Periodic motion:

Periodic Motion

• Introduction to periodic motion.

Resonance

• All objects have a natural frequency at which they naturally vibrate.

• At a natural frequency, a system oscillates without driving or damping force.

• When the frequency of an applied force is close to a natural frequency, the system vibrates with increased amplitude, called resonance.

• Resonance occurs with all types of vibrations or waves.

• Example: Pushing a swing at its resonant frequency maximizes the swing's height.

• Resonant systems generate specific frequencies (musical instruments) or filter frequencies from complex vibrations.

Bridge Collapses and Resonance

• Bridges and buildings can collapse due to resonance.

• Structures oscillate due to traffic, footfall, or machinery.

• If vibrations match the resonance frequency, energy stores at an atomic level.

• When energy exceeds the load limit, structural integrity fails.

• Tacoma Narrows Bridge collapse (1940) was due to mechanical resonance and aeroelastic flutter.

Periodic Motion Defined

• Object repeats its motion along a path about a point in a fixed time interval.

• Also known as oscillatory motion.

• Vibration is another term for mechanical oscillation.

• Examples:

  • Pendulum motion

  • Spring motion

  • Guitar string vibration

  • Earth's rotation and revolution

  • Sun's revolution around the Galaxy center

Oscillation Description

• Oscillation occurs when a restoring force returns the system to equilibrium.

Oscillation Terms

• $F_x$: Force exerted on body in x direction

• $x$: Displacement from equilibrium position

• $m$: Mass

• $a$: Acceleration

Amplitude, Period, Frequency, and Wavelength

• Amplitude (A): Peak-to-peak amplitude of motion (e.g., from A to -A and back to A), which is one complete vibration or cycle. Peak amplitude is the maximum displacement from equilibrium. [Unit: meter or m]

• Period (T): Time for one cycle. [Unit: seconds or s]

• Frequency (f): Number of cycles per second. [Unit: Hertz or Hz]

• Wavelength ($\lambda$): Distance between two consecutive points. [Unit: m]

Frequency and Period Relationship

• $f$ = number of cycles in one second

• $f$ cycles in 1 second

• 1 cycle in $1/f$ seconds

• Period (T) is time for 1 cycle

• $T = 1/f$

Angular Frequency and Velocity

• Angular Frequency omega: Rate at which an object moves through a given angle.

  • $\omega = 2 \pi / T$ [Unit: rad/s]

  • $f = 1/T$

  • $T = 1/f$

  • $\omega = 2 \pi f = 2\pi / T$

• Velocity (V): Displacement in a given interval of time.

  • $V = f \lambda$ [Unit: m/s]

Graphical Representation of Periodic Motion

• Diagram showing displacement over time, identifying:

  • Amplitude (A)

  • Wavelength

  • Crest

  • Trough

  • Equilibrium

Simple Harmonic Motion (SHM)

• SHM is a type of periodic motion where the restoring force is directly proportional to the object's displacement and acts towards the equilibrium position.

Glider and Spring System

• A glider connected to a spring oscillates back and forth when an impulse force is applied.

• This is due to the restoring force.

• When the restoring force is directly proportional to the displacement from equilibrium, it's SHM.

• $F_x \propto -x$

Restoring Force

• Restoring force is always towards the equilibrium, opposite to displacement.

• $F_x = -kx$ … 1

• $k$ is the spring constant [Unit: N/m]

• $k$ = force needed to extend (or compress) the spring by 1m

• Hooke’s Law: An idealized spring exerts a restoring force that obeys Hooke’s law

Acceleration of SHM

• Acceleration: $a = -\frac{k}{m}x$ … 2

• The direction of acceleration is the same as the force, opposite to the displacement.

Equations for SHM

• Hooke's Law is the basic equation for SHM.

Circular Motion and SHM

• Projection of a body's uniform circular motion onto the horizontal axis executes SHM.

• All SHM equations can be derived from this model.

Point Q and Shadow P

• Point Q rotates counter-clockwise in uniform circular motion.

• Its shadow, point P, moves in simple harmonic motion.

Equation for SHM - Displacement

• As point Q moves with constant angular speed, the 'x' component of the phasor at time 't' is:

• $x = A \cos{\theta}$ … 3

• $\theta = \omega t$ (assuming at t=0, $\theta$ =0)

• So: $x = A \cos{\omega t}$

Equation for SHM - Velocity

• v(t)=−Aωsin(ωt+ϕ)

• A is the amplitude

• ω is the angular frequency

• t is time

• ϕ is the phase angle
  
  

Equation for SHM - Acceleration

• a=−ω2x

• ω is the angular frequency

• x is the displacement from the equilibrium position

Acceleration and Displacement

• $a_x = -\omega^2 A \cos{\theta}$

• $a_x = -\omega^2 x$

• Acceleration of point P is directly proportional to displacement x and has the opposite sign.

