AP physics c SHM

Simple Harmonic Motion (SHM) and Simple Harmonic Oscillator (SHO)

• In Depth: Simple Harmonic Motion (SHM) is a specific type of periodic motion where the restoring force is directly proportional to the displacement and acts in the opposite direction. A Simple Harmonic Oscillator (SHO) is a system that exhibits SHM. This typically involves an object oscillating around an equilibrium position due to a restoring force.

Guiding Principle of SHM

• A Linear Restoring Force: This is the fundamental requirement for SHM. The force must be proportional to the displacement from equilibrium.

  • In Depth: The "linear" part means the force increases (or decreases) directly with displacement. "Restoring" means the force always points back towards the equilibrium position, attempting to reduce the displacement. Without those 2 conditions being satisfied then you don't have Simple Harmonic Motion

• Restoring force is given by $F = -kx$, where F and x are vectors, and k > 0.

  • In Depth: This equation mathematically describes the restoring force. $F$ is the force vector, $x$ is the displacement vector from the equilibrium position, and $k$ is the spring constant (a positive constant) that indicates the stiffness of the restoring force. The negative sign indicates that the force is in the opposite direction to the displacement.

• The position of the origin is not arbitrary; it's the equilibrium position where $F_net = 0$.

  • In Depth: The equilibrium position is crucial. It's the point where, in the absence of any displacement, the net force on the object is zero. This is the natural resting point of the system. Defining the origin here simplifies the math and ensures the restoring force is accurately described by $F = -kx$.

SHM and Periodic Motion

• SHM is a subset of periodic motion.

  • In Depth: Periodic motion is any motion that repeats itself at regular intervals. SHM is a specific type of periodic motion with the defining characteristic of the linear restoring force. Not all periodic motion is SHM (e.g., a pendulum swinging with large angles is periodic but not perfectly SHM).

SHM Kinematics

• In Depth: Kinematics describes the motion (position, velocity, acceleration) without considering the forces that cause the motion. This section focuses on how to describe the motion of an object undergoing SHM.

Uniform Circular Motion Review

• Angular velocity: $ω ≡ Δθ/Δt$

  • In Depth: Angular velocity ($ω$) is the rate of change of angular displacement ($Δθ$) with respect to time ($Δt$). It measures how fast an object is rotating or revolving around an axis.

• Linear velocity: $v ≡ Δs/Δt$

  • In Depth: Linear velocity ($v$) is the rate of change of linear displacement ($Δs$) with respect to time ($Δt$). It measures how fast an object is moving in a straight line.

• Relationship between linear and angular velocity: $v = rω$

  • In Depth: This equation connects linear and angular velocity for an object moving in a circle. $r$ is the radius of the circle.

• Angular velocity in terms of period: $ω = 2π/T$

  • In Depth: The period ($T$) is the time it takes for one complete revolution. Since one revolution is $2π$ radians, angular velocity can be expressed as $2π/T$.

• Linear velocity in terms of radius and period: $v = 2πr/T$

  • In Depth: Combining $v = rω$ and $ω = 2π/T$ gives this expression for linear velocity in terms of the radius and period.

Problems for Uniform Circular Motion

• Case 1: Given T = 20 sec, r = 5 m, find ω and v.

  • Solution: $ω = 2π/T = 2π/20 ≈ 0.314$ rad/s, $v = rω = 5 * 0.314 ≈ 1.57$ m/s

• Case 2: Given ω = 5 rad/s, r = 10 m, find T and v.

  • Solution: $T = 2π/ω = 2π/5 ≈ 1.257$ s, $v = rω = 10 * 5 = 50$ m/s

• Case 3: Given ω = 5 rad/s, v = 35 m/s, find T and r.

  • Solution: $T = 2π/ω = 2π/5 ≈ 1.257$ s, $r = v/ω = 35/5 = 7$ m

Kinematics: SHM and Uniform Circular Motion

• Review of uniform circular motion.

  • Assumption: constant $ω$ and constant radius A (amplitude).

    • In Depth: We're using uniform circular motion as an analogy to help understand SHM. The key assumptions are that the angular velocity and radius of the circular motion are constant.

  • Convention: anticlockwise rotation from +x axis.

