Oscillations:2

Fundamental Principles of Simple Harmonic Motion (SHM)

Simple Harmonic Motion (SHM) is defined by three primary characteristics as discussed in previous lectures:

• Vibratory or Oscillatory Nature: The motion is a back-and-forth movement around a central stable equilibrium position.
• Periodicity: The motion repeats itself over regular intervals of time.
• Restoring Force Relationship: The restoring force ($f_r$) is directly proportional to the negative of the displacement ($x$). Mathematically, this is expressed as:     $f_r \propto -x$

Characteristics of Acceleration

The fundamental equation for the acceleration of a body performing SHM is: $a = -\omega^2 \times x$

• Constants: In this equation, $\omega^2$ is a constant for the system.
• Directionality: Acceleration ($a$) is directly proportional to the negative of the displacement. The negative sign indicates that the direction of acceleration and displacement are always opposite (acceleration is always directed toward the mean position).
• Magnitude: As displacement ($x$) increases, the magnitude of acceleration ($a$) increases proportionally.

Mathematical Framework and Kinematics

General Equations of Motion

To calculate the state of a vibrating body at any given time, specific kinematic formulas are used:

• Velocity:$v = \omega \times \sqrt{x_0^2 - x^2}$Note: This formula can be derived either by taking the derivative of the displacement equations or via the projection of uniform circular motion onto a diameter.
• Displacement equations (based on starting conditions):
  1. Starting from the mean position (at time $t = 0$) moving toward positive extreme:         $x = x_0 \times \sin(\omega \times t)$
  2. Starting from the positive extreme position (at time $t = 0$) moving toward the mean position:         $x = x_0 \times \cos(\omega \times t)$

Mass-Spring System Analysis

A standard mass-spring system consists of a mass ($m$) attached to a spring with spring constant ($k$). At the equilibrium (mean) position, the spring exerts no force. Displacement from this point triggers a restoring force.

Force and Acceleration Laws

• Restoring Force (Hooke's Law):$f_r = -k \times x$
  • The negative sign indicates the force opposes the displacement. If the mass is moved to the right ($x > 0$), the force is to the left ($f < 0$). If compressed to the left ($x < 0$), the force is to the right ($f > 0$).
• Newton's Second Law:$f = m \times a$
• Equating Forces:$m \times a = -k \times x$$a = -\frac{k}{m} \times x$

Comparing this to the standard SHM equation ($a = -\omega^2 \times x$), we find: $\omega^2 = \frac{k}{m}$$\omega = \sqrt{\frac{k}{m}}$

Physical Constants and Units

• Spring Constant ($k$): Also known as rigidity or stiffness. It represents the nature of the spring. It is defined as the force required to stretch or compress the spring by a unit length:     $k = \frac{f}{x}$
• Units: The SI unit for $k$ is Newton per meter ($N/m$ or $N\,m^{-1}$).
• Rigidity Comparison: A spring with $k = 200\,N/m$ is more rigid than one with $k = 20\,N/m$ because it requires more force for the same displacement. Examples include ballpoint pen springs (low $k$) versus truck suspension springs (high $k$).
• Effective Mass ($m$): The total mass undergoing the back-and-forth motion.

Time Period and Frequency

• Time Period ($T$):$T = \frac{2\pi}{\omega} = 2\pi \times \sqrt{\frac{m}{k}}$
• Frequency ($f$):$f = \frac{1}{2\pi} \times \sqrt{\frac{k}{m}}$

Mechanical Energy in Simple Harmonic Motion

Energy oscillates between kinetic and potential forms throughout the cycle.

Kinetic Energy (K.E.)

Derived from $K.E. = \frac{1}{2}m v^2$: $K.E. = \frac{1}{2} \times k \times (x_0^2 - x^2)$

• Maximum K.E.: Occurs at the mean position ($x = 0$):     $K.E._{max} = \frac{1}{2} \times k \times x_0^2$
• Minimum K.E.: Occurs at extreme positions ($x = x_0$):     $K.E._{min} = 0$

Potential Energy (P.E.)

Calculated as the work done against the restoring force: $P.E. = \frac{1}{2} \times k \times x^2$

• Maximum P.E.: Occurs at extreme positions ($x = x_0$):     $P.E._{max} = \frac{1}{2} \times k \times x_0^2$
• Minimum P.E.: Occurs at the mean position ($x = 0$):     $P.E._{min} = 0$

Total Energy ($E_{total}$)

$E_{total} = K.E. + P.E.$$E_{total} = \frac{1}{2}k(x_0^2 - x^2) + \frac{1}{2}kx^2 = \frac{1}{2} \times k \times x_0^2$

• Constraint: Total energy is a constant. It does not depend on instantaneous displacement ($x$). It solely depends on the amplitude ($x_0$) and the spring constant ($k$).

The Simple Pendulum

A simple pendulum consists of a mass (bob) suspended by a string of length ($l$).