• This defines SHM.

Angular Frequency ($\omega$) for SHM

• Compare equations 2 and 6:

  • $a = -\frac{k}{m}x$

  • $a = -\omega^2 x$

• $\omega^2 = \frac{k}{m}$

• $\omega = \sqrt{\frac{k}{m}}$

• For SHM:

  • $\omega = \sqrt{\frac{k}{m}}$ … 8

  • $f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$ … 9

  • $T = \frac{1}{f} = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$ … 10

Example

• A spring is mounted horizontally, with its left end fixed. A force of 6 N causes a displacement of 0.03 m. A 0.5 kg body is attached, pulled by 0.02 m, and released.

  • (a) Find the force constant of the spring.

  • (b) Find the angular frequency, frequency, and period of the oscillation.

Solution (a) Force constant of the spring

• Input: $x = 0.03$ m, $F = -6$ N

• Output: $k$

• Equation: $F = -kx$

• Solution: $k = -\frac{F}{x} = -\frac{-6}{0.03} = 200$ N/m

Solution (b) Angular frequency, frequency, and period

• Input: $m = 0.5$ kg, $k = 200$ N/m

• Output: $\omega$, $f$, $T$

• Equations: $\omega^2 = k/m$, $\omega = 2\pi f$, $T = 1/f$

• Solutions:

  • $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}}= 20$ rad/s

  • $f = \frac{\omega}{2\pi} = \frac{20}{2\pi} = 3.2$ Hz

  • $T = \frac{1}{f} = \frac{1}{3.2} = 0.31$ s

Displacement, Velocity and Acceleration with phase angle

• If at $t = 0$ the phasor OQ makes an angle $\phi$ with the positive x axis, then at any later time t this angle is $\theta = \omega t + \phi$.

• We substitute this into Equation 3 to obtain

• $x = A \cos{(\omega t + \phi)}$ … 11

• The constant $\phi$ is called the phase angle

• The value of the cosine function is always between – 1 and 1.

• So in Equation 11, x is always between “– A” and “A” .

Phase angle values and x

• Putting $t = 0$ in Equation 11, we get;

• $x_o = A \cos{\phi}$ … 12

• If $\phi = 0$, $x_o = A \cos{0} = A$

• If $\phi = \pi$, $x_o = A \cos{\pi} = -A$

• If $\phi = \frac{\pi}{2}$, $x_o = A \cos{\frac{\pi}{2}} = 0$

x, v and a with phase angle

• $x = A \cos{(\omega t + \phi)}$

• $v = \frac{dx}{dt} = -\omega A \sin{(\omega t + \phi)}$ … 13

• $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

• $a = -\omega^2 A \cos{(\omega t + \phi)}$ … 14

To find amplitude and phase angle

• If we are given the initial position $x<em>o$ and initial velocity $v</em>o$ for the oscillating body, we can determine the amplitude $A$ and the phase angle $\phi$.

• At $t = 0$,

• $v_o = -\omega A \sin{\phi}$ … 15

Phase angle $\phi$

• x(t)=Acos(ωt+ϕ0​)

• $x_o^2 = A^2 \cos^2{\phi}$ … 17

• $\frac{v_o}{\omega} = [ \frac{-\omega A \sin{\phi}}{\omega}] = -A \sin{\phi}$

• $\frac{v_o^2}{\omega^2} = A^2 \sin^2{\phi}$

Amplitude in SHM

Variations in Simple Harmonic Motion - outline

• How the velocity and acceleration vary during one cycle of SHM.

Variations in Phase Angle

• The angle ‘ϕ ’ is called the phase angle.

• We use ‘x0’ to be position at t = 0.

• In this case:

• xo = Acos(ϕ)

• If ϕ = 0 then xo = Acos(0) = A

• If ϕ = π/4 then xo = Acos(π/4) = 0.71A

• If ϕ = π/2 then xo = Acos(π/2) = 0

• If ϕ = π then xo = Acos(π) = −A

How velocity and acceleration vary during one cycle of SHM

• As the glider undergoes SHM, you can track changes in velocity and acceleration as the position changes between the classical turning points.

