Frequency & Period of SHM (College Board AP® Physics 1: Algebra-Based)

Study Guide

Dan Mitchell-Garnett

Written by: Dan Mitchell-Garnett

Reviewed by: Caroline Carroll

Frequency & period

Relationship between frequency and period

  • A cycle of a system in SHM is when the system performs a complete oscillation and returns to its initial position

One complete cycle of a pendulum

Diagram showing a pendulum's motion in one oscillation, marked at five positions with arrows indicating the direction of movement. Caption reads: "1 Oscillation (T)".
A cycle is complete when the system returns to the initial state. In this case, the initial state is the mass in the equilibrium position while moving left
  • The period of a system in SHM is the time taken for one complete cycle to occur

  • The frequency of a system is the number of cycles which are completed in one second

  • The period and frequency are related by the following equation:

T space equals space 1 over f

  • Where:

    • T = period of an oscillation, measured in straight s

    • f = frequency of the oscillations, measured in Hz

Effect of amplitude on period

  • Amplitude is defined as:

the maximum displacement of the object in SHM from its equilibrium position

  • One characteristic feature of SHM is that amplitude does not affect period

  • Increasing or decreasing the amplitude of a swinging pendulum or mass-spring system, for example, will not alter the period or frequency of that system

Period of an ideal spring

  • An object-ideal spring oscillator can be a horizontal or vertical oscillating system in which an object with mass is attached to an ideal spring

Object-ideal spring systems

Diagram showing a mass (m) attached to a spring (k) in two configurations: vertical on the left and horizontal on the right, with labels "VERTICAL" and "HORIZONTAL".
An object-ideal spring system can oscillate vertically or horizontally
  • The period of an object-ideal spring system is given by:

T subscript s space equals space 2 straight pi square root of m over k end root

  • Where:

    • T subscript s = period of the ideal spring, measured in straight s

    • m = mass of the object, measured in kg

    • k = spring constant of the ideal spring, measured in straight N divided by straight m

  • This equation applies to both vertical and horizontal object-ideal spring systems

Worked Example

Diagram of a mass m connected to two springs in series with constants kx and ky on a horizontal surface.

The horizontal object-ideal spring system above has two springs attached in series to an object with mass m. The springs have spring constants k subscript x and k subscript y.

An effective spring constant is the spring constant of a single spring that would behave the same as the two springs combined.

You may assume that each spring experiences the same force.

Derive an expression for the period of the system in terms of m, k subscript x and k subscript y by finding the system's effective spring constant.

Answer:

Step 1: Analyze the scenario

This system is very similar to a standard horizontal object-ideal spring system, however there are now two springs instead of one.

To answer the question, find an expression for the effective spring constant. The question states that force is evenly distributed between the springs. This means the same value for force can be used for both springs.

Writing the object's displacement in terms of the spring constants will be

Step 2: Find the effective spring constant of this system

  • When the object is at a displacement straight capital delta x from the equilibrium position, this displacement is the sum of the extension of each spring:

straight capital delta x space equals space straight capital delta x subscript x space plus straight capital delta space x subscript y

  • Where:

    • straight capital delta x subscript x is the extension of spring x, measured in straight m

    • straight capital delta x subscript y is the extension of spring y, measured in straight m

  • The extensions can be written in terms of force and spring constants using Hooke's law:

F subscript x space equals space minus k subscript x straight capital delta x subscript x

straight capital delta x subscript x space equals space minus F subscript x over k subscript x

straight capital delta x subscript y space equals space minus F subscript y over k subscript y

  • Recall that force is evenly distributed, so the forces exerted by each spring are equal:

F subscript x space equals space F subscript y space equals space F

straight capital delta x space equals space straight capital delta x subscript x space plus straight capital delta space x subscript y space equals space open parentheses negative F over k subscript x close parentheses space plus space open parentheses negative F over k subscript y close parentheses

  • Recall Hooke's law is:

F space equals space minus k straight capital delta x

  • Write this equation in the format of Hooke's law for a single spring:

straight capital delta x space equals space minus open parentheses 1 over k subscript x space plus space 1 over k subscript y close parentheses F

