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# Simple Harmonic Motion (SHM)

## Introduction

A simple harmonic motion (SHM) is a common type of motion that occurs in various physical systems such as springs, pendulums, and oscillating objects. This motion is due to a repetitive pattern of back-and-forth motion around a central point. Therefore, the force acting on the system is proportional to the displacement from the equilibrium position. In this article, we will delve into the fundamental principles of SHM, its mathematical representation, and some practical examples of its applications.

### What is Simple Harmonic Motion?

Simple harmonic motion (SHM) is a type of motion that occurs when the restoring force acting on a system is proportional to the displacement of the system from its equilibrium position. In other words, the force acting on the system is always directed towards the equilibrium position. Additionally, we can say that the magnitude of the force increases as the system moves farther from the equilibrium position. The most common example of SHM is the motion of a spring-mass system.

We can also define simple harmonic motion as the to and fro motion of a particle about an equilibrium position.

## Types of SHM?

We will explain the two types of simple harmonic motion in form of linear simple harmonic motion(LSHM), and angular simple harmonic motion(ASHM)

### Linear Simple Harmonic Motion (LSHM)

Linear Simple Harmonic Motion (LSHM) is a specific type of simple harmonic motion in which an object oscillates back and forth in a straight line along a single axis. Examples of LSHM include the motion of a mass on a horizontal spring or the motion of a simple pendulum that swings back and forth along a straight line.

We can describe the motion of an object undergoing LSHM by its displacement from its equilibrium position as a function of time. The formula that we can use for this displacement is given by:

x(t) = A*cos(ωt + φ)

Where

x is the displacement of the object from its equilibrium position.

A is the amplitude of the motion (the maximum displacement from the equilibrium position).

ω is the angular frequency of the motion (measured in radians per second).

t is the time, and

φ is the phase angle (a constant that determines the position of the object at time t=0).

#### Velocity of LSHM

Moreover, we can describe the velocity of the object undergoing LSHM by a mathematical expression. The velocity is the first derivative of the displacement with respect to time. Therefore, we can write the formula as

v(t) = -Aωsin(ωt + φ)

Where v is the velocity of the object at time t.

#### Acceleration of LSHM

Similarly, the acceleration of the object can be described by a mathematical expression. The acceleration is the second derivative of the displacement with respect to time, and it is given by:

a(t) = -Aω2cos(ωt + φ)

Where a is the acceleration of the object at time t.

The displacement, velocity, and acceleration of an object undergoing LSHM are all sinusoidal functions with the same angular frequency ω, but different amplitudes and phase angles. The motion of the object is periodic. It can also be characterized by its period (the time it takes to complete one cycle) and its frequency (the number of cycles per unit time).

### Angular Simple Harmonic Motion (ASHM)

Angular Simple Harmonic Motion (ASHM) is a type of simple harmonic motion in which an object rotates back and forth around a fixed axis with a restoring torque proportional to its angular displacement. Examples of ASHM include the motion of a pendulum that swings in a circular arc or the motion of a rotating disk that wobbles around its axis.

#### ASHM as a Function of Time

We can also describe the motion of an object undergoing ASHM by its angular displacement from its equilibrium position as a function of time. The equation for this displacement is

θ(t) = θcos(ωt + φ)

Where θ is the angular displacement of the object from its equilibrium position.

θ is the amplitude of the motion (the maximum angular displacement from the equilibrium position).

ω is the angular frequency of the motion (measured in radians per second).

t is the time, and

φ is the phase angle (a constant that determines the position of the object at time t=0).

#### Formula For ASHM

We can also use mathematical expression to describe the angular velocity of the object undergoing ASHM. The angular velocity is the first derivative of the angular displacement with respect to time, and it is given by:

ω(t) = – θωsin(ωt + φ)

Where ω is the angular velocity of the object at time t.

#### Angular Acceleration

Similarly, the angular acceleration is the second derivative of the angular displacement with respect to time, and it is given by:

α(t) = – θω2cos(ωt + φ)

Where α is the angular acceleration of the object at time t.

The angular displacement, velocity, and acceleration of an object undergoing ASHM are all sinusoidal functions with the same angular frequency ω, but different amplitudes and phase angles. The motion of the object is periodic and we can identify it by its period (the time it takes to complete one cycle) and its frequency (the number of cycles per unit time).