• $x = -A$

• $F = kA$ (to right)

• $a_{max} = \frac{kA}{m}$ (to right)

• $v = 0$

• $K = 0$

• $U_{max} = \frac{1}{2}kA^2$

• $x = 0$

• $F = 0$

• $a = 0$

• $v = maximum$

• $K = maximum$

• $U = 0$

• $x = A$

• $F = -kA$ (to left)

• $a_{max} = -\frac{kA}{m}$ (to left)

• $v = 0$

• $K = 0$

• $U_{max} = \frac{1}{2}kA^2$

• @ x distance

• $x = x$

• $F = -kx$ (to left)

• $a = -\frac{kx}{m}$ (to left)

• $v = v_x$

• $K = \frac{1}{2}mv_x^2$

• $U_{max} = \frac{1}{2}kx^2$

Example

• Consider the system of mass (0.5 kg) and horizontal spring with k = 200 N/m we discussed in the previous example. This time we give the body an initial displacement of +0.015 m and an initial velocity of + 0.4 m/s. Find the period, amplitude, and phase angle of the motion.

Solution

• Period: $T = 2 \pi(\frac{m}{k})^{1/2} = 2 x \pi x (0.5/200)^{1/2} = 0.31 s$

• Amplitude:

  • $\omega = (k/m)^{1/2} = (200/0.5)^{1/2} = 20 rad/s$

  • $A = ( x<em>o^2 + v</em>o^2/ \omega^2 )^{1/2} = [(0.015)^2 + (0.4/20)^2 ]^{1/2} = 0.025 m$

• Phase angle:

  • $\phi = tan^{-1} ( -v<em>o/\omega x</em>o ) = tan^{-1} [ - 0.4 / (20 x 0.015) ] = - 53^o = - 0.93 rad$

• [Note: π radians = 180o]

Energy in Simple Harmonic Motion - Outline

• Kinetic and potential energy in SHM.

• Applying energy conservation law in SHM.

Energy in SHM

• Consider force exerted on an ideal spring

• The kinetic energy of the body is $K = \frac{1}{2}mv^2$

• The potential energy of the spring is $U = \frac{1}{2}kx^2$

• Total mechanical energy E = K + U is conserved

• $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = constant$ … 20

Energy in SHM

• When x = A ( or – A), v = 0. Energy here is entirely potential so;

• $E = \frac{1}{2}kA^2$

• Because E is constant, it is equal to ½ $kA^2$. at any other point. Thus, the total mechanical energy in SHM is;

• $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2 = constant$ … 21

Energy in SHM

• Solving for v for any displacement x;

• $v = \pm \sqrt{\frac{k}{m}} \sqrt{A^2-x^2}$ … 22

• The $\pm$ sign means that at a given value of x the body can be moving in either direction

Energy in SHM

• For example, when $x = \pm \frac{A}{2}$;

• v = $\pm \sqrt{\frac{k}{m}} \sqrt{A^2 - (\pm \frac{A}{2})^2} = \pm \sqrt{\frac{3}{4}} \sqrt{\frac{k}{m}}A$

• Equation 22 also shows that the maximum speed vmax occurs at x =0.

• Using Equation 8 [$\omega = (k/m)^{1/2}$] we find that;

• $v_{max} = A \sqrt{\frac{k}{m}} = \omega A$

• $v_{max}$ oscillates between –$\omega A$ and +$\omega A$

• $x = -A$

• $F = kA$ (to right)

• $a_{max} = \frac{kA}{m}$ (to right)

• $v = 0$

• $K = 0$

• $U_{max} = \frac{1}{2}kA^2$

• $x = 0$

• $F = 0$

• $a = 0$

• $v = maximum$

• $K = maximum$

• $U = 0$

• $x = A$

• $F = -kA$ (to left)

• $a_{max} = -\frac{kA}{m}$ (to left)

• $v = 0$

• $K = 0$

• $U_{max} = \frac{1}{2}kA^2$

• @ x distance

• $x = x$

• $F = -kx$ (to left)

• $a = -\frac{kx}{m}$ (to left)

• $v = v_x$

• $K = \frac{1}{2}mv_x^2$

• $U_{max} = \frac{1}{2}kx^2$

Example

• A spring is mounted horizontally, with its left end held stationary. By attaching a spring balance to the free end and pulling toward the right, we determine that the stretching force is proportional to the displacement and that a force of 6 N causes a displacement of 0.03 m. We remove the spring balance and attach a 0.5 kg body to the end, pull it a distance of 0.02 m, release it, and watch it oscillate. If k = 200 N/m;

  • a) Find the maximum and minimum velocities attained by the oscillating body.

  • b) Compute the maximum acceleration.

  • c) Determine the velocity and acceleration when the body has moved halfway to the centre from its original position.

