# Physics Tutorial: Pendulums. Energy in Simple Harmonic Motion

In this Physics tutorial, you will learn:

• What are damped and sustainable SHM and where do they differ from each other?
• How to calculate energy in SHM?
• What is simple pendulum?
• How to calculate the period of a simple pendulum?
• What are the factors affecting the period of a simple pendulum?
• Which are the parameters used to study the motion of a simple pendulum?
• How to find the angular displacement and angular velocity in a simple pendulum?

## Introduction

How does a wall pendulum clock keeps swinging? Can it swing for long intervals without any human action? Why?

How would you make a pendulum swing at the same amplitude for long periods of time?

Do you know other examples of pendulums application in daily life? Could you provide a few of them?

Obviously, all the above examples have one thing in common: they all need a sustainable source of energy to keep the system operating for long periods. Therefore, energy in SHM is an irreplaceable element if we want to keep objects oscillations at the same parameters.

## Damped and Sustainable Simple Harmonic Oscillations

If we make a system oscillate, the amplitude of oscillations would decrease until it becomes zero if energy is not provided continuously to the system. In this case, the oscillations fade away with time. They are known as damped oscillations whose amplitude decreases with time. Look at the graph below. The "envelope" that surrounds the damped simple harmonic motion graph shows a decreasing exponential function of type x(t) = A0 × e - γ × t where A0 is the initial amplitude and γ is a constant. This is still a simple harmonic motion because the other quantities such as period and frequency don't change with time. This kind of SHM has the equation

x(t) = A0 × e-γ × t cos⁡(ω × t + φ)

As you see, a cosine function is used here instead of the sine function we have used in the previous tutorial because we are interested to point out the initial amplitude A0. Given that cos 0 = 1, it is better to use a cosine function here as it takes the maximum value at the initial instant (t = 0).

On the other hand, in sustainable SHM, the amplitude does not change with time. The envelope shows a horizontal function of the type x(t) = A0.

Therefore, the constant γ is zero as

e-γ × t = e-0 × t = e0 = 1

As a result, we have

A0 × e-γ × t = A0

and therefore, the equation of SHM becomes

x(t) = A0 × cos⁡(ω × t + φ)

This is the same equation as the one obtained when studying sustainable SHM in the previous tutorial. Cosine function is used only for convention but it can be replaced by sine function as well by making the proper arrangements. Thus, we can say that sustainable SHM are a special case of damped SHM.

## Energy in Simple Harmonic Motion

As stated earlier, sustainable SHM cannot exist without a continuous source of energy. Otherwise, oscillations will dump and eventually they fade away.

Any object or system moving in sustainable SHM possesses two kinds of energy: Kinetic Energy KE and Potential Energy PE. As an example, we can take again an oscillating spring with constant k as shown in the figure. Again, it is more suitable to use the cosine function to describe oscillations as initially the spring is stretched at maximum. We consider the duration of event very short, so there are no damping in oscillations. Therefore, the equation of this SHM is

x(t) = A × cos⁡(ω × t + φ)

where A is the amplitude of oscillation or the maximum distance of the object from the equilibrium position.

As explained in the previous topic, velocity is the first derivative of position in respect to the time. Thus, we obtain for v(t)

v(t) = -A × ω × sin⁡(ω × t + φ)

We will use these two equations to describe the energy in SHM. As explained in the Physics tutorial "Elastic Potential Energy", the PE of a stretched spring is calculated by the formula

PE = k × x2/2

or

PE = k × [A × cos ω × t + φ)]2/2
= k × A2 × cos2⁡(ω × t + φ)/2

When the spring is at the positions of maximum extension or compression, the system possesses only PE. Furthermore, this PE takes the maximum value, i.e. the cosine part becomes 1. Hence, we have

PEmax = k × A2/2

On the other hand, kinetic energy KE of the oscillating object is calculated by the formula

KE = m × v2/2

or

KE = m × [-A × ω × sin⁡(ω × t + φ) ]2/2
= m × A2 × ω2 × sin2⁡(ω × t + φ)/2

When the object is at the equilibrium position, the spring is neither stretched nor compressed. Thus, its PE = 0 and KE = KEmax. In this case, the sine part of formula becomes 1 and we obtain

KEmax = m × A2 × ω2/2

We assume the mechanical energy is conserved during a sustainable SHM. Thus, we have

ME = KE + PE = constant

In the first case shown in the above figure (figure a - maximum extension of spring, object at rest), we have

ME = KE + PEmax
= 0 + PEmax
= k × A2/2

In the second figure (figure b - object and spring at equilibrium position, moving due left at maximum speed), we have