### What are the Characteristics of Simple Harmonic Motion?

The key characteristics of SHM are:

Oscillatory Motion: The motion of the system is repetitive, and the system oscillates back and forth around the equilibrium position.

Periodic Motion: The motion of the system is periodic, meaning that it repeats itself after a fixed interval of time called the period.

Amplitude: The maximum displacement of the system from the equilibrium position is called the amplitude.

Frequency: The frequency of the motion is the number of oscillations per unit time, measured in Hertz (Hz).

Phase: The phase of the motion is the position of the system at any given time with respect to its equilibrium position.

## Hooke’s Law

Hooke’s Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, we can express this as:

F = -kx

where F is the restoring force. x is the displacement from the equilibrium position. k is the spring constant.

### Simple Harmonic Motion Formula

Using Hooke’s Law, we can derive the equation of motion for a spring-mass system undergoing SHM. We can now write the simple harmonic formula as:

x = A cos(ωt + φ)

where

x is the displacement of the mass from the equilibrium position.

A is the amplitude.

ω is the angular frequency,

t is time, and

φ is the phase constant.

## Energy in Simple Harmonic Motion

### Kinetic Energy

The kinetic energy of a system undergoing SHM is given by:

K = (1/2)mv2

where m is the mass of the system, and v is the velocity of the system.

### Potential Energy

The potential energy of system undergoing SHM is given by:

U = (1/2)kx2

where k is the spring constant, and x is the displacement of the system from the equilibrium position.

### Total Mechanical Energy

The total mechanical energy of the system is the sum of the kinetic (K) and potential (U) energy. The formula is

E = K + U

## Ten 10 Examples of Simple Harmonic Motion

### 1. Mass-Spring System

One of the most common examples of SHM is the motion of a spring-mass system. In this system, we attach a mass to a spring and allowed to oscillate back and forth. Therefore, the force acting on the system is the force exerted by the spring, which is given by Hooke’s Law. Hence, we determine the period of the motion by the mass of the system and the spring constant.

### 2. Simple Pendulum

Another example of SHM is the motion of a simple pendulum. In this system, we attach the weight of a string and allow it to swing back and forth. The restoring force acting on the system is gravity. Additionally, we can determine the period of the motion by the length of the pendulum and the acceleration due to gravity.

### 3. Simple Harmonic Oscillator

A simple harmonic oscillator is a system that undergoes SHM without any external forces acting on it. Examples of simple harmonic oscillators include a mass attached to a spring with no friction, or a pendulum with no air resistance.

### 4. Tuning fork

A tuning fork is a metal instrument that vibrates at a specific frequency when struck. The oscillatory motion of a tuning fork is also another example of simple harmonic motion.

### 5. Waves on a string

When we pluck or struck a string. It produces waves that propagate back and forth along the length of the string. The motion of waves on a string is an example of simple harmonic motion.

### 7. Swinging door

A swinging door is another example of simple harmonic motion. This is under the condition of door’s ability to swing to and from. As it swings, its displacement from the equilibrium position is proportional to the force of gravity acting on it.

### 8. Bouncing ball

When you drop or throw a ball, the force of gravity makes it to bounces up and down. You can now approximate the motion of the ball as simple harmonic motion. This is because its displacement from its resting position is proportional to the gravitational force acting on it.

### 9. Lateral oscillation of a moving car

When a car is moving on a rough road, it may experience a lateral oscillation due to the bumps in the road. We can now see that the motion of the car undergoes simple harmonic motion.

### 10. Torsional pendulum

A torsional pendulum is a device that consists of a suspended weight that can rotate about a vertical axis. The motion of the pendulum is an example of simple harmonic motion. Thus, its displacement from its resting position is proportional to the restoring torque acting on it.

## What are the Differences Between Simple Harmonic Motion, Period, and Oscillatory Motion

Here is a table to differentiate between SHM, period, and motion

## 5. Simple Harmonic Motion Worked Examples

### Finding the Amplitude, Period, and Frequency

To find the amplitude, period, and frequency of a system undergoing SHM, we can use the equation:

T = 2π/ω

where T is the period, and ω is the angular frequency.

