Comprehensive Study Notes on Simple Harmonic Motion and Waves

Learning Objectives

• Define simple harmonic motion (SHM) and identify its required force condition.
• Recognize and calculate fundamental oscillatory parameters: displacement ($x$), amplitude ($A$), frequency ($f$), period ($T$), and angular frequency ($\omega$).
• Analyze energy transformations between elastic potential energy ($PE$) and kinetic energy ($KE$) throughout an oscillatory cycle.
• Explain how restoring force ($F$), velocity ($v$), and acceleration ($a$) dynamically vary as an object undergoes simple harmonic motion.
• Apply mathematical formulations of SHM to mass-spring systems, simple pendulums, resonance phenomena, and wave intensity metrics.

Fundamentals of Simple Harmonic Motion

• Periodic Motion Definition:
  • Periodic motion is defined as any back-and-forth movement over the exact same path where each complete cycle or oscillation takes an identical amount of time.
  • Examples in physical systems include:
    • A swinging bob in a grandfather clock.
    • A vibrating quartz crystal in a watch.
    • Pistons moving back and forth inside an internal combustion engine.

• Simple Harmonic Motion (SHM):
  • Simple harmonic motion is the most fundamental and elementary form of periodic motion.
  • It serves as the foundational framework for comprehending complex oscillatory and wave behavior in physics, engineering, music, and electrical circuits.
  • Complex real-world phenomena—such as acoustic waves, structural vibrations, and alternating current (AC) circuits—can either be modeled as SHM or decomposed into combinations of simple harmonic oscillations.

Restoring Force and Hooke's Law

• Condition for Simple Harmonic Motion:
  • A system exhibits SHM if and only if it experiences a restoring force that acts opposite to the displacement and is directly proportional to the magnitude of that displacement from the equilibrium position.
  • Any system possessing this exact proportional restoring force is classified as a simple harmonic oscillator.

• Hooke's Law Formulation:
  • For a mass attached to an ideal spring on a horizontal, frictionless surface, the restoring force ($F$) is governed by Hooke's Law:

$F = -k \times x$

• Variables and Units in Hooke's Law:
  • $F$: Restoring force exerted by the spring, measured in Newtons ($N$).
  • $k$: Spring constant, which measures the stiffness of the spring, expressed in Newtons per meter ($N/m$).
  • $x$: Displacement or amount of extension/compression from the undisturbed equilibrium position, measured in meters ($m$).
  • The Negative Sign: Indicates vector direction; the force is always directed opposite to the displacement vector, pulling or pushing the system back toward the equilibrium position ($x = 0$).

Kinematics and Quantitative Parameters of SHM

• Equilibrium Position ($x = 0$):

  • The center position where the system experiences zero net restoring force ($F = 0$).

• Displacement ($x$):

  • The distance and direction of the oscillating object relative to its equilibrium position at any given instant in time.

• Amplitude ($A$):

  • The maximum displacement magnitude of the object on either side of the equilibrium position ($x = +A$ to $x = -A$).

• Period ($T$):

  • The total time required for the system to execute one complete cycle or oscillation (moving from $+A$ to $-A$ and returning to $+A$).
  • Measured in seconds ($s$).

• Frequency ($f$):

  • The total number of complete oscillations or cycles performed per unit of time.
  • Measured in Hertz ($Hz$), where $1\,Hz = 1\text{ cycle/second} = 1\,s^{-1}$.
  • Relationship to period:

$f = \frac{1}{T}$

$T = \frac{1}{f}$

• Angular Frequency ($\omega$):
  • The rate of rotation of the phase angle in radians per second ($rad/s$):

$\omega = 2\pi f = \frac{2\pi}{T}$

• Equations of Motion for a Spring-Mass System:
  • For an object undergoing SHM starting at maximum extension $x = A$ at time $t = 0$:

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

• Frequency and Period Equations for Mass-Spring Oscillators:
  • Frequency:

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

• * Period:

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

• * Angular Frequency:

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

• Maximum Kinematic Values:
  • Maximum Displacement: $x_{\max} = A$
  • Maximum Speed: $v_{\max} = A \omega = A \sqrt{\frac{k}{m}}$
  • Maximum Acceleration: $a_{\max} = A \omega^2 = A \left(\frac{k}{m}\right)$

Dynamics Across the Oscillatory Path

• Extreme Displacement Positions ($x = +A$ and $x = -A$):

  • Velocity: $v = 0\,m/s$ (the mass comes to a instantaneous stop before reversing direction).
  • Restoring Force: Maximum magnitude ($F_{\max} = \mp k A$).
  • Acceleration: Maximum magnitude ($a_{\max} = \mp \frac{k A}{m}$), directed toward equilibrium according to Newton's Second Law ($F = m a$).
  • Potential Energy: Maximum elastic potential energy ($PE_{\max} = \frac{1}{2} k A^2$).
  • Kinetic Energy: Zero ($KE = 0$).

