Chapter 15: Oscillations

Chapter 15: Oscillations

• This chapter will cover systems that oscillate in simple harmonic motion (SHM).

Importance of Simple Harmonic Motion

• SHM is a common and important motion in science and engineering.

• Oscillations and vibrations occur in mechanical, electrical, chemical, and atomic systems.

• Understanding how a system oscillates is crucial for engineering design.

• Complex oscillations can be understood in terms of SHM.

• Oscillations are the sources of waves.

• Simple - returns to same amplitude every time

• Harmonic - if force isn’t proportional to displacement no nice sinusoid wave (force and displacement have to be proportional)

Oscillatory Motion

• Oscillators: Objects that undergo repetitive motion back and forth around an equilibrium position. (Doesn’t mean its simple harmonic)

• Period (T): The time to complete one full cycle or oscillation.

• Frequency (f): The number of cycles per second, measured in Hertz (Hz).

  • $f = \frac{1}{T}$

  • $T = \frac{1}{f}$

  • $1 \text{ Hz} = 1 \text{ cycle per second} = 1 \text{ s}^{-1}$

Simple Harmonic Motion (SHM)

• A particular kind of oscillatory motion.

• Amplitude (A): The object’s maximum displacement from equilibrium.

• Velocity is maximum at 0 or equilibrium

• Closer dots the glider is slowing down

• Towards equilibrium increase in velocity after passing that point it will start to even out and slow down

Simple Harmonic Motion Details

• Position vs. time graph for an object in SHM is sinusoidal.

• Velocity vs. time graph shows velocity is zero at $x = \pm A$ (turning points).

• Maximum speed ($v_{\text{max}}$) is reached at $x = 0$.

• When one is 0 the other is at max

Modeling SHM with Cosine Function

• If released from rest at $t = 0$, motion can be modeled as: $x(t) = A \cos(\omega t)$

• Cosine is a sinusoidal function.

• Angular frequency ($\omega$) is defined as: $\omega = \frac{2\pi}{T}$

• Units of $\omega$ are rad/s: $\omega = 2\pi f$

• W angular frequency

• A amplitude

• wA at equilibrium

Velocity in SHM

• Velocity is the derivative of the position function:

  • $v(t) = \frac{dx}{dt} = -A\omega \sin(\omega t) = -\frac{2\pi A}{T} \sin(\frac{2\pi}{T}t) = -2\pi f A \sin(2\pi f t)$

• Maximum speed: $v_{\text{max}} = A\omega$

Example 15.1: System in Simple Harmonic Motion

• An air-track glider attached to a spring is pulled 20.0 cm to the right and released at $t = 0$ s. It makes 15 oscillations in 10.0 s.

  • What is the period of oscillation?

  • What is the object’s maximum speed?

  • What are the position and velocity at $t = 0.800$ s?

• Model: Object oscillating on a spring is in SHM.

Solution to Example 15.1

• Frequency: $f = \frac{15 \text{ oscillations}}{10.0 \text{ s}} = 1.50 \text{ Hz}$

• Period: 10/15 or$T = \frac{1}{f} = \frac{1}{1.50 \text{ Hz}} = 0.667 \text{ s}$

• Amplitude: $A = 0.200 \text{ m}$

• Maximum speed: $v_{\text{max}} = \frac{2 \pi A}{T} = \frac{2 \pi (0.200 \text{ m})}{0.667 \text{ s}} = 1.88 \text{ m/s}$

• Position at $t = 0.800$ s: $x(t) = A \cos(\frac{2 \pi}{T}t) = (0.200 \text{ m}) \cos(\frac{2 \pi}{0.667 \text{ s}}(0.800 \text{ s})) = (0.200 \text{ m}) \cos(7.54 \text{ rad}) = 0.0625 \text{ m} = 6.25 \text{ cm}$

• Velocity at $t = 0.800$ s:$v(t)=-v_{\text{max}}\sin(\frac{2\pi t}{T})=-(1.88\text{ m/s})\sin(\frac{2\pi\left(0.800s\right)}{0.667\text{ s}})=-1.88\text{ m/s}\sin(7.54\text{ rad})=-1.79\text{ m/s}=-179\text{ cm/s})$

Example 15.2: Finding the Time

• A mass oscillating in SHM starts at $x = A$ and has period T. At what time does the object first pass through $x = \frac{1}{2}A$?

