Simple Harmonic Motion

Simple Harmonic Motion (SHM) is an oscillatory motion in which the acceleration of a particle at any position is directly proportional to its displacement from the mean position and is directed towards the mean position.

There are many examples of Simple Harmonic Motion (SHM) around us.

All simple harmonic motions are both oscillatory and periodic in nature; however, not all oscillatory or periodic motions qualify as SHM. Oscillatory motion is often called harmonic motion, and it is among all types of harmonic motions. The main characteristics of SHM are given below:

• SHM is a periodic and oscillatory motion.
• A restoring force always acts opposite to the displacement from the mean position.
• Acceleration is directly proportional to displacement and directed towards the mean position.
• Velocity is maximum at the mean position and zero at the extreme positions.
• Total mechanical energy remains constant, with continuous conversion between kinetic and potential energy.

Terminology Related to Simple Harmonic Motion

Mean Position

In simple harmonic motion, the position of the object where there is no restoring force acting on it is the mean position. In other words, the point about which the object moves between its extreme positions is called the mean position of the object. The mean position is sometimes referred to as the equilibrium position as well.

Amplitude

The amplitude of a particle in SHM is its maximum displacement from its equilibrium or mean position, and as displacement is a vector quantity, its direction is always away from the mean or equilibrium position. The SI unit of amplitude is the meter, and all the other units of length can also be used for this.

Frequency

The frequency of SHM is the number of oscillations performed by a particle per unit of time. The SI unit of frequency is hertz, or r.p.s. (rotations per second), and is given by:

f =\frac{1}{T}

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

Time Period

For a particle performing SHM, the time period is the amount of time it takes to complete one complete oscillation. As a result, the time period, or simply period, of SHM is the shortest time before the motion repeats itself.

T = \frac{2\pi}{\omega}

 where ω i= Angular frequency & T is the Time period.

Phase

The phase of SHM represents the magnitude and direction of particle displacement at any instant of the motion, which is its state of oscillation.

The expression for a particle's position as a function of time and angular frequency is as follows:

x = A sin (ωt + ϕ)

where (ωt + ϕ) is the phase of particle.

Phase Difference

For two particles performing Simple Harmonic Motion (SHM), the phase difference is defined as the difference between their phase angles. It is denoted by Δϕ. Mathematically, the phase difference is the difference between the phase angles of two particles moving in SHM with respect to the mean position.

For example, if two particles are performing SHM with the same angular frequency ω, and their displacement functions are

x₁ = A sin(ωt + ϕ₁)

x₂ = A sin(ωt + ϕ₂​)

then the phase difference between them is

Δϕ = ϕ₁ − ϕ₂

Two vibrating particles with the same angular frequency are said to be in the same phase if the phase difference between them is an even multiple of π, i.e.,

Δϕ = 2nπ

Where, n = 0, 1, 2, 3, 4, . . . 

Two vibrating particles are said to be in opposite phase if the phase difference between them is an odd multiple of π, i.e.,

Δϕ = (2n+1) π

where , n = 0, 1, 2, ,3. . . .

Types

Linear Simple Harmonic Motion

When a particle moves back and forth along a straight line about a fixed point, the motion is known as linear simple harmonic motion (SHM).

Examples of linear SHM include the oscillation of a liquid column in a U-tube, the motion of a simple pendulum for very small angular displacements, and the small vertical vibrations of a mass suspended from an elastic string.

Conditions for Linear Simple Harmonic Motion

The restoring force or acceleration acting on the particle must always be proportional to the particle's displacement and directed toward the equilibrium position.

F ∝ - X

a ∝ -x

where 

• F is the Restoring Force
• X is the Displacement of Particle from Equilibrium Position
• a is the Acceleration

Angular Simple Harmonic Motion

Angular Simple Harmonic Motion (SHM) occurs when a system oscillates about a fixed axis. In angular SHM the displacement of the particle is measured in terms of angular displacement. A common example of angular SHM is a torsional pendulum.

Conditions for Angular Simple Harmonic Motion

The restoring torque (or) angular acceleration acting on the particle should always be proportional to the particle's angular displacement and oriented towards the equilibrium position.