  • d) Find the total energy, potential energy and kinetic energy at this position.

Solution

• a) Maximum and minimum velocities attained by the oscillating body.

• The velocity v, at any displacement x is given by;

• $v = \pm \sqrt{\frac{k}{m}} \sqrt{A^2 – x^2}$

• The maximum velocity occurs when the body is moving to the right through the equilibrium position, where x = 0.

• $v_{max} = \sqrt{\frac{k}{m}} * A = \sqrt{\frac{200}{0.5}} * 0.02 = 0.4 m/s$

• The minimum (i.e most negative) velocity occurs when the body is moving to the left through x = 0; its velocity –vmax= -0.4 m/s.

Solution (continued)

• b) Maximum acceleration.

• Acceleration is given by;

• $a = -\frac{k}{m} x$

• The maximum (most positive) acceleration occurs at the most negative value of x, x = - A. Hence;

• $a_{max} = - \frac{k}{m} (-A) = -{\frac{200}{0.5}} x (-0.02) = 8 m/s^2$

• The minimum (most negative) acceleration is; – 8 m/s^2, occurring at $x = + A = + 0.02 m$.

Solution (contnued)

• c) Velocity and acceleration when the body has moved halfway to the centre from its original position.

  • At a point halfway to the centre from the initial position, $x = A/2 = 0.01 m$.

  • $v = - \sqrt{\frac{200}{0.5}} x \sqrt{0.02^2 – 0.01^2} = -0.35 m/s$

  • We choose the negative square root because the body is moving from x = A toward x = 0.

  • [Use $a = -(k/m)x$]

  • $a = - \frac{200}{0.5} x (0.01) = - 4 m/s^2$

• At this point the velocity and acceleration have the same sign, so the speed is increasing.

Solution

• d) Find the total energy, potential energy and kinetic energy at this position.

• The total energy has the same value at all points during the motion.

• $E = \frac{1}{2} kA^2 = \frac{1}{2} x 200 x (0.02)^2 = 0.04 J$

• The potential energy is

• $U = \frac{1}{2} kx^2 = \frac{1}{2} x 200 x (0.01)^2 = 0.01 J$

• The kinetic energy is

• $K = \frac{1}{2} mv^2 = \frac{1}{2} x 0.5 x (-0.35)^2 = 0.03 J$

Applications of Simple Harmonic Motion - outline

• Oscillation of a simple pendulum.

The Simple pendulum

• The simple pendulum is an idealized model consisting of a point mass suspended by a massless, unstretchable string. When the point mass is pulled to one side of its straight- down equilibrium position and released, it oscillates about the equilibrium position.

Simple pendulum

• The mass m moves along an arc of circle with radius L equal to the length of the string. The distance x is measured along the arc.

• $x = L\theta$.

• The restoring force;

• $F = - mg \sin{\theta}$

Simple pendulum

• The restoring force is provided by gravity, the tension T merely acts to make the point mass move in an arc.

• If angle $\theta$ is small sin $\theta$ is very nearly equal $\theta$ to in radians.

• $F = -mg\theta = - mg \frac{x}{L}$

• $F = - \frac{mg}{L} x$

• The restoring force is proportional to the displacement and the force constant is $k = mg/L$. The angular frequency $\omega$ of a simple pendulum with small amplitude is;

• $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{mg}{Lm}} = \sqrt{\frac{g}{L}}$

Simple pendulum: Frequency and Period

• The corresponding frequency (f) and period (T) relations are

• $f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$

• $T = \frac{1}{f} = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{g}}$

• The above equations are for a simple pendulum with small amplitude

Mechanical waves - Outline

• Transverse and longitudinal waves

• Relationships between velocity, frequency and period of a periodic wave

• Mathematical description of a wave

Waves

• A wave is a means of transferring energy (disturbance) from one point to another without the source of the wave moving very far in the direction of movement of the wave.

• Two basic types of waves, mechanical waves and non-mechanical waves.

• Mechanical waves are waves that need a medium for propagation. Mechanical waves transport energy without transporting matter. Sound waves, water waves and seismic waves are some examples of mechanical waves.

• Non-mechanical waves are waves that do not need a medium for propagation. The electromagnetic wave is the only non-mechanical wave.

Types of mechanical waves

• Mechanical waves transfer energy by means of vibration (oscillation) of the medium (particles of the medium).

• Differences in the way in which the oscillations of one part of a medium transfer their energy of oscillation to the surrounding medium lead to the description and classification of mechanical waves into two groups.

  • Transverse

  • Longitudinal

Transverse waves

• Waves that cause the medium (particles of the medium) to vibrate at right angles to the direction of motion of the wave energy are called transverse waves.