• The object passes through $x = 0$ at $t = \frac{1}{4}T$.

• SHM graph is not linear between $x = A$ and $x = 0$.

• Use $x(t) = A \cos(\frac{2\pi t}{T})$.

Solution to Example 15.2

• $\frac{1}{2}A = A \cos(\frac{2\pi t}{T})$

• $\frac{1}{2} = \cos(\frac{2\pi t}{T})$

• $\frac{2\pi t}{T} = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$

• $t = \frac{T}{6}$

Simple Harmonic Motion and Circular Motion

• A “shadow movie” of a ball moving in uniform circular motion demonstrates SHM.

• The shadow moves with SHM.

• A block on a spring also moves with SHM.

The Phase Constant

• If an object in SHM is not initially at rest at $x = A$ when $t = 0$, use a phase constant $\phi_0$

• $x(t) = A \cos(\omega t + \phi_0)$

• $v(t) = -A\omega \sin(\omega t + \phi_0)$

• $v_{\text{max}} = A\omega$

Example 15.3: Using Initial Conditions

• An object on a spring oscillates with a period of 0.80 s and an amplitude of 10 cm. At $t = 0$ s, it is 5.0 cm to the left of equilibrium and moving to the left. What are its position and direction of motion at $t = 2.0$ s?

• Model: Object oscillating on a spring is in simple harmonic motion.

Solution to Example 15.3

• Find the phase constant $\phi_0$ from the initial condition $x(0) = -5.0 \text{ cm}$.

• $-5.0 \text{ cm} = A \cos(\phi<em>0) = (10 \text{ cm}) \cos(\phi</em>0)$

• $\cos(\phi_0) = -\frac{1}{2}$

• $\phi_0 = \pm \frac{2\pi}{3} \text{ rad} = \pm 120^\circ$

• Since the oscillator is moving to the left at $t = 0$, $\phi_0 = \frac{2\pi}{3} \text{ rad}$.

• Angular frequency: $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.80 \text{ s}} = 7.85 \text{ rad/s}$

• Position at $t = 2.0$ s:

  • $x(t) = A \cos(\omega t + \phi_0) = (10 \text{ cm}) \cos((7.85 \text{ rad/s})(2.0 \text{ s}) + \frac{2\pi}{3} \text{ rad}) = (10 \text{ cm}) \cos(17.8 \text{ rad}) = 5.0 \text{ cm}$

• Velocity at $t = 2.0$ s:

  • $v(t) = -A\omega \sin(\omega t + \phi_0) = -(10 \text{ cm})(7.85 \text{ rad/s}) \sin((7.85 \text{ rad/s})(2.0 \text{ s}) + \frac{2\pi}{3} \text{ rad}) = 68 \text{ cm/s}$

Energy in Simple Harmonic Motion

• An object of mass m on a frictionless horizontal surface attached to a spring with spring constant k.

• Energy transforms between kinetic energy ($K$) and potential energy ($U$), but the mechanical energy $E = K + U$ is constant.

• $E = K + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

• $E = U = \frac{1}{2}kA^2$ (at $x = \pm A$)

• $E = K = \frac{1}{2}mv_{\text{max}}^2$ (at $x = 0$)

Frequency of Simple Harmonic Motion

• In SHM, when K is maximum, $U = 0$, and when U is maximum, $K = 0$.