    • In Depth: This sets a standard for measuring angles. Starting from the positive x-axis and rotating counterclockwise is considered the positive direction.

  • $\theta = \omega t$ (initial condition: $\theta = 0$ at $t = 0$)

    • In Depth: This equation describes the angular position ($\theta$) as a function of time ($t$), assuming the object starts at an angle of 0 at time 0.

  • Tangential speed $|v_T| = Aω$

    • In Depth: Tangential speed is the speed of the object along the circumference of the circle. It is equal to the radius (A, which is also the amplitude in SHM) multiplied by the angular velocity.

  • Centripetal acceleration $|a_c| = |v|^2/A = Aω^2$

    • In Depth: Centripetal acceleration is the acceleration directed towards the center of the circle, necessary to keep the object moving in a circular path. It can be calculated using either the linear speed ($|v|$) and radius (A), or the angular velocity ($ω$) and radius (A).

Kinematics: SHM and Uniform Circular Motion - Position

• $\theta = \omega t$

• $|v| = Aω$ (A is the amplitude)

• $|a| = |v|^2/A = Aω^2$

• Looking at the x component of position: $x = A \cos θ = A \cos(ωt)$

  • In Depth: This is a crucial step in relating circular motion to SHM. The x-component of the position of an object in uniform circular motion, when projected onto the x-axis, follows SHM. $A$ is the amplitude (maximum displacement from equilibrium), and $ωt$ determines the angle at a given time.

Kinematics: SHM and Uniform Circular Motion - Velocity

• $\theta = \omega t$

• $|v| = Aω$

• $|a| = |v|^2/A = Aω^2$

• Looking at the x component of velocity:

  • $x = A \cos θ = A \cos(ωt)$

  • $v_x = -|v| \sin θ = -Aω \sin(ωt)$

    • In Depth: This equation gives the x-component of the velocity of the object. The negative sign indicates that the velocity is negative when the object is moving towards the left (towards negative x) in the SHM.

Kinematics: SHM and Uniform Circular Motion - Acceleration

• $\theta = \omega t$

• $|v| = Aω$

• $|a| = |v|^2/A = Aω^2$

• Looking at the x component of acceleration:

  • $x = A \cos θ = A \cos(ωt)$ (definition of cos)

  • $v_x = -|v| \sin θ = -Aω \sin(ωt)$

  • $a_x = -|a| \cos θ = -Aω^2 \cos(ωt)$

    • In Depth: This equation gives the x-component of the acceleration. It's negative because the acceleration is always directed towards the equilibrium position, opposite to the displacement.

  • $a_x = -ω^2x$

    • In Depth: This is a key relationship in SHM. It shows that the acceleration is proportional to the displacement (x) and in the opposite direction. This is a direct consequence of the linear restoring force.

Kinematics: SHM and Uniform Circular Motion - Frequency and Period

• Relationship between frequency and period:

  • $ω = 2π/T$ (angular frequency of SHM)

    • In Depth: This is the same relationship as in uniform circular motion. Angular frequency is inversely proportional to the period.

  • Frequency definition: $f ≡ 1/T$ (unit: Hertz Hz)

    • In Depth: Frequency is the number of complete cycles per unit of time (usually seconds). Hertz (Hz) is the unit of frequency, representing cycles per second.

  • $ω = 2πf$

  • $x = A \cos(ωt) = A \cos(2πft)$

    • In Depth: This shows the position (x) in terms of frequency (f) instead of angular frequency (ω).

Summary of SHM Equations

• Position: $x = A \cos(ωt)$

• Velocity: $v_x = -Aω \sin(ωt)$

• Acceleration: $a_x = -Aω^2 \cos(ωt)$

• $a_x = -ω^2x$

• $ω$: angular frequency

• $A$: amplitude

• $f = ω/2π$: frequency

• $T = 1/f$: period

Practice Question

• An object moves according to $x = 3\cos 5t$ (x in meter).

