AP Exams: Physics C Mech Unit 6 Oscillations Simple Harmonic Motion Differential Equation Solutions xAcosωtφ

What Is This?

Simple Harmonic Motion (SHM) is a periodic motion where the restoring force is proportional to the displacement from the equilibrium position. This topic appears in exams as it tests your understanding of the underlying physics and mathematical modeling.

Why It Matters

This topic is frequently tested in undergraduate physics exams, carrying around 20-30% of the total marks. It assesses your ability to apply mathematical models to real-world problems, demonstrating a deep understanding of the underlying physics. The examiner wants to see if you can derive the differential equation, solve it, and interpret the results.

Core Concepts

To tackle SHM questions, you must own the following foundational ideas:

• Displacement is a vector quantity, but in SHM, we often consider it as a scalar quantity, measuring the distance from the equilibrium position.
• Restoring force is proportional to the displacement from the equilibrium position, given by Hooke's Law: F = -kx, where k is the spring constant and x is the displacement.
• Angular frequency (ω) is related to the frequency (f) and period (T) of the motion by ω = 2πf = 2π/T.

Prerequisites

Before tackling SHM, you must already understand:

• Kinematics: basic concepts of motion, such as displacement, velocity, and acceleration.
• Differential equations: basic concepts of solving differential equations, including separation of variables and integrating factors.
• Trigonometry: basic concepts of trigonometry, including sine, cosine, and tangent functions.

The Rule-Book (How It Works)

The primary rule for SHM is given by the differential equation:

x''(t) + (k/m)x(t) = 0

where x(t) is the displacement at time t, k is the spring constant, and m is the mass.

Sub-rules and exceptions:

• The differential equation assumes a simple harmonic motion with no damping or external forces.
• The solution to the differential equation is given by x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle.

A simple visual pattern or mnemonic:

Imagine a simple pendulum swinging back and forth. The displacement of the pendulum from its equilibrium position is proportional to the restoring force, which is given by Hooke's Law.

Exam / Job / Audit Weighting

Frequency: 20-30% Difficulty Rating: Intermediate Question Type or Real-World Task Type: Mathematical modeling, problem-solving

Difficulty Level

Intermediate

Must-Know Rules, Formulas, Standards, or Principles

The three most important rules for SHM are:

1. Hooke's Law: F = -kx, where k is the spring constant and x is the displacement.
2. Differential equation: x''(t) + (k/m)x(t) = 0, where x(t) is the displacement at time t, k is the spring constant, and m is the mass.
3. Solution to the differential equation: x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle.

Worked Examples (Step-by-Step)

Example 1: Easy

A simple pendulum has a length of 1 m and a mass of 0.5 kg. It is displaced by 30° from its equilibrium position and released. Find the angular frequency of the pendulum.

Question: Find the angular frequency of the pendulum.
Solution:

• The differential equation for SHM is x''(t) + (k/m)x(t) = 0.
• The solution to the differential equation is x(t) = A cos(ωt + φ).
• The angular frequency is given by ω = √(k/m).
• The spring constant k can be found using the equation k = mg/L, where m is the mass, g is the acceleration due to gravity, and L is the length of the pendulum.
• Substituting the values, we get k = 0.5 kg × 9.8 m/s² / 1 m = 4.9 N/m.
• The angular frequency is then ω = √(4.9 N/m / 0.5 kg) = 2.2 rad/s.

Example 2: Medium

A mass of 2 kg is attached to a spring with a spring constant of 10 N/m. The mass is displaced by 0.5 m from its equilibrium position and released. Find the velocity of the mass at time t = 0.5 s.

Question: Find the velocity of the mass at time t = 0.5 s.
Solution:

• The differential equation for SHM is x''(t) + (k/m)x(t) = 0.
• The solution to the differential equation is x(t) = A cos(ωt + φ).
• The velocity of the mass is given by v(t) = -Aω sin(ωt + φ).
• The angular frequency is given by ω = √(k/m).
• The spring constant k is given as 10 N/m.
• The mass m is given as 2 kg.
• Substituting the values, we get ω = √(10 N/m / 2 kg) = 1.4 rad/s.
• The velocity of the mass at time t = 0.5 s is then v(0.5 s) = -A × 1.4 rad/s × sin(1.4 rad/s × 0.5 s + φ).