• $K<em>{\text{max}} = U</em>{\text{max}}$

• $\frac{1}{2}mv_{\text{max}}^2 = \frac{1}{2}kA^2$

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

• $v_{\text{max}} = A\omega = 2\pi f A = \frac{2\pi A}{T}$

• $\omega = \sqrt{\frac{k}{m}}$

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

• $T = 2\pi\sqrt{\frac{m}{k}}$

Example 15.4: Using Conservation of Energy

• A 500 g block on a spring is pulled a distance of 20 cm and released. The oscillations have a period of 0.80 s.

  • At what position(s) is the block’s speed 1.0 m/s?

  • What is the spring constant?

• Model: The motion is SHM. Energy is conserved.

Solution to Example 15.4

• $E = U = \frac{1}{2}kA^2$

• $\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2$

• $x = \pm \sqrt{A^2 - \frac{mv^2}{k}}$

• $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.80 \text{ s}} = 7.85 \text{ rad/s}$

• $x = \pm \sqrt{A^2 - \frac{v^2}{\omega^2}} = \pm \sqrt{(0.20 \text{ m})^2 - \frac{(1.0 \text{ m/s})^2}{(7.85 \text{ rad/s})^2}} = \pm 0.15 \text{ m} = \pm 15 \text{ cm}$

• $k = m\omega^2 = m(\frac{2\pi}{T})^2 = (0.50 \text{ kg})(\frac{2\pi}{0.80 \text{ s}})^2 = 31 \text{ N/m}$

Simple Harmonic Motion Diagram

• Shows motion to the right and to the left.

• At $x = 0$, the object’s speed is maximum, but acceleration is zero.

Acceleration in Simple Harmonic Motion

• Acceleration is the time-derivative of the velocity:

  • $a = \frac{dv_x}{dt} = -A\omega^2 \cos(\omega t)$

• $a_x = -\omega^2 x$

Dynamics of Simple Harmonic Motion

• Consider a mass m oscillating on a horizontal spring with no friction.

• Spring force: $F_{\text{sp}} = -k \Delta x$

• Newton’s second law: $F<em>{\text{net}} = F</em>{\text{sp}} = -kx = ma_x$

• $a_x = -\frac{k}{m}x$

• $\omega = \sqrt{\frac{k}{m}}$

Vertical Oscillations

• Motion for a mass hanging from a spring is the same as for horizontal SHM, but the equilibrium position is affected.

• $k \Delta L = mg$

The Simple Pendulum

• Consider a mass m attached to a string of length L, free to swing back and forth.

• Newton’s second law for the tangential component of gravity:

  • $F<em>{\text{net}} = F</em>G = -mg \sin(\theta) = ma_t = m\frac{d^2s}{dt^2} = m L \frac{d^2\theta}{dt^2}$

Small Angle Approximation for Simple Pendulum

• For small angles (\theta < 10^\circ), $\sin(\theta) \approx \theta$

  • $F_{\text{net}} = -mg\theta = -mg \frac{s}{L}$

• Angular frequency:

  • $\omega = \sqrt{\frac{g}{L}}$

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

Example 15.7: The Maximum Angle of a Pendulum

• A 300 g mass on a 30-cm-long string oscillates as a pendulum. It has a speed of 0.25 m/s as it passes through the lowest point. What maximum angle does the pendulum reach?

• Model: Assume the angle remains small, so the motion is simple harmonic motion.

Solution to Example 15.7

• $\omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.8 \text{ m/s}^2}{0.30 \text{ m}}} = 5.72 \text{ rad/s}$