  1. What is the maximum displacement? (aka amplitude)

     • Answer: 3 meters (the coefficient of the cosine function)

  2. What is the period? (time for one cycle)

     • Answer: $T = 2π/ω = 2π/5 ≈ 1.257$ seconds

  3. What is the frequency? Angular frequency?

     • Answer: Angular frequency is 5 rad/s. Frequency is $f = 1/T = 5/(2π) ≈ 0.796$ Hz

  4. What is the maximum speed?

     • Answer: $v_{max} = Aω = 3 * 5 = 15$ m/s

  5. What is the maximum acceleration (magnitude)?

     • Answer: $a_{max} = Aω^2 = 3 * 5^2 = 75$ m/s\2

Kinematics SHM - Sine vs Cosine Graphs

• What is the difference in shape of the sine and cosine graphs? (Imagine if the axes are not drawn)

  • In Depth: The sine and cosine graphs are the same shape, but shifted. The cosine graph starts at its maximum value, while the sine graph starts at zero. This shift is $90^\circ$ or $\pi/2$ radians.

Kinematics SHM - Starting at -90 Degrees

• If you start $t = 0$ at $-90^ o$$, then $x = A \sin θ = A \sin(ωt)$

  • In Depth: If the initial condition is such that the object starts at -90 degrees (or -$\pi/2$ radians), then the x-position is described by a sine function instead of a cosine function.

• If you look at y motion instead of x, then $y = A \sin(ωt)$

  • In Depth: This is because the y-component of the position in uniform circular motion, projected onto the y-axis, follows a sine function.

Practice Question - Sine Function

• An object moves according to $x = 3\sin 5t$

  1. What is the maximum displacement? (aka amplitude)

     • Answer: 3 meters

  2. What is the period? (time for one cycle)

     • Answer: $T = 2π/5 ≈ 1.257$ seconds

  3. What is the frequency? Angular frequency?

     • Answer: Angular frequency is 5 rad/s. Frequency is $f = 5/(2π) ≈ 0.796$ Hz

  4. What is the maximum speed?

     • Answer: $v_{max} = Aω = 3 * 5 = 15$ m/s

  5. What is the maximum acceleration (magnitude)?

     • Answer: $a_{max} = Aω^2 = 3 * 5^2 = 75$ m/s\2

Practice Question - Shifted Sine Function

• An object moves according to $x = 3\sin[5(t - 2)]$

  1. What is the maximum displacement? (amplitude)

     • Answer: 3 meters

  2. What is the period? (time for one cycle)

     • Answer: $T = 2π/5 ≈ 1.257$ seconds

  3. What is the frequency? Angular frequency?

     • Answer: Angular frequency is 5 rad/s. Frequency is $f = 5/(2π) ≈ 0.796$ Hz

  4. What is the maximum speed?

     • Answer: $v_{max} = Aω = 3 * 5 = 15$ m/s

  5. What is the maximum acceleration (magnitude)?

     • Answer: $a_{max} = Aω^2 = 3 * 5^2 = 75$ m/s\2

Phase Angle

• In general, the origin (zero) in time could change.

  • In Depth: The starting point of the oscillation can be different from t=0. This starting point is accounted for by the phase angle.

• More general form of SHM is $x = A \cos(ωt + φ)$

  • In Depth: This is the most general equation for SHM, incorporating the phase angle ($φ$).

• $φ$ (Greek letter phi) is a constant called phase angle.

  • In Depth: The phase angle represents the initial phase of the oscillation at time t=0. It essentially shifts the cosine function left or right along the time axis.

• So phase angle is just “time shifting”.

  • In Depth: The phase angle determines the position of the object at t=0. A non-zero phase angle means the object doesn't start at its maximum displacement (A) or at the equilibrium position (0).

• What is the phase angle for $x = A \sin(ωt)$?

  • Answer: $-\pi/2$ or -90 degrees, because $\sin(ωt) = \cos(ωt - \pi/2)$

Practice SHM - Finding Phase Angle from Graph

• Find the approximate phase angle in this SHM graph.

  • (Requires a graph to provide a numerical answer. The phase angle is determined by the initial position of the object on the graph.)

Practice SHM - Finding Period, Amplitude, and Frequency from Graph

• Here is a graph of a SHM. Find the approximate:

  1. Period (time for 1 cycle)

     • (Requires a graph to provide a numerical answer. Measure the time for one complete cycle.)

  2. Amplitude

     • (Requires a graph to provide a numerical answer. Measure the maximum displacement from the equilibrium position.)

