Period, Sinusoidal Nature, and Pendulums in Simple Harmonic Motion

Sinusoidal Nature of Simple Harmonic Motion

• Projection of Circular Motion:
  • Simple harmonic motion (SHM) can be understood by projecting the x-component of an object moving in a circle of radius $a$ at a constant speed $v_{max}$.
  • The x-component of velocity for such an object varies as $v = \text{plus or minus } v_{max} \times \frac{1}{\text{square root of } x^2 - a^2}$. Note: Effectively, this circular projection mirrors the motion of a mass on a spring.

Mathematical Functions of Motion

• Position as a Function of Time:
  • The position of an object in SHM can be expressed using the cosine function: $x = a\text{cos}(\text{omega } t)$.
  • The choice between sine and cosine depends on the starting point of the oscillation:
    • Cosine Function: Used if the motion starts at the maximum displacement (the zero point of time is at maximum amplitude).
    • Sine Function: Used if the motion starts at the zero point (equilibrium), which represents a shift of a quarter period relative to the cosine curve.
• Angular Frequency ($\omega$):
  • Defined as $2\times\text{pi}$ times the regular frequency ($f$).
  • $\text{omega} = 2\times\text{pi} \times f$.
  • It can also be written in terms of the period ($T$) as $\text{omega} = \frac{2\times\text{pi}}{T}$.
• Velocity as a Function of Time:
  • Calculated as $v = -v_{max}\text{sin}(\text{omega } t)$.
• Acceleration as a Function of Time:
  • Derived from Newton's Second Law ($F = ma$) and Hooke's Law:
    • $a = \frac{F}{m}$
    • $a = -\frac{kx}{m}$
    • Substituting the position function: $a = -\frac{ka}{m} \text{cos}(\text{omega } t)$.

Period and Frequency of a Mass-Spring System

• The Period ($T$):
  • The time required for one complete cycle.
  • Formula: $T = 2\times\text{pi}\times\text{sqrt}(\frac{m}{k})$.
• The Frequency ($f$):
  • The number of cycles per unit time (the inverse of the period).
  • Formula: $f = \frac{1}{T} = \frac{1}{2\times\text{pi}}\times\text{sqrt}(\frac{k}{m})$.

Concept Tests: Mass and Spring Constant Variations

• Question 1: Impact of Doubling Mass

  • Scenario: A glider with a spring attached to each end oscillates. If the mass of the glider is doubled, what happens to the period?
  • Answer: The period will increase.
  • Reasoning: The period is proportional to the square root of the mass ($T \text{ is proportional to } \text{sqrt}(m)$). Therefore, an increase in mass leads to an increase in the period.

• Question 2: Impact of Doubling Amplitude

  • Scenario: What happens to the period if the amplitude is doubled?
  • Answer: The period remains unchanged. (Based on the formula $T = 2\times\text{pi}\times\text{sqrt}(\frac{m}{k})$, amplitude is not a factor).

• Question 3: Adding Springs in Parallel

  • Scenario: If identical springs are added in parallel to the original glider, what happens to the period?
  • Answer: The period will decrease.
  • Reasoning: Adding springs in parallel makes the system act like a stronger spring, effectively increasing the spring constant ($k$). Since the period is inversely proportional to the square root of the spring constant ($T \text{ is proportional to } \frac{1}{\text{sqrt}(k)}$), an increase in $k$ results in a decrease in $T$.

The Simple Pendulum

• Definition:
  • A simple pendulum consists of a mass (bob) at the end of a lightweight cord.
• Assumptions for SHM:
  • The cord does not stretch.
  • The mass of the cord is negligible (massless).
  • The restoring force in a pendulum is $F = mg\text{sin}(\theta)$. Because this is proportional to $\text{sin}(\theta)$ rather than the displacement ($\theta$) itself, it is only true SHM under the Small Angle Approximation.
• Small Angle Approximation:
  • When the angle ($\theta$) is small and measured in radians, the following approximations hold true:
    • $\text{sin}(\theta) \text{ is approximately } \theta$
    • $\text{tan}(\theta) \text{ is approximately } \theta$
    • $\text{cos}(\theta) \text{ is approximately } 1$
  • This approximation is generally accurate for angles up to approximately $30$ degrees when using radians.
• Equations for a Simple Pendulum:
  • In the context of the pendulum, the effective spring constant $k$ is represented as $\frac{mg}{L}$.
  • Period ($T$): $T = 2\times\text{pi}\times\text{sqrt}(\frac{L}{g})$.
  • Frequency ($f$): $f = \frac{1}{2\times\text{pi}}\times\text{sqrt}(\frac{g}{L})$.
• Mass Independence:
  • As long as the cord is considered massless and the amplitude (angle) remains small, the period of a simple pendulum does not depend on the mass attached to it.

Concept Tests: Pendulum Mass and Length

• Question 1: Comparing Mass

  • Scenario: Two pendula have the same length but different masses. How do their periods compare?
  • Answer: The period is the same for both cases.
  • Reasoning: The period depends only on length ($L$) and gravitational acceleration ($g$), not mass.

• Question 2: Comparing Length

  • Scenario: Two pendula have different lengths: one is length $L$ and the other is length $4L$. How do their periods compare?
  • Answer: The period of the pendulum with length $4L$ is two times that of the pendulum with length $L$.
  • Reasoning: The period varies with the square root of the length ($T \text{ is proportional to } \text{sqrt}(L)$). Since $\text{sqrt}(4) = 2$, making the pendulum four times longer doubles the period.

Example Problems and Calculations

• Scenario: A pendulum makes $28$ oscillations in $50$ seconds.
• Problem A: Find the Period ($T$)
  • Calculation: $T = \frac{50\text{ seconds}}{28\text{ cycles}} = 1.8\text{ seconds}$.
• Problem B: Find the Frequency ($f$)
  • Calculation: $f = \frac{28\text{ cycles}}{50\text{ seconds}} = 0.56\text{ Hz}$.
  • Note: This can also be calculated as the inverse of the period ($\frac{1}{1.8} \text{ seconds}$).