• T ∝ -θ
•  α ∝ -θ

where 

• T is Torque
• θ is the Angular Displacement
• α is the Angular Acceleration

Linear and Angular SHM

Equations for Simple Harmonic Motion

Let's consider a particle of mass (m) doing simple harmonic motion along a path A'OA; the mean position is O. Let the speed of the particle be V₀ when it is at position P (at some distance from point O)

At the time, t = 0, the particle at P (moving towards point A)

At the time t, the particle is at Q (at a distance X from point O). At this point, if velocity is V, then:

The force F acting on a particle at point p is given as,

F = -K X [where, K = positive constant]

We know that,

F = m a [where, a = Acceleration at Q]

⇒ m a = -K x

⇒ a = -(K/m) x

As K/m = ω² 

Thus, a = -ω²x

Also, we know a = dX/dt. 

Therefore, d²x/d²t = -ωx.

d²x/d²t + ω²x = 0

which is the differential equation for linear simple harmonic motion.

Solutions of Differential Equations

The solutions to the differential equation for simple harmonic motion are as follows:

Equation of SHM is, d²x/d²t + ω²x = 0

Multiply by 2 \frac{dx}{dt}, to get

2 \frac{dx}{dt}\cdot \frac{d^2x}{dt^2}+2 \omega^2 x \frac{dx}{dt}=0

\Rightarrow \frac{d}{dt}\left(\left(\frac{dx}{dt}\right)^2+\omega^2 x^2\right)=0

After integration, we get a separable equation

\left(\frac{dx}{dt}\right)^2+\omega^2 x^2=C^2,

\Rightarrow \frac{dx}{\sqrt{A^2-x^2} \cdot dt}=\omega 

\Rightarrow \frac{dx}{\sqrt{A^2-x^2}}=\omega dt

Integrating,

\sin^{-1}\left(\frac{x}{A}\right)=\omega t+\phi

\Rightarrow \frac{x}{A} = \sin (\omega t+\phi)

\bold{\Rightarrow x= A\sin (\omega t+\phi)}

This is the required Solution of the SHM Equation.

Different Cases of the Solution

For particle is in its mean position at point (O) [ϕ =0], displacement function becomes

 x = A sin ωt.

For t = 0, when object is at rest, displacement function becomes

x = A sin ϕ

For particle in any position throughout the SHM (any time t), displacement function becomes

x = A sin (ωt+ϕ)

Energy in Simple Harmonic Motion

A system performing SHM is called a harmonic oscillator. The energy of the particle performing the SHM is discussed below in the particle.

Let's take a particle of mass (m) performing linear SHM with angular frequency (ω) and the amplitude of the particle is (A)

Now we know that the displacement of the particle at any time is given using the SHM equation.

x = A sin (ωt + Φ)...(i)

where Φ is the phase difference.

Differentiating eq(i) wrt time, we get

v = A ω.cos (ωt + Φ)

v = ω. A cos (ωt + Φ)

v = ω.√(A² - x)... (ii)

Again, differentiating eq(ii) wrt time, we get

a = -ω². A sin (ωt + Φ)

a = -ω²x

The restoring force acting on the body is,

F = -kx

where, k = mω²

Now for the energy of the SHM particle.

Kinetic Energy of Particles

Kinetic Energy (K.E.) = 1/2 mv². Here, (v² = ω²(A² - x²))

\text{K.E.} = \frac{1}{2} m \omega^2 (A^2 - x^2)

Also, the kinetic energy of the particle in SHM is,

\text{K.E.} = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t + \Phi)

Potential Energy of Particles

For the potential energy we know that,

Potential Energy (P.E.) = - Work Done

P.E = -F.dx

P.E = kxdx                                   (As dx is also negative)

Integrating from o to x

P.E = (kx²)/2

We know that k = mω²

P.E = (mω²x²)/2

We know that, {x = Asin(ωt + Φ)}

\text{P.E.} = \frac{m \omega^2}{2} A^2 \sin^2(\omega t + \Phi)

Total Mechanical Energy of the Particle

Total Energy (E) = Kinetic Energy (K.E.) + Potential Energy (P.E.)

E = \frac{1}{2} m \omega^2 (A^2 - x^2) + \frac{m \omega^2 x^2}{2}

E = \frac{1}{2} m \omega^2 A^2

This is the total energy of the particle in SHM.

Solved Questions

Question 1: A particle performs SHM with amplitude 4 m and angular frequency 5 rad/s. Find the maximum velocity and maximum acceleration.