ME = KEmax + PE
= KEmax + 0
= KEmax
= m × A2 × ω2/2

In the third figure (figure c - maximum compression of spring, object at rest), we have

ME = KE + PEmax
= 0 + PEmax
= k × (-A)2/2
= k × A2/2

In the fourth figure (figure d- object and spring at equilibrium position, moving due right at maximum speed), we have

ME = KEmax + PE
= KEmax + 0
= KEmax
= m × A2 × ω2/2

and in the fifth (and last) figure (figure e - maximum extension of spring, object at rest), we have

ME = KE + PEmax
= 0 + PEmax
= k × (-A)2/2
= k × A2/2

In all the other positions, the mechanical energy of the system is

ME = KE + PE
= m × A2 × ω2 × sin2⁡(ω × t + φ)/2 + k × A2 × cos2⁡(ω × t + φ)/2

### Example 1

A 400 g object is attached at the end of a spring of constant 4000 N/m. Initially the spring is stretched by 20 cm and then it is released. As a result, the system object-spring starts performing simple harmonic motion. Calculate:

1. The mechanical energy of the system
2. Speed and position of the object at t = 2 s
3. Kinetic energy of the object when it is at halfway between the equilibrium position and maximum compression, moving due left
4. Potential energy of the spring when it is at 5 cm from equilibrium position moving due right

### Solution 1

Clues:

m = 400 k = 0.4 kg

k = 4000 N/m

A = xmax = 20 cm = 0.2 m

a We can calculate the system's ME by finding PEmax as they are equal. Thus,

ME = PEmax
= k × A2/2
= 4000 × 0.22/2
= 80J

b Now, it is better to use the sine equation

x(t) = A × sin ⁡(ω × t + φ)

for position and the cosine equation

v(t) = A × ω × cos⁡(ω × t + φ)

for the velocity of the object because the object is initially at the maximum extension, i.e. φ = 1/4 of a cycle = 1/4 × 2π = π/2 This means that when t = 0 we have

x(0) = A × sin (ω × 0 + φ)
= A × sin φ
= A × sin π/2
= A × 1
= A

Thus, giving that

t = 2 s
ω = √k/m
= √4000/0.4
= √10 000

and

we obtain by substituting the above values:

x(2) = 0.2 × sin (100 × 2 + 1.57)
= 0.2 × sin 201.57
= 0.2 × 0.486
= 0.097 m
= 9.7 cm

This result means the object is 9.7 cm on the right of the equilibrium position as shown in the figure Also,

v(2) = -A × ω × cos (ω × t + φ)
v(2) = -0.2 × 100 × cos (100 × 2 + π/2)
= -20 × 0.874
= -17.47 m/s

c We can take x = -10 cm = - 0.1 m as the object is halfway between the equilibrium position and maximum compression. Thus, since the potential energy in this position is

PE = k × x2/2
= 4000 × 0.12/2
= 20 J

Hence, the kinetic energy at the given position is

KE = ME - PE
= 80 J - 20J
= 60J

d Now, we have x = 5 cm = 0.05 m for the position. Thus,

PE = k × x2/2
= 4000 × 0.052/2
= 5J

## Angular SHM. Simple Pendulum

As stated in the previous tutorial, there is another type of SHM besides the linear SHM. It is known as Angular Simple Harmonic Motion and its simplest case known as "simple pendulum".

A simple pendulum consists of a small mass m otherwise known as bob, attached on a light and non-elastic string (thread) of length L tied to a fixed and rigid support as in the figure below. When the bob is as in the position shown in the figure, the pendulum is not in equilibrium. The bob attracted by gravity tries to stay as low as possible and therefore, the pendulum tends to get a stable vertical position. That's why it swings several times before stopping at vertical position. Therefore, it is appropriate to study the time needed for one swing (period) in order to obtain useful information about the characteristics of this periodical motion. Let's denote by θ the initial angle formed by the string and the vertical position. There are two forces acting on the bob: the gravitational force Fg = m × g and the tension force T of the string. Since these two forces have not opposite directions, they don't cancel each other. Thus, a non-zero resultant force will appear in the swinging direction as in the figure below: This non-zero resultant force

F = - m × g × sin θ

is known as restoring force. It is responsible for the swinging process. (The sign minus indicates that the restoring force is in the opposite direction to the displacement x. It is not written in the figure above to avoid confusion).

The angle θ is taken very small for two reasons:

1. Because the effect of air resistance becomes negligible and the situation becomes very similar to harmonic motion. (As we know, harmonic motion is periodical as well).
2. Because for small angles sin θ ≈ θ (in radians). This helps us to simplify the study of pendulum.