Restoring Force and Small Angle Approximation

When the bob is displaced by an angle ($\theta$) from the vertical:

• The weight ($mg$) is resolved into components: $mg \times \cos(\theta)$ (opposing tension) and $mg \times \sin(\theta)$ (the restoring force).
• Restoring Force: $f_r = -mg \times \sin(\theta)$.
• Acceleration: From $ma = -mg \times \sin(\theta)$, we get $a = -g \times \sin(\theta)$.
• Small Angle Constraint: For the motion to be considered SHM, $\theta$ must be small ($\leq 10^{\circ}$). In this range, $\sin(\theta) \approx \theta$ (in radians).
• By substitution ($\theta = \frac{x}{l}$):     $a = -\frac{g}{l} \times x$

Period and Frequency of Pendulum

• Angular Frequency: $\omega = \sqrt{\frac{g}{l}}$
• Time Period: $T = 2\pi \times \sqrt{\frac{l}{g}}$
• Frequency: $f = \frac{1}{2\pi} \times \sqrt{\frac{g}{l}}$
• Second's Pendulum: A pendulum with a time period of exactly $2\,s$. Its length is approximately $0.99\,m$ or $1\,m$.

Advanced Spring Configurations and Effective Constants

When multiple springs are used, they combine into an effective spring constant ($k_{eq}$).

Series Combination

Springs connected end-to-end (similar to resistors in parallel): $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + ...$

• For two springs: $k_{eq} = \frac{k_1 \times k_2}{k_1 + k_2}$

Parallel Combination

Springs connected side-by-side or supporting the same mass (similar to resistors in series): $k_{eq} = k_1 + k_2 + ...$

Effects of Cutting a Spring

If a spring of constant $k$ is cut into $n$ equal parts, each part becomes more rigid. The new spring constant for each part is: $k_{new} = n \times k$

Force-Displacement ($f-x$) Graphs

• Slope: The slope of an $f$ vs $x$ graph represents the spring constant ($k$).     $\text{Slope} = \frac{\Delta f}{\Delta x} = k$

Numerical Examples and Procedural Applications

1. Energy at Half-Amplitude

Question: A particle has total energy $E$. Find its K.E. and P.E. at $x = \frac{x_0}{2}$.

• P.E.: $P.E. = \frac{1}{2}k(\frac{x_0}{2})^2 = \frac{1}{4}(\frac{1}{2}kx_0^2) = \frac{E}{4}$.
• K.E.: $K.E. = E - P.E. = E - \frac{E}{4} = \frac{3}{4}E$.

2. Time Taken for Half-Amplitude ($T_{total} = 24\,s$)

• Starting from Mean position to $x = \frac{x_0}{2}$:     Use $x = x_0\sin(\omega t)$. This results in $\omega t = \frac{\pi}{6}$. Since $\omega = \frac{2\pi}{T}$, we get $t = \frac{T}{12}$. For $T=24$, $t = 2\,s$.
• Starting from Extreme position to mean-centered $x = \frac{x_0}{2}$:     Use $x = x_0\cos(\omega t)$. This results in $\omega t = \frac{\pi}{3}$, leading to $t = \frac{T}{6}$. For $T=24$, $t = 4\,s$.
• Explanation: Velocity is maximum at the mean position and decreases toward the extreme. Therefore, the first half-displacement from the mean is covered faster ($2\,s$) than the second half-displacement reaching the extreme ($4\,s$).

3. Length of a Pendulum with Specific values

Question: Length $l = \frac{39.2}{\pi^2}$ m, $g = 9.8\,m/s^2$. Find $T$. $T = 2\pi \times \sqrt{\frac{39.2}{\pi^2 \times 9.8}}$$T = 2\pi \times \frac{1}{\pi} \times \sqrt{4} = 2 \times 2 = 4\,s$.

Questions & Discussion

Q: At what point do Kinetic Energy and Potential Energy become equal? A: Equate $\frac{1}{2}k(x_0^2 - x^2) = \frac{1}{2}kx^2$. Solving for $x$ gives $2x^2 = x_0^2 \rightarrow x = \pm \frac{x_0}{\sqrt{2}}$. This corresponds to approximately $0.707 \times x_0$.

Q: What is the relationship between length and time period cycles? A: If the length ($l$) is increased to $16l$, the time period ($T$) increases to $4T$ (since $T \propto \sqrt{l}$). Conversely, if $T$ becomes $2T$, the length was quadrupled ($4l$).

Homework Assignments:

1. Calculate K.E. and P.E. at $1/4$ the amplitude and $3/4$ the amplitude and verify the sum equals total energy $E$.
2. Review the concepts of Resonance and Damping (Critical, Under-damping, Over-damping) as specific reading assignments from the text.
3. Solve the DPP (Daily Practice Problems) provided with the lecture.