Solution: Given

• Aplitude, A = 4 m
• Angular frequency, ω = 5 rad/s

Maximum Velocity

v_max = Aω

v_max = 4 × 5

v_max = 20 m/s

Maximum Acceleration

a_max = Aω²

a_max = 4 × 5²

a_max = 4 × 25

a_max = 100 m/s²

Maximum velocity = 20 m/s

Maximum acceleration = 100 m/s²

Question 2: Find the total energy of a particle of mass 2 kg performing SHM with amplitude 0.5 m and angular frequency 6 rad/s.

Solution: E = \frac{1}{2} m \omega^2 A^2

Given

m = 2 kg

ω = 6 rad/s

A = 0.5 m

E = \frac{1}{2} \times 2 \times 6^2 \times (0.5)^2

E = 1 \times 36 \times 0.25

E = 9 \, \text{J}

Question 3: A spring with a spring constant of 1200 N m⁻¹ is mounted on a horizontal table. A 3 kg mass is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 m before being released. Determine the following:

1. The frequency of oscillations,
2. Maximum acceleration of the mass, and
3. The maximum speed of the mass.

Solution: Given:

• Spring Constant, k = 1200 N/m.
• Mass of Object, m = 3 kg.
• Displacement, x = 2 m.

(1) Frequency of Oscillation:

We know that frequency (f) = 1/Time period (T)          T = 2π/ω and ω = √k/m]

Therefore,

f = (1/2π)√k/m

  = (1/2 × 3,14) √1200/3 = 3.18 Hz.

(2) Maximum Acceleration:

Maximum Acceleration (a) = ω²x

where, ω = Angular frequency = √k/m

Therefore, a = x(k/m)

a = 2 × (1200/3) 

a= 800 m/s².

(3) Maximum Speed:

Maximum Speed (V) = ωx 

Put, ω = √k/m.

Therefore, V = x(√k/m)

V = 2 × (√1200/3) 

V = 40 m/s.

Question 4: A mass of 2 kg is attached to the end of the spring with a spring constant of 50 N/m. What is the period of the resulting simple harmonic motion? (π = 3.14)

Solution: Formula for time period is

T = 2π√(k/m)

where,

• m is the mass
• k is the spring constant

Thus, T = 2π√(50/2) 

⇒ T = 2π√(25) 

⇒ T = 2π/5 

⇒ T ≈ 1.26 s

So, the time period of the SHM is approximately 1.26 s.

Question 5: A block of mass 0.5 kg is attached to the end of the spring (spring constant = 100 N/m). If the block is displaced 0.1 m from its equilibrium position, then what is the maximum speed of the block during its motion?

Solution: The maximum speed of the block is given by:

v_max = Aω

where, 

• A is Amplitude of Motion
• ω is Angular Frequency

Also, angular frequency ω is given by:

ω = √(k/m)

where, 

• m is the mass
• k is the spring constant

Given: 

• Amplitude(A) = 0.1 m
• k = 100 N/m
• m = 0.5 Kg

⇒ v_max  = 0.1 × √(100/0.5) 

⇒ v_max  = 0.1 × √(1000/5) 

⇒ v_max  = 0.1 × √(200) 

⇒ v_max  = √2 

So, the maximum speed of the block during its motion is √2  m/s.

Unsolved Problems

Problem 1. A damped harmonic oscillator has a frequency of 5 oscillations per second. For every 10 oscillations, the amplitude of the oscillator drops to half. Find the time taken to drop the amplitude to 1/1000 of the original value.

Problem 2. If the length of a simple pendulum in SHM is increased by 21%, then what is the percentage increase in the time period of the pendulum of the increased length

Problem 3. It is given that the ratio of maximum acceleration to maximum velocity in an SHM is 10 seconds⁻¹ and at t = 0, the displacement is 5 m. What is the maximum acceleration? Given that the initial phase is π/ 4

Problem 4. If a child is swinging in a sitting position and then he stands up, then how will the time period of the swing be affected?

Problem 5. The displacement of a particle in simple harmonic motion is given by x(t) = Asin(πt/90). Find the ratio of kinetic energy to the potential energy at t = 210 seconds

Related Articles:

• Frequency
• Angular Velocity
• Angular Momentum