📐 Level 2 - Topic 14: Harmonic Motion & Simple Harmonic Oscillators - Rhythmic Motion Explained 🎵🔄

Example 1: Calculating Parameters from a Description

A pendulum completes 20 full swings in 10 seconds. The maximum angle it reaches from the vertical is \( 15^\circ \).

1. Calculate Period (T): Time for 1 swing = Total time / Number of swings = \( \frac{10 \text{ seconds}}{20 \text{ swings}} = 0.5 \text{ seconds/swing} \). So, Period \( T = 0.5 \text{ s} \).

   \( T = 0.5 \text{ s} \)

2. Calculate Frequency (f): Frequency \( f = \frac{1}{T} = \frac{1}{0.5 \text{ s}} = 2 \text{ Hz} \).

   \( f = 2 \text{ Hz} \)

3. Amplitude (A): The maximum angular displacement is given as \( 15^\circ \). So, Amplitude \( A = 15^\circ \) (or in radians, \( 15^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{12} \text{ radians} \)).

   \( A = 15^\circ \text{ or } \frac{\pi}{12} \text{ radians} \)

4. Angular Frequency (ω): \( \omega = 2\pi f = 2\pi \times 2 \text{ Hz} = 4\pi \text{ rad/s} \).

   \( \omega = 4\pi \text{ rad/s} \)

5. Phase Shift (φ): Information about initial position is needed to determine phase shift. If we assume it starts at maximum displacement at \( t=0 \), phase shift could be 0. Without initial conditions, phase shift is undetermined. For simplicity, often assume \( \phi = 0 \) if starting at maximum displacement for cosine, or \( \phi = 0 \) for sine if starting at equilibrium and moving in positive direction.

Solution: Period \( T = 0.5 \text{ s} \), Frequency \( f = 2 \text{ Hz} \), Amplitude \( A = 15^\circ \) (or \( \frac{\pi}{12} \) rad), Angular frequency \( \omega = 4\pi \text{ rad/s} \). Phase shift is not determined from the given information.

Example 2: Writing Equation of SHM

A spring-mass system oscillates with simple harmonic motion. It has an amplitude of 5 cm and a frequency of 4 Hz. At \( t = 0 \), the mass is at its maximum displacement from equilibrium.

1. Identify given parameters: Amplitude \( A = 5 \text{ cm} \). Frequency \( f = 4 \text{ Hz} \). Initial condition: maximum displacement at \( t = 0 \).
2. Calculate Angular Frequency (ω): \( \omega = 2\pi f = 2\pi \times 4 \text{ Hz} = 8\pi \text{ rad/s} \).
3. Determine Phase Shift (φ): Since it starts at maximum displacement, cosine function is appropriate, and we can set phase shift \( \phi = 0 \) for simplicity. Using \( \cos(0) = 1 \) (maximum value). If we used sine, we'd need \( \phi = \frac{\pi}{2} \) because \( \sin(t + \frac{\pi}{2}) = \cos(t) \). For cosine, \( \phi = 0 \) is simpler.
4. Write the equation of motion: Using cosine form, \( x(t) = A\cos(\omega t + \phi) \) with \( A = 5 \), \( \omega = 8\pi \), \( \phi = 0 \).

   \( x(t) = 5\cos(8\pi t) \text{ cm} \)

Solution: The equation of simple harmonic motion is \( x(t) = 5\cos(8\pi t) \) cm.

Example 3: Interpreting Equation of SHM

Consider the SHM described by \( x(t) = 3\sin(2t + \frac{\pi}{4}) \) meters. Find amplitude, frequency, period, and phase shift.

1. Compare with standard form \( x(t) = A\sin(\omega t + \phi) \): By comparing, we can directly read off the parameters.
2. Amplitude (A): Coefficient of sine function, \( A = 3 \text{ meters} \).

   \( A = 3 \text{ m} \)

3. Angular Frequency (ω): Coefficient of \( t \) inside sine function, \( \omega = 2 \text{ rad/s} \).

   \( \omega = 2 \text{ rad/s} \)

4. Frequency (f): Use \( \omega = 2\pi f \Rightarrow f = \frac{\omega}{2\pi} = \frac{2}{2\pi} = \frac{1}{\pi} \text{ Hz} \).

   \( f = \frac{1}{\pi} \text{ Hz} \)

5. Period (T): \( T = \frac{1}{f} = \pi \text{ seconds} \). Or directly, \( T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \text{ seconds} \).

   \( T = \pi \text{ s} \)

6. Phase Shift (φ): The constant term added inside the sine function, \( \phi = \frac{\pi}{4} \text{ radians} \).

   \( \phi = \frac{\pi}{4} \text{ radians} \)

Solution: Amplitude \( A = 3 \text{ m} \), Angular frequency \( \omega = 2 \text{ rad/s} \), Frequency \( f = \frac{1}{\pi} \text{ Hz} \), Period \( T = \pi \text{ s} \), Phase shift \( \phi = \frac{\pi}{4} \text{ radians} \).

Example 4: Spring-Mass System

Consider a mass of 0.5 kg attached to a spring. When displaced and released, it oscillates with SHM. If the spring constant \( k = 20 \text{ N/m} \), find the angular frequency and period of oscillation.

1. Formula for Angular Frequency of Spring-Mass System: For a spring-mass system, angular frequency is given by \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is spring constant and \( m \) is mass.
2. Substitute given values: \( k = 20 \text{ N/m} \), \( m = 0.5 \text{ kg} \).

   \( \omega = \sqrt{\frac{20 \text{ N/m}}{0.5 \text{ kg}}} = \sqrt{40} = 2\sqrt{10} \text{ rad/s} \)

3. Calculate Period (T): \( T = \frac{2\pi}{\omega} = \frac{2\pi}{2\sqrt{10}} = \frac{\pi}{\sqrt{10}} \text{ seconds} \).

   \( T = \frac{\pi}{\sqrt{10}} \approx \frac{3.1415}{3.162} \approx 0.993 \text{ seconds} \)

Solution: Angular frequency \( \omega = 2\sqrt{10} \text{ rad/s} \approx 6.32 \text{ rad/s} \), Period \( T \approx 0.993 \text{ seconds} \).

Example 5: Simple Pendulum (Small Angle Approximation)

A simple pendulum has a length of 1 meter. Assuming small angle oscillations, find the period of oscillation. (Use acceleration due to gravity \( g \approx 9.8 \text{ m/s}^2 \)).

1. Formula for Period of Simple Pendulum (Small Angle Approximation): For small angles, period of a simple pendulum is approximately \( T = 2\pi\sqrt{\frac{L}{g}} \), where \( L \) is length and \( g \) is acceleration due to gravity.
2. Substitute given values: \( L = 1 \text{ meter} \), \( g \approx 9.8 \text{ m/s}^2 \).

   \( T = 2\pi\sqrt{\frac{1 \text{ m}}{9.8 \text{ m/s}^2}} = 2\pi\sqrt{\frac{1}{9.8}} \text{ seconds} \)

3. Evaluate: \( \sqrt{\frac{1}{9.8}} \approx \sqrt{0.102} \approx 0.319 \).

   \( T \approx 2\pi \times 0.319 \approx 2 \times 3.1415 \times 0.319 \approx 2.004 \text{ seconds} \)

Solution: Period of the simple pendulum is approximately 2.004 seconds.