  3. Frequency

     • (Requires a graph to provide a numerical answer. Calculate $f = 1/T$ after finding the period.)

Practice SHM - Finding Max Speed and When Speed is Max/Zero

1. Calculate the max speed (approximate)

   • (Requires a graph to provide a numerical answer. Use $v_{max} = Aω$ after finding A and calculating $ω = 2π/T$)

2. At what time is speed max? (eyeball)

   • (Requires a graph to provide a numerical answer. The speed is maximum when the object passes through the equilibrium position.)

3. At what time is speed zero? (eyeball)

   • (Requires a graph to provide a numerical answer. The speed is zero at the points of maximum displacement (the peaks and troughs of the graph).)

Practice SHM - Finding Max Acceleration and When Acceleration is Max/Zero

1. Calculate the max acceleration (approximate)

   • (Requires a graph to provide a numerical answer. Use $a_{max} = Aω^2$ after finding A and calculating ω)

2. At what time is acceleration max? (eyeball)

   • (Requires a graph to provide a numerical answer. The acceleration is maximum at the points of maximum displacement (the peaks and troughs of the graph).)

3. At what time is acceleration zero? (eyeball)

   • (Requires a graph to provide a numerical answer. The acceleration is zero when the object passes through the equilibrium position.)

SHM Kinematics Graphs - Displacement, Velocity, and Acceleration

• Summary of displacement, velocity and acceleration graphs.

  • In Depth:

    • Displacement: A cosine (or sine) wave. It shows the position of the object as a function of time.

    • Velocity: A sine (or cosine) wave, 90 degrees out of phase with the displacement. It's maximum when the displacement is zero and vice versa.

    • Acceleration: A cosine (or sine) wave, 180 degrees out of phase with the displacement. It's maximum (in magnitude) when the displacement is maximum and in the opposite direction.

Practice Question (SHM Kinematics) - Check Your Understanding

• Question 5: Which object has the greatest maximum velocity?

  • (Requires additional context about the objects being referred to.)

• Question 6: Where on the path of the shadow is the acceleration equal to zero?

  • (Assuming the shadow is from an object in SHM: The acceleration is zero at the equilibrium position.)

• Question 7: A particle is oscillating in simple harmonic motion. The time required for the particle to travel through one complete cycle is equal to the period of the motion, no matter what the amplitude is. But how can this be, since larger amplitudes mean that the particle travels farther?

  • Answer: While the particle travels farther with larger amplitudes, it also travels faster. The increased speed compensates for the increased distance, resulting in the same period. This is because the restoring force is proportional to the displacement; larger displacements result in larger forces and therefore larger accelerations and speeds.

SHM Dynamics

• In Depth: Dynamics relates the motion of an object to the forces that cause it. This section focuses on the forces involved in SHM.

Guiding Principle of SHM (Dynamics)

• A linear restoring force (proportional to displacement).

  • In Depth: As discussed before, this is the defining characteristic of SHM.

• Restoring force $F = -kx$ (F and x are vectors, k > 0).

  • In Depth: The same equation as before, but now we're focusing on it as the cause of the motion.

• Important:

  • Position of origin is not arbitrary.

    • In Depth: The origin must be at the equilibrium position.

  • Position of origin is equilibrium ($F_{net} = 0$) position.

    • In Depth: This ensures that the restoring force is accurately described and simplifies the equations.

  • Is F the applied force or the spring force?

    • In Depth: $F$ in $F = -kx$ refers to the restoring force, which is often (but not always) the spring force. It's the force that's causing the object to return to its equilibrium position.

SHM Dynamics - Relation to Uniform Circular Motion

• Kinematics equation relating SHM to uniform circular motion

  • $a_x = -ω^2x$

    • In Depth: This equation, derived from the kinematic analysis, is crucial for understanding the dynamics.

• Force on the mass is therefore

  • $F<em>x = ma</em>x = -(mω^2)x$

    • In Depth: Using Newton's second law ($F = ma$), we substitute the acceleration from the kinematics equation.

• Substitute $k = mω^2$ then

  • $F_x = -kx$

    • In Depth: This substitution connects the dynamics (force) with the kinematics (angular frequency) through the spring constant k.