• Examples include S and L earthquake waves, vibrating waves in a piano or a guitar string

Longitudinal waves

• Waves that cause the medium (particles of the medium) to vibrate back and forth along the direction of the line of motion of the wave energy are called longitudinal waves.

• Examples include P earthquake waves, sound/acoustic waves

Periodic waves

• Wave patterns which repeat at regular intervals (in space and time) are said to be periodic waves.

Periodic waves

• Repeating pattern

• Velocity
  v = distance/time

• $v = \frac{\lambda}{T}$

• Period:

• $T = \frac{1}{f}$

• Frequency:

• $f = \frac{1}{T}$

• $v = f \lambda$

• For light waves write v = c

• $c = f \lambda$

Mathematical description of a wave

• Consider sinusoidal waves

• $y = y (x, t)$

• At $x = 0$, where the wave originates

• $y(x = 0, t) = A \cos{\omega t}$

• [ Note: $\omega = 2\pi f$ ]

• $y(x = 0, t) = A \cos{2\pi ft}$

• Particle oscillates with SHM

  • Amplitude = A

  • Frequency = f

  • Angular frequency:

  • $\omega = 2\pi f$

  • [Note: At t = 0 the particle at x = 0 is at its maximum positive displacement (y = A) and is instantaneously at rest (because the value of y is a maximum)].

Wave function

• $y(x,t) = A \cos{\omega t} = A \cos{\omega (t – \frac{x}{v})}$

• Since $\cos(-\theta) = \cos{\theta}$, we can rewrite $y (x,t) = A \cos{\omega (\frac{x}{v} –t)}$ = $A \cos{2\pi f (\frac{x}{v} –t)}$

• We can express equation in various forms.

• Recall: $T = \frac{1}{f}$ and $\lambda = \frac{v}{f}$

Wave function

• $y(x,t) = A \cos{2\pi(\frac{x}{\lambda} – \frac{t}{T})}$

• Equation for sinusoidal wave moving in the + x direction

• In terms of wave number (k)

• $k = \frac{2\pi}{\lambda}$

Wave equation

• Substitute $\lambda = \frac{2\pi}{k}$ and $f = \frac{\omega}{2\pi}$ into $v = f\lambda$

  • $v = \frac{\omega}{2\pi} x \frac{2\pi}{k}$

  • $\omega = vk$ (periodic wave)

• Re-write wave equation

• $y(x,t) = A \cos{(2\pi \frac{x}{\lambda} – 2\pi \frac{t}{T})}$ as

• $y(x,t) = A \cos{(kx – \omega t)}$ in +ve x direction

• $y(x,T) = A \cos{(kx + \omega t)}$ in –ve x direction

Wave equation

• The quantity $(kx \pm \omega t)$ is called the phase.

• For a wave travelling in the +x direction, that means

• $kx – \omega t = constant$.

• Taking the derivative with respect to t, we find

• $k \frac{dx}{dt} = \omega$, or $\frac{dx}{dt} = \frac{\omega}{k}$

• or $v = \frac{\omega}{k}$

Example

• One end of a clothesline is wiggled up and down sinusoidally with frequency 2 Hz and amplitude 0.075 m.

• The wave speed is v = 12 m/s.

• At time t = 0 s the end has positive displacement and is instantaneously at rest.

  • a) Find angular frequency, period, wavelength and wave number of the wave.

  • b) Write the wave function of the wave

Solution

• a) Angular frequency, period, wavelength and wave number of the wave.

  • Angular frequency:

  • $\omega = 2\pi f = 2\pi x 2 = 12.6 rad/s$

  • Period:

  • $T = \frac{1}{f} = \frac{1}{2} = 0.5 s$

  • Wavelength:

  • $\lambda = \frac{v}{f} = \frac{12}{2} = 6 m$

  • Wave number:

  • $k = \frac{2\pi}{\lambda} = \frac{2\pi}{6} = 1.05 rad/m$
    or

  • $k = \frac{\omega}{v} = \frac{4\pi}{12} = 1.05 rad/m$

Solution

• b) Write the wave function of the wave

  • $y(x,t) = A \cos{2\pi (\frac{x}{\lambda} - \frac{t}{T})}$

    • $= (0.075) \cos{2\pi (\frac{x}{6} - \frac{t}{0.5})}=(0.075) \cos{[1.05x – 12.6t)}$

Velocity and acceleration in waves - outline

• How the velocity and acceleration changes in a wave.

Particle velocity (v) and acceleration (a) in a wave