$): The magnitude of the maximum displacement of the particle from its mean position.
• Angular Frequency ($\omega$): The rate of change of phase, related to the period $T$ by:$$\omega = \frac{2\pi}{T}$$The SI unit of angular frequency is radians per second ( rad/s ).
• Phase ($ \omega t + \phi $): The time-dependent argument of the cosine function that determines the state of motion (position and velocity) at any instant $t$.
• Phase Constant ($\phi$): The initial phase angle at $t = 0$, determined by the initial position and velocity of the particle.

1. Consider a reference particle $P$ moving with constant angular speed $\omega$ on a circle of radius $A$.
2. Let the initial position vector $\vec{OP}$ at $t = 0$ make an angle $\phi$ with the positive x-axis.
3. At any time $t$, the position vector $\vec{OP}$ makes an angle $(\omega t + \phi)$ with the positive x-axis.
4. The projection of $P$ onto the x-axis (point $P'$) has a position coordinate given by:$$x(t) = A \cos(\omega t + \phi)$$
5. The projection of $P$ onto the y-axis (point $P''$) has a position coordinate given by:$$y(t) = A \sin(\omega t + \phi)$$

This proves that the projection of uniform circular motion on any diameter is mathematically identical to linear simple harmonic motion.

The instantaneous velocity $v(t)$ is the time derivative of displacement:$$v(t) = \frac{d}{dt}[A \cos(\omega t + \phi)] = -\omega A \sin(\omega t + \phi)$$

• The maximum speed (velocity amplitude) is $v_m = \omega A$, occurring at the mean position ($x = 0$).
• The velocity is zero at the extreme positions ($x = \pm A$).
• The velocity function has a phase difference of $\pi/2$ relative to the displacement function.

The instantaneous acceleration $a(t)$ is the time derivative of velocity:$$a(t) = \frac{d}{dt}[-\omega A \sin(\omega t + \phi)] = -\omega^2 A \cos(\omega t + \phi)$$Substituting the displacement equation $x(t) = A \cos(\omega t + \phi)$ yields:$$a(t) = -\omega^2 x(t)$$

• The acceleration is directly proportional to the displacement but opposite in direction (always directed towards the mean position).
• The maximum acceleration amplitude is $a_m = \omega^2 A$, occurring at the extreme positions ($x = \pm A$).
• The acceleration is zero at the mean position ($x = 0$).
• The acceleration function has a phase difference of $\pi$ relative to the displacement function.

The force acting on a particle of mass $m$ executing SHM is:$$F(t) = ma(t) = -m\omega^2 x(t)$$Defining the force constant (or spring constant) as $k = m\omega^2$, the force law becomes:$$F(t) = -kx(t)$$

• This is a linear restoring force, directly proportional to the displacement and directed opposite to it.
• The angular frequency of the system is determined by its inertia ($m$) and stiffness ($k$):$$\omega = \sqrt{\frac{k}{m}}$$
• The time period of the linear harmonic oscillator is:$$T = 2\pi\sqrt{\frac{m}{k}}$$

The kinetic energy of the particle at any instant $t$ is:$$K = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t + \phi) = \frac{1}{2}kA^2 \sin^2(\omega t + \phi)$$Kinetic energy is maximum at the mean position ($x = 0$) and zero at the extreme positions ($x = \pm A$).

The potential energy of the particle, associated with the conservative restoring force $F = -kx$, is:$$U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \cos^2(\omega t + \phi)$$Potential energy is zero at the mean position ($x = 0$) and maximum at the extreme positions ($x = \pm A$).

The total mechanical energy of the system is the sum of its kinetic and potential energies:$$E = K + U = \frac{1}{2}kA^2 [\sin^2(\omega t + \phi) + \cos^2(\omega t + \phi)]$$Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$, we get:$$E = \frac{1}{2}kA^2$$

• The total mechanical energy of a harmonic oscillator is constant and independent of time.
• Energy continuously transforms between kinetic and potential forms during the cycle.
• Both $K$ and $U$ oscillate with a frequency of $2\nu$ (period $T/2$).

1. Let the pendulum be displaced by a small angle $\theta$ from the vertical.
2. The forces acting on the bob are:
   • Tension $\vec{T}$ along the string.
   • Gravitational force $m\vec{g}$ acting vertically downwards.
3. Resolve $mg$ into two components: $mg \cos\theta$ along the string (balanced by tension) and $mg \sin\theta$ perpendicular to the string (providing the restoring force).
4. The restoring torque $\tau$ about the support point is:$$\tau = -L(mg \sin\theta)$$
5. Using Newton's rotational law $\tau = I\alpha$, where $I = mL^2$ is the moment of inertia:$$mL^2 \alpha = -mgL \sin\theta \implies \alpha = -\frac{g}{L} \sin\theta$$
6. For small angular displacements ($\theta \ll 1$ radian), we apply the small-angle approximation $\sin\theta \approx \theta$:$$\alpha \approx -\frac{g}{L} \theta$$
7. Comparing this with the SHM angular acceleration equation $\alpha = -\omega^2 \theta$, we find:$$\omega = \sqrt{\frac{g}{L}}$$
8. The time period of a simple pendulum is:$$T = 2\pi\sqrt{\frac{L}{g}}$$