We can find the amplitude by measuring the maximum displacement of the system from the equilibrium position.

Therefore, to find the frequency we can use the formula:

f = 1/T

where f is the frequency.

### Finding the Maximum Velocity and Acceleration

The maximum velocity and acceleration of a system undergoing SHM can be found using the equations:

vmax = Aω

amax = Aω2

where vmax is the maximum velocity, amax is the maximum acceleration. A is the amplitude, and ω is the angular frequency.

### Finding the Total Energy and Potential Energy

The formula for total energy and potential energy of a system undergoing SHM is:

E = (1/2)mv2 + (1/2)kx2

U = (1/2)kx2

where E is the total mechanical energy, m is the mass of the system. v is the velocity of the system, k is the spring constant. Additionally, x is the displacement of the system from the equilibrium position.

#### Solving Problems

##### Question 1

A spring makes 60 revolutions in 15 seconds. Find the period and the frequency of the vibration

###### Solution

Data

Time = 15 s

Number of vibrations = 60

Period, T = Time / Number of vibrations = 15 / 60 = 0.25 s

The frequency, f = 1 / T = 1 / 0.25 = 4 Hz

Therefore, the period of the motion is 0.25 seconds, while frequency is 4 Hz.

##### Question 2

A mass of 0.5 kg is attached to a spring with a spring constant of 50 N/m. If the mass is displaced 0.2 m from equilibrium and released, what is the maximum speed attained by the mass during the resulting motion?

###### Solution

Using the equation for the maximum speed of simple harmonic motion:

vmax = Aω

where A is the amplitude and ω is the angular frequency.

The angular frequency can be found using the equation:

ω = √(k/m)

where k is the spring constant and m is the mass.

Substituting in the given values:

ω = √(50 N/m / 0.5 kg)

ω = 10 rad/s

The amplitude is given as 0.2 m, so:

vmax = 0.2 m * 10 rad/s

We will now have vmax= 2 m/s

Therefore, the maximum speed attained by the mass during the motion is 2 m/s.

##### Question 3

A mass of 1 kg is attached to a spring with a spring constant of 80 N/m. If the mass is displaced 0.1 m from equilibrium and released, what is the displacement of the mass after 1 second?

###### Solution

Using the equation for the displacement of simple harmonic motion:

x(t) = A * cos(ωt + φ)

where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase angle.

The angular frequency can be found using the equation:

ω = √(k/m)

where k is the spring constant and m is the mass.

Substituting in the given values:

ω = √(80 N/m / 1 kg)

ω = 8.94 rad/s

The amplitude is given as 0.1 m, so:

x(t) = 0.1 m x cos(8.94 rad/s x t + φ)

To find the phase angle, we need to use the initial conditions. At t = 0, the mass is displaced 0.1 m from equilibrium and moving in the negative direction, so:

x(0) = – 0.1 m

v(0) = 0 m/s

Differentiating the displacement equation to find the velocity:

v(t) = -Aωsin(ωt + φ)

v(0) = -0.1 m x 8.94 rad/s x sin(φ) = 0 m/s

sin(φ) = 0

φ = 0 or π

Since the mass is initially moving in the negative direction, the correct phase angle is:

φ = π

Substituting in the values for t = 1 second:

x(1) = 0.1 m x cos(8.94 rad/s x 1 s + π)

x(1) = -0.054 m

Therefore, the displacement of the mass after 1 second is -0.054 m.

##### Question 4

A mass of 2 kg is attached to a spring with a spring constant of 200 N/m. If the mass is displaced 0.05 m from equilibrium and released, what is the maximum acceleration of the mass during the resulting motion?

###### Solution

Using the equation for the maximum acceleration of simple harmonic motion:

amax = Aω2

where A is the amplitude and ω is the angular frequency.

The angular frequency can be found using the equation:

ω = √(k/m)

where k is the spring constant and m is the mass.

Substituting in the given values:

ω = √(200 N/m / 2 kg)

ω = 10 rad/s

The amplitude is given as 0.05 m, so:

amax = 0.05 m x 102 rad/s2

amax = 5 m/s2

Therefore, the maximum acceleration of the mass during the motion is 5 m/s2.