• $v<em>{\theta</em>{\text{max}}} = \omega A</p><ul><li><p>$v_{\text{max}}$at lowest point = 0.25m/s</p></li><li><p>Amplitude</p></li><li><p>$A = s = L \theta_{\text{max}}$</p></li></ul></li><li><p>arc length:$A = v_{\text{max}} / \omega = \frac{0.25 \text{ m/s}}{5.72 \text{ rad/s}} = 0.0437 m$</p></li><li><p>$\theta_{\text{max}} = A /L= 0.0437 \text{ m} /0.30 \text{ m} = 0.146 \text{ rad} = 8.3^\circ$</p></li></ul><h3 collapsed="false" seolevelmigrated="true">Simple Harmonic Motion Model</h3><ul><li><p>For any system with a restoring force that’s linear or can be approximated as linear.</p><ul><li><p>Motion is SHM around the equilibrium position.</p></li><li><p>Frequency and period are independent of the amplitude.</p></li><li><p>Equation of motion:$\frac{d^2u}{dt^2} = -Cu$</p><ul><li><p>where$C$is a collection of constants.</p></li></ul></li><li><p>Angular frequency:$\omega = \sqrt{C}$</p></li><li><p>Position and velocity:</p><ul><li><p>$u = A \cos(\omega t + \phi_0)$</p></li><li><p>$vu = -v{\text{max}} \sin(\omega t + \phi_0)$</p></li><li><p>where A and${\phi_0}$are determined by the initial conditions.</p></li></ul></li><li><p>Mechanical energy is conserved.</p></li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Damped Oscillations</h3><ul><li><p>An oscillation that runs down and stops.</p></li><li><p>Damping is due to energy dissipation.</p></li><li><p>Drag force due to air resistance is a common cause.</p></li><li><p>Linear drag model:$F_{\text{drag}} = -bv$</p></li></ul><h4 collapsed="false" seolevelmigrated="true">Damped Oscillations Solution</h4><ul><li><p>When a mass on a spring experiences Hooke’s law and a linear drag force:</p><ul><li><p>$x(t) = Ae^{-\frac{b}{2m}t} \cos(\omega t + \phi_0)$</p><ul><li><p>where the angular frequency is:</p></li><li><p>$\omega = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}} = \sqrt{\omega_0^2 - \frac{b^2}{4m^2}}$</p></li><li><p>$\omega_0 = \sqrt{\frac{k}{m}}$: angular frequency of the undamped oscillator ($b = 0$).</p></li></ul></li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Position-versus-time graph for a damped oscillator</h3><ul><li><p>$x_{\text{max}}= Ae^{-\frac{b}{2m}t}$</p></li></ul><h3 collapsed="false" seolevelmigrated="true">Mathematical Aside: Exponential Decay</h3><ul><li><p>Exponential decay occurs in a vast number of physical systems.</p></li><li><p>Mechanical vibrations, electric circuits, and nuclear radioactivity.</p></li><li><p>$v = A e^{-\frac{t}{\tau}} = A \exp(-\frac{t}{\tau})$</p></li></ul><h3 collapsed="false" seolevelmigrated="true">Energy in Damped Systems</h3><ul><li><p>Due to the drag force, mechanical energy is no longer conserved.</p></li><li><p>$E(t) = E_0 e^{-\frac{t}{\tau}}$</p></li></ul><h4 collapsed="false" seolevelmigrated="true">Time Constant</h4><ul><li><p>$\tau = \frac{m}{b}$: The decay constant of the mechanical energy.</p></li></ul><h5 collapsed="false" seolevelmigrated="true">The oscillator’s mechanical energy decays exponentially with time constant</h5><h3 collapsed="false" seolevelmigrated="true">Driven Oscillations and Resonance</h3><ul><li><p>An oscillating system with a natural frequency$f0$subjected to a periodic external force of driving frequency$f{\text{ext}}$.</p></li><li><p>Amplitude is generally not very high if$f{\text{ext}}$differs much from$f0$.</p></li><li><p>As$f{\text{ext}}$approaches$f0$, the amplitude rises dramatically.</p></li><li><p>Resonance: When$f{\text{ext}} = f0$$, the amplitude is maximum.

• A singer or musical instrument can shatter a crystal goblet by matching the goblet’s natural oscillation frequency.

Significance of the Damping Constant

• The smaller the damping constant:

  • broader response.

  • the wider the peak in the resonating amplitude.