Also, we know that the arc x from the initial position of the bob to the vertical position (which in this case represents the displacement), is

x = L × θ

Hence, we have

F = -m × g × sin θ
= -m × g × θ
= - m × g × x/L
= -(m × g/L) × x

This equation is similar to the Hooke's Law (spring equation) F = - k × x. The only difference is that here the role of x is played by mg/L.

From the equation of angular frequency in a SHM

ω = /T
= √k/m

we obtain for the period T of oscillations in a SHM

T = 2π × √m/k

Hence, we substitute k = mg/L in simple pendulum and thus, we obtain for the period of simple pendulum

T = 2π × √mmgL
= 2π × √L/g

From the last equation, we draw the following conclusions:

1. A pendulum with a shorter thread (L is small) has a smaller period of oscillation (a smaller numerator decreases the value of fraction). This means the bob in these conditions swings faster than if the pendulum had a longer thread.
2. If the pendulum had a higher starting point, the initial angle θ would be greater. Consequently, the displacement x would be greater as well. Also, the restoring force would be greater. However, in absence of air resistance, this would not affect the period of pendulum as period does not depend on the angle θ or displacement x.
3. If the bob had a greater weight (or mass), this would not affect the period of pendulum as period is independent from mass of the bob.
4. In the above explanation, we have considered the case when the bob is released without any initial speed. If it is pushed at a certain speed, the restoring force increases as we have used an additional force besides the gravity. Thus, the bob makes a greater arc (the angle θ increases). However, this doesn't affect the period of pendulum.

## Equation of Motion in a Simple Pendulum

The equation of SHM for a simple pendulum considers the rotational parameter of angular displacement θ as dependent variable and the time t as independent one, i.e. it is an equation of type

θ(t) = θ0 × cos⁡(ω × t)

where θ0 is the amplitude, i.e. the initial angle to the vertical direction, ω = 2π / T the angular frequency θ(t) is the angle to the vertical direction at a given instant t.

Obviously, we must use the same procedure in determining the angular velocity and angular acceleration as in the other cases of SHM, i.e. by taking the first derivative of θ(t) as equation of velocity v(t) and the second derivative of θ(t) [or the first derivative of v(t)] for the acceleration. Hence, we obtain

ω(t) = /dt
= d[θ0 × cos (ω × t)]/dt
= -θ0 × ω × sin (ω × t)

for angular velocity, and

α(t) = dv/dt
= d[θ0 × ω × sin⁡(ω × t) ]/dt
= -θ0 × ω2 × cos⁡(ω × t)
= -ω2 × θ(t)

for the angular acceleration in a simple pendulum.

## Energy in a Simple Pendulum

To know the maximum speed and the highest position of the bob, we use the law of energy conservation. Thus, if considering the bob at position 1 at rest, it is obvious that its kinetic energy is zero. Therefore, since h = L - L × cos θ, we have Mechanical Energy (1)
= Gravitational Potential Energy (1) + Kinetic Energy (1)
= Gravitational Potential Energy (1) + 0
= Gravitational Potential Energy (1)
= m × g × h
= m × g × (L - L × cos θ)
= m × g × L (1 - cos θ)

On the other hand, the kinetic energy is at maximum at the lowest point (position 2 or vertical position). In this position, gravitational potential energy is zero and all mechanical energy is due to kinetic energy. Thus, we can write

Mechanical Energy (2)
= Gravitational Potential Energy (2) + Kinetic Energy (2)
= 0 + Kinetic Energy (2)
= m × v2/2

These two energies are equal from the law of mechanical energy conservation. Thus, we can write

m × g × L (1 - cos θ) = m × v2/2

Simplifying mass we obtain,

g × L (1 - cos θ) = v2/2

Thus,

v2 = 2 × g × L × (1 -cos θ)

Or

v = √2 × g × L × (1 - cos θ )

After calculating v for known values of L and θ, we can find h as well, giving that

h = L × (1 - cos θ)

### Example 2

A 2 kg bob is attached at the end of a 1 m thread as shown in the figure. Calculate:

1. Initial height h0 of the bob
2. Period of this simple pendulum
3. Height of the bob at t = 3.2 s
4. Speed of the bob at t = 3.2 s

Take cos 200 = 0.94, sin 200 = 0.34 and g = 10 m/s2.

