Frequency and Period of SHM

Noah Garcia

8 min read

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Study Guide Overview

This study guide covers Simple Harmonic Motion (SHM), focusing on oscillations and the relationship between restoring force and displacement. Key concepts include period, frequency, and angular frequency, along with their relationships. It explores SHM in spring-mass systems and simple pendulums, including their respective period formulas and how mass, spring constant, length, and gravity affect these. The guide also emphasizes connections between these systems and provides practice questions covering formula application, conceptual understanding, and graphical analysis, with a final exam focus including energy in SHM.

#AP Physics C: Mechanics - Simple Harmonic Motion (SHM) 🎢

Hey there, future AP Physics C master! Let's dive into Simple Harmonic Motion (SHM). Think of it as the physics of things that go back and forth, like a spring or a swing. This guide is designed to make sure you're not just memorizing formulas, but truly understanding how SHM works. Let's get started!

#What is Simple Harmonic Motion?

At its core, SHM is all about oscillations where the restoring force is directly proportional to the displacement. Imagine a spring – the more you stretch or compress it, the harder it pulls back. That's SHM in action!

Restoring Force: The force that brings the object back to its equilibrium position.
Proportionality: The restoring force is directly proportional to the displacement from equilibrium.

#Key Concepts: Frequency, Period, and Angular Frequency

These three amigos are essential for describing any SHM system. Let's break them down:

Period (T): The time it takes for one complete oscillation (back-and-forth motion). Think of it as the duration of one full swing. Measured in seconds (s).
Frequency (f): The number of complete oscillations per second. It's how often the motion repeats. Measured in Hertz (Hz), where 1 Hz = 1 cycle per second.
Angular Frequency (ω): Measures the rate of change of the angle in radians per second. It's like frequency but in angular terms. Measured in radians per second (rad/s).

Key Concept

These three are related by:

$T = \frac{1}{f} = \frac{2\pi}{\omega}$

Memory Aid

Remember this: "The faster it wiggles, the shorter the period!" (High frequency = short period, low frequency = long period)

#SHM in Action: Spring-Mass Systems

#Period of a Spring-Mass System

For a mass attached to a spring, the period of oscillation ( $T_s$ ) is given by:

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

m: Mass (kg) attached to the spring.
k: Spring constant (N/m), a measure of the spring's stiffness.

Quick Fact

Heavier mass = longer period. Stiffer spring = shorter period. 💡

Exam Tip

Notice that the period of a spring-mass system doesn't depend on gravity! It only cares about mass and spring stiffness.

#Visualizing Spring-Mass SHM

Caption: A visual representation of a spring-mass system undergoing SHM. The mass oscillates back and forth, demonstrating the periodic motion.

#SHM in Action: Simple Pendulums

#Period of a Simple Pendulum

The period of a simple pendulum ( $T_p$ ) is given by:

$T_p = 2\pi\sqrt{\frac{l}{g}}$

l: Length of the pendulum (m), measured from the pivot point to the center of mass of the bob.
g: Acceleration due to gravity (approximately 9.8 m/s² on Earth) 🌍.

Quick Fact

Longer pendulum = longer period. Stronger gravity = shorter period. 💡

Exam Tip

The period of a simple pendulum doesn't depend on the mass of the bob! It only cares about the length and gravity.

#Visualizing Pendulum SHM

Caption: A visual representation of a simple pendulum undergoing SHM. The pendulum swings back and forth, demonstrating the periodic motion.

#Connections Between Spring and Pendulums

AP loves to test how well you can connect concepts. Notice that both spring-mass systems and pendulums have periods that are proportional to the square root of a physical parameter (mass for springs, length for pendulums) and inversely proportional to the square root of another (spring constant for springs, gravity for pendulums).

Common Mistake

Don't confuse the formulas for spring-mass systems and pendulums. Spring period depends on mass and spring constant, while pendulum period depends on length and gravity.