• Equilibrium Position ($x = 0$):

  • Restoring Force: Zero ($F = 0\,N$).
  • Acceleration: Zero ($a = 0\,m/s^2$).
  • Velocity: Maximum speed in either positive or negative direction ($v = \pm v_{\max}$).
  • Potential Energy: Zero ($PE = 0$).
  • Kinetic Energy: Maximum ($KE_{\max} = \frac{1}{2} m v_{\max}^2$).

Energy Transformations in Simple Harmonic Motion

• Elastic Potential Energy ($PE$):
  • Energy stored in the deformation of the spring:

$PE = \frac{1}{2} k x^2$

• Kinetic Energy ($KE$):
  • Energy associated with the velocity of the oscillating mass:

$KE = \frac{1}{2} m v^2$

• Total Mechanical Energy ($E$):
  • In an ideal system without friction or non-conservative forces, total mechanical energy is strictly conserved:

$E = PE + KE = \frac{1}{2} k x^2 + \frac{1}{2} m v^2$

• Conservation Expressions:
  • At maximum displacement ($x = A, v = 0$):

$E = \frac{1}{2} k A^2$

• * At equilibrium ($x = 0, v = v_{\max}$):

$E = \frac{1}{2} m v_{\max}^2$

• * Equating these gives the relationship between amplitude and maximum velocity:

$\frac{1}{2} k A^2 = \frac{1}{2} m v_{\max}^2 \implies v_{\max} = A \sqrt{\frac{k}{m}}$

The Simple Pendulum

• Physical Setup and Assumptions:

  • Consists of a point mass (bob) of mass $m$ attached to a light, inextensible string or rod of length $L$ suspended from a frictionless pivot.
  • Assumes zero air resistance and zero mechanical friction.

• Restoring Force Derivation:

  • The gravitational force ($m g$) breaks into components along and perpendicular to the arc path:
    • Radial component along string: $m g \cos(\theta)$, balanced by tension.
    • Tangential restoring force along trajectory: $F = -m g \sin(\theta)$.
  • For small displacement angles (typically $\theta < 10^\circ$ or $\theta < 0.174\text{ rad}$), the small-angle approximation applies: $\sin(\theta) \approx \theta$.
  • Since arc length displacement $s = L \theta \implies \theta = \frac{s}{L}$, the restoring force becomes:

$F \approx -\left(\frac{m g}{L}\right) s$

• Equivalence to Spring Constant:
  • Comparing $F = -\left(\frac{m g}{L}\right) s$ to Hooke's Law $F = -k x$ demonstrates that the effective constant is:

$k_{\text{effective}} = \frac{m g}{L}$

• Period of a Simple Pendulum ($T$):
  • Substituting $k_{\text{effective}} = \frac{m g}{L}$ into the spring-mass period formula $T = 2\pi \sqrt{\frac{m}{k}}$ yields:

$T = 2\pi \sqrt{\frac{L}{g}}$

• * $T$: Period in seconds ($s$).

  • $L$: String length in meters ($m$).
  • $g$: Acceleration due to gravity ($9.80\,m/s^2$ on Earth).
  • Key Property: The period of a simple pendulum is independent of the bob mass $m$ and depends solely on string length $L$ and local gravitational acceleration $g$.

• Frequency of a Simple Pendulum ($f$):

$f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$

Resonance

• Definition:

  • Resonance is the physical phenomenon occurring when the frequency of an externally applied periodic force matches the natural frequency of an system.
  • This driving matching condition yields a dramatically increased amplitude of oscillation as energy is efficiently transferred into the system.

• Resonant Frequency Criteria:

  • Length or structural mass/stiffness dictates natural frequency.
  • Examples include:
    • Pushing a playground swing in rhythm with its natural period increases swing height.
    • Adjusting an antenna length or circuit components so its natural radio frequency matches incoming broadcast waves.

Practical Applications and Mechanical Systems

• Automotive Suspension Systems:
  • A car suspension unit incorporates coil springs and shock absorber dampers connected between the wheel mass and car body mass.
  • Modeled as a spring-damper-mass system to absorb road deflections, controlling oscillatory motion and rapidly dissipating vibrational energy.

Wave Motion and Sound Intensity Formulas

• Transverse Wave Speed on a Stretched String:

$v = \sqrt{\frac{F_T}{m/L}} = \sqrt{\frac{F_T}{\mu}}$

• * $F_T$: Tension force in string ($N$).