Example 3: Hard

A simple pendulum has a length of 2 m and a mass of 1 kg. It is displaced by 60° from its equilibrium position and released. Find the phase angle φ.

Question: Find the phase angle φ.
Solution:

• The differential equation for SHM is x''(t) + (k/m)x(t) = 0.
• The solution to the differential equation is x(t) = A cos(ωt + φ).
• The phase angle φ can be found using the equation φ = -arctan(ω/ω₀), where ω is the angular frequency and ω₀ is the initial angular frequency.
• The initial angular frequency ω₀ is given by ω₀ = g/L, where g is the acceleration due to gravity and L is the length of the pendulum.
• The angular frequency ω is given by ω = √(k/m).
• The spring constant k can be found using the equation k = mg/L, where m is the mass, g is the acceleration due to gravity, and L is the length of the pendulum.
• Substituting the values, we get k = 1 kg × 9.8 m/s² / 2 m = 4.9 N/m.
• The angular frequency is then ω = √(4.9 N/m / 1 kg) = 2.2 rad/s.
• The initial angular frequency is then ω₀ = 9.8 m/s² / 2 m = 4.9 rad/s.
• The phase angle φ is then φ = -arctan(2.2 rad/s / 4.9 rad/s) = -22.5°.

Common Exam Traps & Mistakes

Trap 1: Incorrectly assuming a simple harmonic motion

• Wrong answer: x(t) = A sin(ωt)
• Correct approach: Check if the motion is indeed simple harmonic by verifying that the restoring force is proportional to the displacement.

Trap 2: Forgetting to include the phase angle

• Wrong answer: x(t) = A cos(ωt)
• Correct approach: Include the phase angle φ in the solution to the differential equation.

Trap 3: Misinterpreting the direction of the velocity

• Wrong answer: v(t) = Aω cos(ωt + φ)
• Correct approach: Remember that the velocity is given by v(t) = -Aω sin(ωt + φ).

Trap 4: Incorrectly solving the differential equation

• Wrong answer: x(t) = A exp(-ωt)
• Correct approach: Use the correct solution to the differential equation, which is x(t) = A cos(ωt + φ).

Trap 5: Forgetting to include the amplitude

• Wrong answer: x(t) = cos(ωt + φ)
• Correct approach: Include the amplitude A in the solution to the differential equation.

Trap 6: Misinterpreting the units of the angular frequency

• Wrong answer: ω = 2π/T
• Correct approach: Remember that the angular frequency is given by ω = √(k/m).

Shortcut Strategies & Exam Hacks

Hack 1: Use the formula ω = √(k/m) to find the angular frequency

• This formula can be used to quickly find the angular frequency of a simple harmonic motion.

Hack 2: Use the formula x(t) = A cos(ωt + φ) to find the displacement

• This formula can be used to quickly find the displacement of a simple harmonic motion.

Hack 3: Use the formula v(t) = -Aω sin(ωt + φ) to find the velocity

• This formula can be used to quickly find the velocity of a simple harmonic motion.

Hack 4: Use the formula φ = -arctan(ω/ω₀) to find the phase angle

• This formula can be used to quickly find the phase angle of a simple harmonic motion.

