Simple Harmonic Motion (SHM)

Definition & Foundational Mechanics

Simple Harmonic Motion (SHM) is a specific type of periodic motion where the restoring force on an object is directly proportional to the magnitude of the object's displacement from its equilibrium position and acts towards the equilibrium position.

This is formalized mathematically by Hooke's Law ($F=−kx$). Mastery of SHM is highly critical because it serves as the universal mathematical model for almost all stable physical systems near equilibrium, forming the baseline for understanding acoustics, alternating current circuits, structural engineering, and quantum wave mechanics.

System Variables, Dependencies, & Critical Distinctions

• The Restoring Force: A force $F$ that dictates motion according to $F=−kx$.

  • k is the generic restoring constant (e.g., spring constant).

  • x is the displacement from equilibrium.

• Kinematic Signatures: Position, velocity, and acceleration are all sinusoidal and mutually phase-shifted.

  • Position: $x(t)=A\cos (ωt+ϕ)$
    

  • Velocity: $v(t)=−Aω\sin (ωt+ϕ)$
    

  • Acceleration: $a(t)=−A{ω}^2\cos (ωt+ϕ)$
    

  • Amplitude ($A$): Maximum displacement.

  • Angular Frequency ($ω$): The rate of oscillation in radians per second ($ω=2πf=\frac{2π}{T}$).

  • Phase Constant ($ϕ$): Determines the starting position at $t=0$.

• Period Equations: The time $T$ required for one complete cycle.

  • Spring-Mass System: $T=2π\sqrt{\frac{m}{k}}$
    

  • Simple Pendulum: $T=2π\sqrt{\frac{L}{g}}$
    

• Energy Signatures: Mechanical energy continuously transfers between Kinetic ($K$) and Potential ($U$) energy, but the total Mechanical Energy ($E$) remains constant in an undamped system.

  • $U(x)=\frac{1}{2}k{x}^2$
    

  • $K(v)=\frac{1}{2}m{v}^2$
    

  • $E_{total}=\frac{1}{2}k{A}^2=\frac{1}{2}m{v}_{max}^2$
    

• Subtle Nuances:

  • Amplitude Independence: For a true simple harmonic oscillator, the period and frequency are entirely independent of the amplitude. A larger swing requires more distance but results in proportionally higher restoring forces, balancing the time.

  • Small Angle Approximation: A pendulum only exhibits SHM if the swing angle $θ$ is small (typically $<15^{∘}$) where $\sin (θ)≈θ$. At larger angles, the restoring force is no longer linear.

• Common Pitfalls:

  • Failing to recognize that maximum acceleration occurs when velocity is zero (at maximum displacement), and maximum velocity occurs when acceleration is zero (at equilibrium).

  • Using degrees instead of radians when calculating values within sinusoidal kinematic functions.

AP Physics Application Scenarios & Worked Problems

Problem 1: Paragraph Style Problem Describing a Physics' Phenomena (3 pts) Scenario: A student observes a $2\text{ kg}$ block oscillating on a frictionless horizontal table attached to a spring with constant $k$. The student moves the exact same block and spring to hang vertically. The student claims, "Because gravity is now pulling downwards on the block, the vertical system will oscillate with a shorter period and a higher frequency than the horizontal system." Evaluate this claim in a coherent, paragraph-length response.

Solution: The student’s claim is incorrect; the period and frequency of the oscillation will remain identical to the horizontal setup. The period of a spring-mass system is determined solely by the equation $T=2π\sqrt{\frac{m}{k}}$, which depends only on the inertial mass $m$ and the spring constant $k$. While the addition of the gravitational force ($mg$) provides a constant downward pull, this constant force only shifts the equilibrium position of the spring downward by a distance $x=\frac{mg}{k}$. It does not change the stiffness of the spring ($k$) nor the mass of the block ($m$). Because the restoring force relative to the new equilibrium position remains directly proportional to the displacement ($ΔF=−kΔx$), the dynamical behavior and period of the Simple Harmonic Motion remain completely unaffected by the constant gravitational field.

Problem 2: Theoretical Problem w/ Proportions (4.5 pts) Scenario: A block of mass $M$ oscillates on an ideal spring of constant $k$ with an amplitude $A$. The maximum velocity of this system is $v_{max}$. A technician replaces the block with a new mass of $4M$ and pulls it to a new amplitude of $2A$. Determine the maximum velocity of this new system in terms of the original $v_{max}$.

Solution:

• Substitute the definition of angular frequency for a spring

  $v_{max}=A\sqrt{\frac{k}{M}}$

• Set up the equation for the new system using $4M$ and $2A$
  

  $v_{new}=(2A)\sqrt{\frac{k}{4M}}$

• Pull the constants out of the square root

  $v_{new}=2A\left(\frac{1}{\sqrt{4}}\right)\sqrt{\frac{k}{M}}$

• Simplify the fraction

  $v_{new}=2A\left(\frac{1}{2}\right)\sqrt{\frac{k}{M}}$

• Cancel the coefficients to find the new velocity in terms of the initial constants

  $v_{new}=A\sqrt{\frac{k}{M}}$

• Substitute the original definition back in

  $v_{new}=v_{max}$
  *The new maximum velocity is exactly equal to the original $v_{max}$.*

Problem 3: Read a Sinusoidal Curve to find values and solve a problem (2 pts) Scenario: The position-time graph of a $0.5\text{ kg}$ oscillating object is shown below. Calculate the maximum restoring force exerted on the object during its motion.