  • $\mu = \frac{m}{L}$: Linear mass density of the string ($kg/m$).

• Universal Wave Equation:

$v = f \lambda$

• * $v$: Wave propagation velocity ($m/s$).

  • $f$: Frequency ($Hz$).
  • $\lambda$: Wavelength ($m$).

• Sound Intensity in Three-Dimensional Media ($I$):

  • Assuming uniform spherical expansion from a point source:

$I = \frac{P}{A} = \frac{P}{4\pi r^2}$

• * $P$: Acoustic power output in Watts ($W$).

  • $A$: Surface area of sphere ($4\pi r^2$) in square meters ($m^2$).
  • $r$: Distance from source in meters ($m$).
  • Units of intensity: $W/m^2$

• Inverse Square Law Ratios:

  • Intensity vs. Distance:

$\frac{I_1}{I_2} = \frac{(r_2)^2}{(r_1)^2}$

• * Amplitude vs. Distance:

$\frac{A_2}{A_1} = \frac{r_1}{r_2}$

• Sound Intensity Level in Decibels (Level of Loudness, $\beta$ or $L$):

$\beta = 10 \log_{10}\left(\frac{I}{I_0}\right)$

• * $\beta$: Sound intensity level in decibels ($dB$).
  • $I$: Measured sound intensity ($W/m^2$).
  • $I_0$: Standard reference intensity / threshold of human hearing ($I_0 = 10^{-12}\,W/m^2$).

Conceptual Review and Self-Assessment Questions

• Question 1: In a periodic process, the number of cycles per unit of time is called?

  • Answer: Frequency ($f$).

• Question 2: In a periodic process, the time required to complete one full cycle is called?

  • Answer: Period ($T$).

• Question 3: A mass on a spring undergoes SHM. The maximum displacement from equilibrium is called?

  • Answer: Amplitude ($A$).

• Questions 4–7 (Referencing horizontal mass-spring diagram with equilibrium $0$ and bounds $-A$ to $+A$):

• Question 4: When the mass reaches point $x = +A$, its instantaneous velocity is?
  • Answer: Zero.
• Question 5: When the mass reaches point $x = 0$, its instantaneous velocity is?
  • Answer: Maximum and can be positive or negative.
• Question 6: When the mass reaches point $x = +A$, its elastic potential energy is?
  • Answer: Maximum.
• Question 7: When the mass reaches point $x = 0$, its elastic potential energy is?
  • Answer: Zero.
• Question 8: When the mass reaches point $x = +A$, its instantaneous acceleration is?
  • Answer: Maximum and negative (because restoring force points toward $-x$).
• Question 9: When the mass reaches point $x = 0$, its instantaneous acceleration is?
  • Answer: Zero.

Step-by-Step Worked Problems

Problem 1: Simple Pendulum Period and Frequency

• Problem Statement:
  • A simple pendulum consists of a small $0.50\,kg$ bob attached to a $1.20\,m$ light string. It is displaced by a small angle ($< 10^\circ$) and released. Calculate (1) the period of oscillation ($T$) and (2) the frequency of oscillation ($f$). Assume $g = 9.80\,m/s^2$.
• Step 1: Calculate Period ($T$):

$T = 2\pi \sqrt{\frac{L}{g}}$

$T = 2\pi \sqrt{\frac{1.20\,m}{9.80\,m/s^2}}$

$T = 2\pi \sqrt{0.1224\,s^2}$

$T = 2\pi (0.3499\,s) \approx 2.20\,s$

• Step 2: Calculate Frequency ($f$):

$f = \frac{1}{T}$

$f = \frac{1}{2.20\,s} \approx 0.455\,Hz$

Problem 2: Determining Gravitational Acceleration ($g$) on Another Planet

• Problem Statement:
  • An astronaut on a newly discovered planet sets up a simple pendulum of length $0.800\,m$. The astronaut observes that the pendulum completes $20$ full oscillations in $36.0\text{ seconds}$ at small displacement angles. Calculate (1) the period $T$ and (2) acceleration due to gravity $g$ on this planet.
• Step 1: Calculate Period ($T$):

$T = \frac{\text{Total Time}}{\text{Number of Oscillations}} = \frac{36.0\,s}{20} = 1.80\,s$

• Step 2: Algebraic Derivation and Calculation for $g$:

$T = 2\pi \sqrt{\frac{L}{g}} \implies T^2 = 4\pi^2 \left(\frac{L}{g}\right) \implies g = \frac{4\pi^2 L}{T^2}$

$g = \frac{4\pi^2 (0.800\,m)}{(1.80\,s)^2} = \frac{4(9.8696)(0.800)}{3.24} = \frac{31.583}{3.24} \approx 9.75\,m/s^2$