### Solution 2

a The initial height h0 of the bob is

h0 = L - L × cos θ
= L(1 - cos θ )
= 1 × (1-cos 200)
= 1 - 0.94
= 0.06 m

b The period of this pendulum is

T = 2π × √L/g
= 2 × 3.14 × √1/10
≈ 2 s

c To calculate the height of the bob at a given time we must calculate the angle formed by the thread to the vertical at that specific time. for this, we use the equation of SHM for pendulums,

θ(t) = θ0 × cos⁡(ω × t)

where t = 3.2 s, θ0 = 200 = π/9rad and ω = /T = /2 = π rad/s. Thus, we obtain

θ(3.2) = π/9 × cos (3.2π)
= π/9 × (-0.81)

Thus, at t = 3.2 s, the object is at

h = L - L × cos (-0.283)
= 1 - 1 × 0.96
= 0.04 m

d The speed of the bob at t = 3.2 s is calculated through the energetic method. Thus, since initially (at h0 = 0.06 m) the bob had only gravitational potential energy, the amount of mechanical energy of the bob during the entire motion is

GPE0 + KE0 = GPE(3.2) + KE(3.2)
m × g × h0 + 0 = m × g × h + m × v2/2
g × h0 = g × h + v2/2
10 × 0.06 = 10 × 0.04 + v2/2
0.6 = 0.4 + v2/2
v = √2 × (0.6 - 0.4)
= √0.4
= 0.63 m/s

## Summary

Damped oscillations are a kind of SHM whose amplitude decreases with time by a factor of e - γ × t (the envelope). The equation of damped oscillations is

x(t) = A0 × e - γ × t cos⁡(ω × t + φ)

On the other hand, in sustainable SHM, the amplitude does not change with time. The envelope shows a horizontal function of the type x(t) = A0. Therefore, the equation of sustainable SHM is

x(t) = A0 × cos⁡(ω × t + φ)

Cosine function is used only for convention but it can be replaced by sine function as well by making the proper arrangements.

Sustainable SHM's are a special case of damped SHM's. They cannot exist without a continuous source of energy. Otherwise, oscillations will dump and eventually they fade away.

Any object or system moving in sustainable SHM possesses two kinds of energy: Kinetic Energy KE and Potential Energy PE. The equation for the mechanical energy of an oscillating spring in SHM is

ME = KE + PE
= m × A2 × ω2 × sin2 (ω × t + φ)/2 + k × A2 × cos2⁡(ω × t + φ)/2

The maximum values for potential and kinetic energy in an oscillating spring (when the other term of the ME formula is zero), are

PEmax = k × A2/2

and

KEmax = m × A2 × ω2/2

There is another type of SHM besides the linear SHM. It is known as Angular Simple Harmonic Motion and its simplest case known as "simple pendulum".

A simple pendulum consists of a small mass m otherwise known as bob which is hung on a light and non-elastic string (thread) of length L attached to a fixed and rigid support.

The period T of oscillations in a spring doing SHM is

T = 2π × √m/k

In a simple pendulum we substitute k = mg/L and thus, we obtain for the period of simple pendulum

T = 2π × √L/g

The quantity

F = -m × g sin⁡θ

is known as the restoring force in a simple pendulum.

The equation of SHM for a simple pendulum considers the rotational parameter of angular displacement θ as dependent variable and the time t as independent one, i.e. it is an equation of type

θ(t) = θ0 × cos⁡(ω × t)

where θ0 is the amplitude, i.e. the initial angle to the vertical direction, ω = 2π / T the angular frequency θ(t) is the angle to the vertical direction at a given instant t.

Using the derivation rules, we obtain

ω(t) = /dt = θ0 × ω × sin⁡(ω × t)

for angular velocity, and

α(t) = dv/dt
= -θ0 × ω2 × cos⁡(ω × t)
= -ω2 × θ(t)

for the angular acceleration in a simple pendulum.

Mechanical energy in a simple pendulum is

ME = GPE + KE
= m × g × L (1 - cos θ) + m × v2/2

## Pendulums. Energy in Simple Harmonic Motion Revision Questions

1. The period of the pendulums K and P in the figure are 8s and 12s respectively. If both bobs are released simultaneously from the rest at the positions shown in the figure, what will their position be after 6s? 2. The thread of the simple pendulum shown in the figure forms initially an angle of 300 to the horizontal direction. When a 3 kg bob is released, it starts to swing in SHM. What is the bob's speed after 1.4 s?

1. 0
2. 2.32 m/s
3. 8.64 m/s
4. 0.45 m/s

3. Which statement(s) below is (are) correct regarding simple pendulum?

I. In absence of gravity the period of pendulum decreases but it doesn't become zero

II. No matter how heavy the bob is, the pendulum will swing equally for a fixed angle if the angle is small

III. In absence of air the period of pendulum depends only by thread's length and gravity

1. I only
2. I and III
3. II only
4. II and III