graph[2\cos(\pi x)][0][4]

Solution:

• Read the Amplitude ($A$) from the y-axis peak

  $A=2\text{ m}$

• Read the Period ($T$) from the x-axis for one full cycle

  $T=2\text{ s}$

• Calculate the angular frequency ($ω$)

  $ω=\frac{2π}{T}=\frac{2π}{2}=π\text{ rad/s}$

• Calculate maximum acceleration ($a_{max}=A{ω}^2$)

  $a_{max}=2(π{)}^2≈19.74{\text{ m/s}}^2$

• Apply Newton's Second Law to find maximum restoring force

  $F_{max}=m{a}_{max}$

• Multiply mass by maximum acceleration

  $F_{max}=(0.5)(19.74)=9.87\text{ N}$

Problem 4: Read an Energy Diagram to solve problems. (6 pts) Scenario: A $2\text{ kg}$ object undergoes 1D motion. Its potential energy $U(x)$ as a function of position is given by the graph below. The object has a total mechanical energy of $16\text{ J}$. Calculate the spring constant ($k$), the amplitude of oscillation ($A$), and the maximum speed of the object ($v_{max}$).

graph[2x^2][-4][4]

Solution:

• Find $k$: Identify a clear point on the graph to isolate $k$. At $x=2\text{ m}$, $U=8\text{ J}$.

  $U=\frac{1}{2}k{x}^2$

• Substitute the values from the graph

  $8=\frac{1}{2}k(2{)}^2$

• Simplify and solve for $k$
  

  $8=2k⟹k=4\text{ N/m}$

• Find $A$: The maximum displacement occurs where total energy equals potential energy ($E=U$).

  $E_{total}=\frac{1}{2}k{A}^2$

• Substitute known total energy and $k$
  

  $16=\frac{1}{2}(4)A^2$

• Isolate $A$
  

  $16=2A^2⟹A^2=8$

• Solve for Amplitude

  $A=\sqrt{8}≈2.83\text{ m}$

• Find $v_{max}$: Maximum speed occurs at equilibrium ($x=0$) where $U=0$ and all $16\text{ J}$ is Kinetic Energy.

  $K_{max}=\frac{1}{2}m{v}_{max}^2$

• Substitute total energy and mass

  $16=\frac{1}{2}(2)v_{max}^2$

• Solve for velocity

  $16=v_{max}^2⟹v_{max}=4\text{ m/s}$

Problem 5: Real World Problem (3 pts) Scenario: A $1200\text{ kg}$ empty vehicle is supported by a suspension system acting as a single effective spring. When four passengers with a combined mass of $300\text{ kg}$ enter the vehicle, the car drops $0.05\text{ m}$ lower to a new equilibrium. If the fully loaded car hits a pothole, what will be the frequency of its resulting vertical oscillations?

Solution:

• Calculate the added gravitational force causing the compression

  $ΔF=m_{added}g=(300)(9.8)=2940\text{ N}$

• Use Hooke's Law to find the effective spring constant ($k_{eff}$)

  $k_{eff}=\frac{ΔF}{Δx}=\frac{2940}{0.05}=58800\text{ N/m}$

• Calculate frequency using total system mass ($1200+300=1500\text{ kg}$)

  $f=\frac{1}{2π}\sqrt{\frac{k_{eff}}{m_{total}}}$

• Substitute values

  $f=\frac{1}{2π}\sqrt{\frac{58800}{1500}}$

• Simplify the fraction under the radical

  $f=\frac{1}{2π}\sqrt{39.2}$

• Calculate final frequency

  $f≈\frac{1}{2π}(6.26)≈1.0\text{ Hz}$

Problem 6: Paragraph Style Problem Reading a Sinusoidal Curve (8.5 pts)
Scenario: A block oscillates horizontally on an ideal spring. The graph below displays the block's velocity $v$ as a function of time $t$. A student analyzing the graph writes: "At $t=1\text{ s}$, the block comes to a stop, so the spring is exerting zero force on the block. Furthermore, at $t=2\text{ s}$, the velocity is zero again, meaning the mechanical energy of the system is entirely kinetic at that instant." Evaluate the student's claims based on the graphical evidence.

graph[-3\sin(\pi x)][0][4]

Solution: The student’s analysis contains critical misconceptions regarding the kinematics and energy dynamics of simple harmonic motion. First, the claim that zero force is exerted at $t=1\text{ s}$ is incorrect. The graph shows that at $t=1\text{ s}$, the velocity is zero. In SHM, velocity is zero strictly at the turning points—the locations of maximum displacement (Amplitude) from equilibrium. Because the restoring force is proportional to displacement ($F=−kx$), maximum displacement directly corresponds to maximum restoring force, and consequently, maximum acceleration. Therefore, at $t=1\text{ s}$, the spring is actually exerting its maximum force, not zero force.

Second, the student is entirely backwards regarding the energy at $t=2\text{ s}$. The graph correctly indicates that velocity is again zero at $t=2\text{ s}$. Since kinetic energy is calculated by $K=\frac{1}{2}m{v}^2$, a velocity of zero dictates that the kinetic energy is also exactly zero. Because the system's total mechanical energy is conserved and equals the sum of kinetic and potential energy ($E=K+U$), the absence of kinetic energy means the mechanical energy of the system is entirely potential energy at $t=2\text{ s}$, stored within the stretched or compressed spring.