F space equals space minus open parentheses fraction numerator 1 over denominator 1 over k subscript x space plus space 1 over k subscript y end fraction close parentheses straight capital delta x space equals space minus open parentheses fraction numerator k subscript x k subscript y over denominator k subscript x space plus space k subscript y end fraction close parentheses straight capital delta x

  • From this equation, it can be determined that the effective spring constant, k subscript e, is:

k subscript e space equals space fraction numerator k subscript x k subscript y over denominator k subscript x space plus space k subscript y end fraction

Step 3: Substitute this into the equation for the period of an object-ideal spring system

  • Find the equation for the period of this system from the equation sheet:

T subscript s space equals space 2 straight pi square root of m over k end root

  • Substitute k subscript e in place of k in this equation to produce the final equation:

T subscript s space equals space 2 straight pi square root of fraction numerator m over denominator fraction numerator k subscript x k subscript y over denominator k subscript x space plus space k subscript y end fraction end fraction end root

  • This can be simplified to the following:

T subscript s space equals space 2 straight pi square root of fraction numerator m open parentheses k subscript x italic space plus space k subscript y close parentheses over denominator k subscript x k subscript y end fraction end root

Examiner Tips and Tricks

This equation for this system's period appears on your equation sheet:

T subscript s space equals space 2 straight pi square root of m over k end root

You do not need to know the derivation. You may, however, need to rearrange it to find frequency or use it to derive new equations.

Additionally, this equation looks similar to the period for a pendulum, make sure you know which equation applies to which system. This equation features the spring constant, which may help to remind you.

Period of a simple pendulum

  • A pendulum consists of a mass suspended from a string which oscillates side to side

  • The period of an oscillating pendulum is given by:

T subscript p space equals space 2 straight pi square root of l over g end root

  • Where:

    • T subscript p = the period of one cycle of the pendulum, measured in straight s

    • l = length of the string, measured in straight m

    • g = acceleration due to gravity at Earth’s surface, measured in straight m divided by straight s squared

  • This equation applies to a pendulum which is displaced by a small angle only

Worked Example

Which equation correctly shows the frequency of a pendulum of length l oscillating through a small angle?

A: f subscript p space equals space 2 straight pi square root of g over l end root

B: f subscript p space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of l over g end root

C: f subscript p space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of g over l end root

D: None of the above

Answer: C

Step 1: Analyze the scenario

  • A pendulum of length l is oscillating through a small angle

  • This means it can be modeled as SHM

Step 2: Recall the relevant equations

  • From the equation sheet, the relevant equation for a pendulum's period is:

T subscript p space equals space 2 straight pi square root of l over g end root

  • Recall that frequency is the reciprocal of period:

f space equals space 1 over T

Step 3: Apply the specific conditions

  • Substitute the period of a pendulum into this equation for frequency:

f subscript p space equals space fraction numerator 1 over denominator 2 straight pi square root of l over g end root end fraction

  • Simplify this expression:

f subscript p space equals space fraction numerator 1 over denominator 2 straight pi end fraction times fraction numerator 1 over denominator square root of l over g end root end fraction space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of g over l end root

  • Therefore the correct answer is C

Examiner Tips and Tricks

Again, this equation is given on the equation sheet and you will not be expected to derive it in your exam. You may, however, be required to demonstrate that the restoring torque in this system arises from a component of weight acting perpendicular to the string.

The weight force of the mass in a pendulum system acts downwards. A component of this acts perpendicular to the line of the string at the extreme end of the oscillation. This has the value m g sin θ.

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Dan Mitchell-Garnett

Author: Dan Mitchell-Garnett

Expertise: Physics Content Creator

Dan graduated with a First-class Masters degree in Physics at Durham University, specialising in cell membrane biophysics. After being awarded an Institute of Physics Teacher Training Scholarship, Dan taught physics in secondary schools in the North of England before moving to Save My Exams. Here, he carries on his passion for writing challenging physics questions and helping young people learn to love physics.

Caroline Carroll

Author: Caroline Carroll

Expertise: Physics Subject Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.