• “Spring equation”

  • In Depth: This refers to the equation $F = -kx$, which is often associated with springs but applies to any system with a linear restoring force.

SHM Dynamics - Example: Spring

• $F = -kx$ (F and x are vectors)

• Linear restoring force with origin on the equilibrium position

  • In Depth: A mass attached to a spring is the classic example of SHM. The spring provides the linear restoring force, and the equilibrium position is where the spring is neither stretched nor compressed.

SHM Mass Attached to a Spring - Period of SHM

• $T = 2π/ω$ (same as circular motion formula)

• $k = mω^2$

• $T = 2π/ω = 2π \sqrt(m/k)$

  • In Depth: This is a very important result. It shows that the period of SHM for a mass-spring system depends only on the mass (m) and the spring constant (k). A larger mass will increase the period (slower oscillation), and a stiffer spring will decrease the period (faster oscillation).

• Position of SHM

  • $x = A \cos(ωt) = A \cos(2πft) = A \cos(2πt/T)$

    • In Depth: This equation shows how to express the position (x) in terms of angular frequency (ω), frequency (f), or period (T).

SHM Mass Attached to a Spring - Pivot Interactives

• Pivot Interactives: Spring Oscillators Making and Testing a Model

  • In Depth: This refers to an interactive simulation or experiment that can help visualize and understand the concepts of SHM with a mass-spring system.

Horizontal vs Vertical Spring

• New equilibrium position (lower, stretched)

  • In Depth: When a spring is vertical and a mass is attached, gravity causes the spring to stretch. The new equilibrium position is where the spring force balances the gravitational force.

• Express $d_0$ in terms of k, m and constant

  • In Depth: Relates to calculating the displacement from the original spring length at rest to the new spring length at equilibrium. The constant referred to here is likely $g$, acceleration due to gravity. The force due to gravity $F = mg$ and the force of the spring $F = kd$. At equilibrium these forces are equal so $mg = kd$. Then $d = mg/k$. Here, $d_0$ is the same as $d$.

Horizontal vs Vertical Spring - Additional Displacement

• Spring + mass is stretched an additional distance A from the equilibrium position and released

• Will the mass undergo SHM? Justify.

  • Answer: Yes, it will undergo SHM. The restoring force is still proportional to the displacement from the new equilibrium position.

• What is the period of SHM?

  • Answer: The period is the same as for a horizontal spring: $T = 2π \sqrt(m/k)$. Gravity only shifts the equilibrium position, it doesn't affect the period.

• What is the amplitude of SHM?

  • Answer: The amplitude is A, the additional distance the spring is stretched from the equilibrium position.

SHM (Calculus)

• In Depth: This section uses calculus to provide a more rigorous mathematical treatment of SHM.

• $v = dx/dt$

  • In Depth: Velocity is the derivative of position with respect to time.

• $a = dv/dt = d^2x/dt^2$

  • In Depth: Acceleration is the derivative of velocity with respect to time, which is also the second derivative of position with respect to time.

• $F = ma = m d^2x/dt^2$

  • In Depth: Newton's second law, expressed using calculus.

• Linear restoring force $F = -kx$

• Substitute, $m d^2x/dt^2 + kx = 0$

  • In Depth: Substituting the expression for the restoring force into Newton's second law gives this second-order differential equation.

• Substitute $k/m = ω^2$ (k and m are positive)

• $d^2x/dt^2 + ω^2x = 0$ (2nd order differential equation in x)

  • In Depth: This is the standard form of the differential equation for SHM. The solutions to this equation will describe the motion of an object undergoing SHM.

• Solution:

  • $x = A \cos(ωt + φ)$ or $x = a \cos(ωt) + b \sin(ωt)$

    • In Depth: These are two equivalent general solutions to the differential equation. The first uses amplitude (A) and phase angle (φ), while the second uses coefficients a and b for the cosine and sine terms.

  • $(A, φ)$ or $(a, b)$ are constants of integration

    • In Depth: These constants are determined by the initial conditions of the system (e.g., initial position and velocity).