Question-Type Taxonomy

Format 1: Multiple-choice questions

• Example: What is the angular frequency of a simple harmonic motion with a spring constant of 10 N/m and a mass of 2 kg?
• Answer: 1.4 rad/s

Format 2: Short-answer questions

• Example: Find the displacement of a simple harmonic motion with an amplitude of 0.5 m, an angular frequency of 2 rad/s, and a phase angle of 30° at time t = 0.5 s.
• Answer: x(t) = 0.25 m

Format 3: Long-answer questions

• Example: A simple pendulum has a length of 2 m and a mass of 1 kg. It is displaced by 60° from its equilibrium position and released. Find the phase angle φ.
• Answer: φ = -22.5°

Format 4: Problem-solving questions

• Example: A mass of 2 kg is attached to a spring with a spring constant of 10 N/m. The mass is displaced by 0.5 m from its equilibrium position and released. Find the velocity of the mass at time t = 0.5 s.
• Answer: v(t) = -0.5 m/s

Practice Set (MCQs)

Question 1: Easy

What is the angular frequency of a simple harmonic motion with a spring constant of 5 N/m and a mass of 1 kg?

A) 1.4 rad/s B) 2.2 rad/s C) 3.1 rad/s D) 4.5 rad/s

Correct answer: B) 2.2 rad/s Explanation: The angular frequency is given by ω = √(k/m) = √(5 N/m / 1 kg) = 2.2 rad/s.
Why the distractors are tempting: The distractors are tempting because they are close to the correct answer, but they are not the correct answer.

Question 2: Medium

Find the displacement of a simple harmonic motion with an amplitude of 0.5 m, an angular frequency of 2 rad/s, and a phase angle of 30° at time t = 0.5 s.

A) 0.25 m B) 0.5 m C) 0.75 m D) 1.0 m

Correct answer: A) 0.25 m Explanation: The displacement is given by x(t) = A cos(ωt + φ) = 0.5 m × cos(2 rad/s × 0.5 s + 30°) = 0.25 m.
Why the distractors are tempting: The distractors are tempting because they are close to the correct answer, but they are not the correct answer.

Question 3: Hard

A simple pendulum has a length of 2 m and a mass of 1 kg. It is displaced by 60° from its equilibrium position and released. Find the phase angle φ.

A) -15° B) -22.5° C) -30° D) -45°

Correct answer: B) -22.5° Explanation: The phase angle φ can be found using the equation φ = -arctan(ω/ω₀), where ω is the angular frequency and ω₀ is the initial angular frequency. The initial angular frequency ω₀ is given by ω₀ = g/L, where g is the acceleration due to gravity and L is the length of the pendulum. The angular frequency ω is given by ω = √(k/m), where k is the spring constant and m is the mass. Substituting the values, we get ω = √(4.9 N/m / 1 kg) = 2.2 rad/s and ω₀ = 9.8 m/s² / 2 m = 4.9 rad/s. The phase angle φ is then φ = -arctan(2.2 rad/s / 4.9 rad/s) = -22.5°.
Why the distractors are tempting: The distractors are tempting because they are close to the correct answer, but they are not the correct answer.

30-Second Cheat Sheet

• The differential equation for SHM is x''(t) + (k/m)x(t) = 0.
• The solution to the differential equation is x(t) = A cos(ωt + φ).
• The angular frequency is given by ω = √(k/m).
• The phase angle φ can be found using the equation φ = -arctan(ω/ω₀), where ω is the angular frequency and ω₀ is the initial angular frequency.
• The initial angular frequency ω₀ is given by ω₀ = g/L, where g is the acceleration due to gravity and L is the length of the pendulum.

Learning Path

1. Beginner foundation: Understand the basic concepts of kinematics, differential equations, and trigonometry.
2. Core rules: Learn the differential equation for SHM, the solution to the differential equation, and the formula for the angular frequency.
3. Practice: Practice solving SHM problems using the formula and the differential equation.
4. Timed drills: Practice solving SHM problems under timed conditions to improve your speed and accuracy.
5. Mock tests: Take mock tests to assess your knowledge and identify areas for improvement.

Related Topics

• Damped oscillations: This topic is related to SHM because it involves the study of oscillations that are damped by external forces.
• Forced oscillations: This topic is related to SHM because it involves the study of oscillations that are driven by external forces.
• Wave motion: This topic is related to SHM because it involves the study of waves that propagate through a medium.