##### Question 5

A mass of 0.2 kg is attached to a spring with a spring constant of 40 N/m. If the mass is displaced 0.1 m from equilibrium and released, what is the kinetic energy of the mass when it passes through the equilibrium position?

###### Solution

We will apply the formula for conservation of energy:

E = KE + PE

where E is the total energy, KE is the kinetic energy, and PE is the potential energy.

At the equilibrium position, the potential energy is zero and the total energy is equal to the kinetic energy:

KE = E

The total energy can be found using the equation E = (1/2) k A2

where k is the spring constant and A is the amplitude of the motion,

and

E = (1/2) mv2

where m is the mass and v is the velocity of the mass when it passes through the equilibrium position.

The amplitude can be found using the equation:

A = xmax

where xmax is the maximum displacement from equilibrium.

Substituting in the given values:

A = 0.1 m

k = 40 N/m

m = 0.2 kg

v = ?

Using the equation for the velocity of simple harmonic motion:

v(t) = -Aωsin(ωt + φ)

At the equilibrium position, t = 0 and sin(φ) = 0, so:

v = 0 m/s

Substituting in the values for E:

E = (1/2)kA2

Will give us

E = 1/2 x 40 x (0.1)2

Which is equivalent to

E = 0.2 J

Therefore, the kinetic energy of the mass when it passes through the equilibrium position is 0.2 J.

## Summary

Simple harmonic motion is a common type of motion that occurs in various physical systems. The motion is due to the repetitive pattern of back-and-forth motion around a central point, where the force acting on the system is proportional to the displacement from the equilibrium position. We can use Hooke’s Law to derive the equation of motion for a spring-mass system undergoing SHM. Additionally, we can express the energy in the system in terms of kinetic and potential energy. Examples of systems undergoing SHM include spring-mass systems, pendulums, and simple harmonic oscillators.

## Frequently Asked Questions (FAQs)

### Question: What is the difference between simple harmonic motion and periodic motion?

Answer: Simple harmonic motion is a type of periodic motion where the motion of the system is due to the repetitive pattern of back-and-forth motion around a central point. Additionally, the force acting on the system is proportional to the displacement from the equilibrium position.

### Question: Can simple harmonic motion occur without a restoring force?

Answer: No, simple harmonic motion requires a restoring force that is proportional to the displacement from the equilibrium position.

### Question: How is the period of a system undergoing simple harmonic motion affected by the mass of the system?

Answer: The period of a system undergoing simple harmonic motion is inversely proportional to the square root of the mass of the system. Thus, this means that as the mass of the system increases, the period of the motion will also decrease.

### Question: What is the role of the spring constant in a spring-mass system undergoing simple harmonic motion?

Answer: The spring constant determines the strength of the restoring force acting on the system. Hence, it is directly proportional to the displacement from the equilibrium position. Moreover, a higher spring constant results in a stronger restoring force, and therefore a shorter period of the motion.

### Question: Can the amplitude of a system undergoing simple harmonic motion be negative?

Answer: Yes, the amplitude of a system undergoing simple harmonic motion can be negative if the system starts from a position below the equilibrium position. In this case, the system oscillates below the equilibrium position rather than above it.

### Question: How does damping affect simple harmonic motion?

Answer: Damping, or the dissipation of energy from a system, can cause the amplitude of simple harmonic motion to decrease over time. This results in a decrease in the period and frequency of the motion. In heavily damped systems, the motion can cease altogether.

### Question: What is the relationship between the period and frequency of a system undergoing simple harmonic motion?

Answer: The period and frequency of a system undergoing simple harmonic motion are inversely proportional. That is, as the period of the motion increases, the frequency decreases, and vice versa.

### Question: What is the equation for the velocity of a system undergoing simple harmonic motion?

Answer: The formula for velocity of a system undergoing simple harmonic motion is

v(t) = -Aωsin(ωt + φ)

where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase angle.

### Question: What is the equation for the acceleration of a system undergoing simple harmonic motion?

Answer: The equation for acceleration of a system undergoing simple harmonic motion is

a(t) = -Aω2cos(ωt + φ)

where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase angle.

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