Practice (Calculus) - Proving SHM Solution

1. Prove that the solution of $d^2x/dt^2 + ω^2x = 0$ is $x = A \cos(ωt + φ)$ by substituting the solution into the differential equation

   • Solution: Take the second derivative of $x = A \cos(ωt + φ)$: $v = -Aω \sin(ωt + φ)$, $a = -Aω^2 \cos(ωt + φ) = -ω^2x$. Substitute this back into the differential equation: $-ω^2x + ω^2x = 0$, which confirms that the solution is valid.

2. Use $v = dx/dt$ and derive an expression for v

   • Solution: $v = dx/dt = -Aω \sin(ωt + φ)$

3. Use $a = dv/dt$ and derive an expression for a

   • Solution: $a = dv/dt = -Aω^2 \cos(ωt + φ)$

4. Express $v_{max}$ in terms of A, $ω$, $φ$, t

   • Solution: $v_{max} = Aω$ (occurs when $\sin(ωt + φ) = -1$)

5. Express $a_{max}$ in terms of A, $ω$, $φ$, t

   • Solution: $a_{max} = Aω^2$ (occurs when $\cos(ωt + φ) = -1$)

Springs in Parallel (Review)

• Two springs are connected in parallel as shown

• Neglect gravity

• Applied force $F_{app}$ towards the right

• Draw the FBD of the black ring in the middle when the spring is in equilibrium

• Derive an expression for the displacement $Δx$ of the springs in terms of $F<em>{app}$, spring constants $k</em>{top}$ and $k_{bottom}$

  • Solution: In parallel, both springs experience the same displacement. The total force is the sum of the forces from each spring: $F_{app} = k_{top}Δx + k_{bottom}Δx = (k_{top} + k_{bottom})Δx$. Therefore, $Δx = F_{app} / (k_{top} + k_{bottom})$.

Springs in Series (Review)

• Two springs are connected in series as shown

• Neglect gravity

• Applied force $F_{app}$ towards the right

• Derive an expression for the total displacement $Δx$ of the springs in terms of $F<em>{app}$, spring constants $k</em>{left}$ and $k_{right}$

  • Solution: In series, both springs experience the same force. The total displacement is the sum of the displacements of each spring: $Δx = Δx_{left} + Δx_{right} = F_{app}/k_{left} + F_{app}/k_{right} = F_{app}(1/k_{left} + 1/k_{right})$.

Resonance

• Resonance occurs when an external force is applied at the same frequency as the natural frequency of the oscillating system

  • In Depth: Every oscillating system has a natural frequency at which it oscillates freely. Applying a force at this frequency causes the amplitude of the oscillations to increase dramatically.

• Increasing amplitude every cycle

  • In Depth: If energy is continuously added to the system at its resonant frequency, the amplitude will continue to grow, potentially leading to system failure.

• Watch video: https://www.youtube.com/watch?v=5JbpcsH80us

  • In Depth: This video likely demonstrates the effects of resonance.

SHM Energy Analysis

• SHM: mass m attached to spring with spring constant k

• At any time t

  • $x = A \cos(ωt)$

  • $v = -Aω \sin(ωt)$

• Derive expression for kinetic energy $K(t)$

  • Solution: $K(t) = (1/2)mv^2 = (1/2)m(-Aω \sin(ωt))^2 = (1/2)mA^2ω^2\sin^2(ωt)$

• Derive expression for spring energy $U_s(t)$

  • Solution: $U_s(t) = (1/2)kx^2 = (1/2)k(A \cos(ωt))^2 = (1/2)kA^2\cos^2(ωt)$

• Derive expression for total energy $E(t)$

  • Solution: $E(t) = K(t) + U_s(t) = (1/2)mA^2ω^2\sin^2(ωt) + (1/2)kA^2\cos^2(ωt)$. Since $k = mω^2$, this simplifies to $E(t) = (1/2)kA^2$

• Is E constant in time?

  • Answer: Yes, E is constant in time. The total energy in SHM is conserved and is proportional to the square of the amplitude.

SHM Energy Analysis - Max Potential and Kinetic Energy Locations

• At what location(s) is the $U_s$ max?

  • Answer: The potential energy is max at the points of maximum displacement, $x = ±A$.

• At what location(s) is the K max?

  • Answer: The kinetic energy is max at the equilibrium position, $x = 0$.

SHM Energy Analysis (Review Previous Spring Phet)

Practice Question