Problem 3: Trapeze Performers as a Simple Pendulum

• Problem Statement:
  • A pair of trapeze performers at the circus is swinging from ropes attached to a large elevated platform. Treat the performers as a simple pendulum with a length of $16\,m$. Determine the period for one complete back-and-forth cycle.
• Step-by-Step Solution:

$T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{16\,m}{9.80\,m/s^2}} = 2\pi \sqrt{1.6326\,s^2} = 2\pi (1.2778\,s) \approx 8.03\,s$

Problem 4: Vertical Suspension to Horizontal Spring Oscillator

• Problem Statement:
  • A spring stretches $0.150\,m$ when a $0.300\,kg$ mass is gently suspended vertically from it. The spring is then set up horizontally with the $0.300\,kg$ mass resting on a frictionless table. The mass is pulled so the spring stretches $0.100\,m$ from equilibrium and released from rest. Determine (a) the spring constant $k$ and (b) the amplitude $A$ of horizontal oscillation.
• Step 1: Find Spring Constant $k$ using vertical equilibrium:

$F_g = F_{\text{spring}} \implies m g = k x_{\text{vertical}}$

$(0.300\,kg)(9.80\,m/s^2) = k (0.150\,m)$

$2.94\,N = k (0.150\,m)$

$k = \frac{2.94\,N}{0.150\,m} = 19.6\,N/m$

• Step 2: Determine Horizontal Amplitude $A$:
  • Since the mass is pulled $0.100\,m$ away from equilibrium and released from rest, its initial displacement equals its maximum displacement:

$A = 0.100\,m$

Problem 5: Horizontal Oscillator Parameters from Half-Cycle Measurement

• Problem Statement:
  • An object on a horizontal, frictionless surface attached to a spring is displaced $0.120\,m$ from equilibrium and released with zero initial speed. After $0.800\,s$, its displacement is $0.120\,m$ on the opposite side (passing equilibrium once). Find (a) amplitude $A$, (b) period $T$, and (c) frequency $f$.
• Solution:
  • (a) Amplitude: $A = 0.120\,m$
  • (b) Period: Moving from $+A$ to $-A$ corresponds to half a full cycle ($t = \frac{T}{2}$):

$\frac{T}{2} = 0.800\,s \implies T = 2 \times 0.800\,s = 1.60\,s$

• * (c) Frequency:

$f = \frac{1}{T} = \frac{1}{1.60\,s} = 0.625\,Hz$

Problem 6: Calculating Mass for a Target Frequency

• Problem Statement:
  • A spring stretches by $0.018\,m$ when a $2.8\,kg$ object is suspended from its end. How much mass should be attached to this spring so its vibration frequency is $f = 3.0\,Hz$?
• Step 1: Calculate Spring Constant $k$:

$k = \frac{m_1 g}{x_1} = \frac{(2.8\,kg)(9.8\,m/s^2)}{0.018\,m} = \frac{27.44\,N}{0.018\,m} \approx 1524.44\,N/m$

• Step 2: Solve for target mass $m_2$ given $f = 3.0\,Hz$:

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m_2}} \implies f^2 = \frac{k}{4\pi^2 m_2} \implies m_2 = \frac{k}{4\pi^2 f^2}$

$m_2 = \frac{1524.44\,N/m}{4\pi^2 (3.0\,s^{-1})^2} = \frac{1524.44}{4\pi^2 (9.0)} = \frac{1524.44}{355.305} \approx 4.29\,kg$

Problem 7: Acoustic Intensity from Megaphone Output

• Problem Statement:
  • A megaphone emits sound with a power output of $45\,W$. If the megaphone is operating $14.25\,m$ away from a person, what is the sound intensity on the person?
• Solution:

$I = \frac{P}{4\pi r^2}$

$I = \frac{45\,W}{4\pi (14.25\,m)^2}$

$I = \frac{45}{4\pi (203.0625)} = \frac{45}{2551.76} \approx 0.018\,W/m^2$

Problem 8: Level of Loudness at Threshold of Hearing

• Problem Statement:
  • What is the level of loudness (intensity level $\beta$) of a sound of intensity $I = 10^{-12}\,W/m^2$?
• Solution:

$\beta = 10 \log_{10}\left(\frac{I}{I_0}\right)$

$\text{Given } I_0 = 10^{-12}\,W/m^2:$

$\beta = 10 \log_{10}\left(\frac{10^{-12}\,W/m^2}{10^{-12}\,W/m^2}\right) = 10 \log_{10}(1)$

$\text{Since } \log_{10}(1) = 0:$

$\beta = 10 \times 0 = 0\,dB$