#Final Exam Focus

Okay, here's the lowdown on what to focus on for the exam:

Master the Formulas: Know the period equations for both spring-mass systems and pendulums inside and out. Practice applying them in different scenarios.
Understand the Relationships: Be able to explain how changes in mass, spring constant, length, and gravity affect the period and frequency of SHM.
Graphical Analysis: Be prepared to analyze graphs of displacement, velocity, and acceleration versus time for SHM. Understand the phase relationships between these quantities.
Energy in SHM: Remember that energy is constantly being exchanged between kinetic and potential energy in SHM. The total energy remains constant (if no friction).
Conceptual Questions: Expect questions that test your understanding of the underlying principles of SHM, not just your ability to plug numbers into formulas.

Exam Tip

When solving SHM problems, always start by identifying the type of system (spring-mass or pendulum) and then choose the appropriate formulas. Draw diagrams to visualize the motion. Pay close attention to units.

#Practice Questions

Practice Question

Multiple Choice Questions

A mass m is attached to a spring with spring constant k. If the mass is doubled and the spring constant is halved, what happens to the period of oscillation? (A) It is halved. (B) It is doubled. (C) It is quadrupled. (D) It is multiplied by $\sqrt{2}$ . (E) It is multiplied by 2 $\sqrt{2}$ .
A simple pendulum has a length L and a period T on Earth. If the pendulum is taken to a planet with twice the gravitational acceleration, what is the new period? (A) $T/2$ (B) $T/\sqrt{2}$ (C) $T$ (D) $\sqrt{2}T$ (E) $2T$

Free Response Question

A 0.5 kg mass is attached to a spring with a spring constant of 20 N/m. The mass is pulled 0.1 m from its equilibrium position and released. Assume there is no friction.

(a) Calculate the period of oscillation of the mass-spring system. (2 points) (b) Calculate the maximum speed of the mass during its oscillation. (3 points) (c) Determine the total energy of the system. (2 points) (d) If the same mass is used to form a pendulum, what should be the length of the pendulum to have the same period as the spring-mass system? (3 points)

Answer Key

Multiple Choice

(E)
- The period of a spring is given by (T = 2\pi\sqrt{\frac{m}{k}}). If mass is doubled and spring constant is halved, the new period is (T' = 2\pi\sqrt{\frac{2m}{k/2}} = 2\pi\sqrt{\frac{4m}{k}} = 2\pi\sqrt{\frac{m}{k}}\times 2 = 2T). Therefore, the period is multiplied by 2. 2. (B)
- The period of a pendulum is given by (T = 2\pi\sqrt{\frac{L}{g}}). If the gravitational acceleration is doubled, the new period is (T' = 2\pi\sqrt{\frac{L}{2g}} = \frac{1}{\sqrt{2}}2\pi\sqrt{\frac{L}{g}} = \frac{T}{\sqrt{2}}).

Free Response Question

(a) Period of oscillation:

*   (T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.5}{20}} = 2\pi\sqrt{0.025} = 2\pi(0.158) = 0.99 s \approx 1.0 s) (2 points)

(b) Maximum speed:

*   The potential energy at maximum displacement is equal to the kinetic energy at equilibrium position. ( \frac{1}{2} k A^2 = \frac{1}{2} m v^2 \). (1 point)
*   (v = \sqrt{\frac{k}{m}}A = \sqrt{\frac{20}{0.5}}(0.1) = \sqrt{40}(0.1) = 0.63 m/s) (2 points)

*   (E = \frac{1}{2} k A^2 = \frac{1}{2} (20)(0.1)^2 = 0.1 J) (2 points)

(d) Length of the pendulum:

*   The period of the pendulum is (T = 2\pi\sqrt{\frac{l}{g}}). (1 point)
*   We want the period of the pendulum to be equal to the period of the spring. So (1 = 2\pi\sqrt{\frac{l}{9.8}}). (1 point)
*   (l = \frac{9.8}{(2\pi)^2} = 0.25 m) (1 point)

Keep up the great work, and you'll ace this exam! You've got this! 💪