Physics Week 2: Simple Harmonic Motion (SHM) and Pendulums

Simple Harmonic Motion (SHM) Fundamentals

The Physics Behind SHM

• Definition: Simple Harmonic Motion (SHM) describes the oscillatory motion of an object under a restoring force directly proportional to the displacement and opposite in direction.

• Mass-Spring System: Consider a mass attached to a horizontal spring. When the spring is stretched or compressed, the mass experiences a net restoring force.

• Hooke's Law and Newton's Second Law: The restoring force F is given by Hooke's Law -kx, where k is the spring constant and x is the displacement from equilibrium. By Newton's Second Law (F = ma), we have:
  F = ma = -kx

• Differential Equation of SHM: Since acceleration (a) is the second derivative of position (x) with respect to time (ẍ),
  mẍ = -kx
  ẍ = -\frac{k}{m}x

• Nature of the Solution: The function x(t) describing the position in SHM must be a function whose second derivative is equal to the original function multiplied by some negative constant (-\frac{k}{m}).

  • Examples of functions that satisfy this property (where the second derivative is a constant times the original function) include functions like A\sin(bx) , A\cos(bx) , and Ae^{bx}.

    • For f(x) = \sin(3x), f''(x) = -9\sin(3x). (Works)

    • For f(x) = \cos(2x), f''(x) = -4\cos(2x). (Works)

    • For f(x) = e^{5x}, f''(x) = 25e^{5x}. (Works)

    • For f(x) = \tan(4x), f''(x) = 2\tan(4x)\sec^2(4x). (Does NOT work, as it's not a constant times the original function).

The Math Behind SHM

• General Solution: Sine, cosine, or exponential functions can describe SHM. For simplicity, we typically use sine and cosine functions. They are related by the identity: \sin\theta = \cos(\theta - \frac{\pi}{2}).

• Position Function: The most general form of the position function for SHM using cosine is: x(t) = A\cos(\omega t + \phi)

  • A: Amplitude, the maximum displacement from equilibrium (units of length, e.g., meters).

  • \omega: Angular frequency, a measure of how quickly the oscillations occur (units of radians per second, rad/s).

  • \phi: Phase constant (or initial phase angle), determines the initial position of the oscillator at t=0 (units of radians).

  • (\omega t + \phi): The entire expression within the cosine function is called the phase (units of radians).

• Derivatives of the Position Function:

  • Velocity Function: Taking the first derivative of x(t) yields the velocity function:
    v(t) = ẋ(t) = -A\omega\sin(\omega t + \phi)

  • Acceleration Function: Taking the second derivative of x(t), or the first derivative of v(t), yields the acceleration function:
    a(t) = ẍ(t) = -A\omega^2\cos(\omega t + \phi)

  • Verification: This confirms that the second derivative (a(t)) is equal to a constant (-\omega^2) times the original position function (x(t)), thus satisfying the SHM differential equation (a(t) = -\omega^2 x(t)).

• Solving for Variables in the Phase: Due to the periodic nature of trigonometric functions, solving for variables within the phase \omega t + \phi typically yields an infinite number of solutions. To find a unique solution, additional constraints (e.g., smallest positive value) are often provided.

  • Example: To solve 1/2 = \cos(5t + \frac{\pi}{2}) for the smallest positive value of t:

    • The inverse cosine of 1/2 yields principal values of \frac{\pi}{3} and \frac{5\pi}{3} (using the CAST rule for positive cosine).

    • Case 1: 5t + \frac{\pi}{2} = \frac{\pi}{3} \Rightarrow 5t = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6} \Rightarrow t = -\frac{\pi}{30} (negative time, not smallest positive).

    • Case 2: 5t + \frac{\pi}{2} = \frac{5\pi}{3} \Rightarrow 5t = \frac{5\pi}{3} - \frac{\pi}{2} = \frac{10\pi}{6} - \frac{3\pi}{6} = \frac{7\pi}{6} \Rightarrow t = \frac{7\pi}{30} (smallest positive time).

Sample Problem: Graphing SHM

• Problem: Sketch a position versus time graph for the motion of an object described by x(t) = 2.0\text{m } \cos(\pi t + \frac{\pi}{6}).

• Amplitude: The amplitude is A = 2.0\text{m}, so the maximum and minimum position values are \pm 2.0\text{m}.

• Initial Position (t=0):
  x(0) = 2.0\text{m } \cos(0 + \frac{\pi}{6}) = 2.0\text{m } \cos(\frac{\pi}{6}) = 2.0\text{m } (\frac{\sqrt{3}}{2}) \approx 1.732\text{m}. The phase constant \phi = \frac{\pi}{6} determines the object's initial position, velocity, and acceleration.

• Angular Frequency: The angular frequency is \omega = \pi\text{ rad/s}.

• Period (T): The period is related to angular frequency by \omega = \frac{2\pi}{T}.
  \pi = \frac{2\pi}{T} \Rightarrow T = 2.0\text{s}.

• Time Intercepts (Zeros, x(t)=0):

  • Set x(t)=0:
    0 = 2.0\text{m } \cos(\pi t + \frac{\pi}{6})

  • This implies the phase must be an odd multiple of \frac{\pi}{2} (e.g., \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, etc.).

  • First zero: \pi t + \frac{\pi}{6} = \frac{\pi}{2} \Rightarrow \pi t = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \Rightarrow t = \frac{1}{3}\text{s}.

  • Second zero: \pi t + \frac{\pi}{6} = \frac{3\pi}{2} \Rightarrow \pi t = \frac{3\pi}{2} - \frac{\pi}{6} = \frac{9\pi}{6} - \frac{\pi}{6} = \frac{8\pi}{6} = \frac{4\pi}{3} \Rightarrow t = \frac{4}{3}\text{s}.

  • These two zeros are separated by half a period (T/2 = (4/3) - (1/3) = 1\text{s}), confirming T=2.0\text{s}. Other zeros occur at \dots, -\frac{2}{3}\text{s}, \frac{7}{3}\text{s}, \dots (separated by periods).

Determining Angular Frequency (\omega)

• Relating Physics to Math: By equating the physical acceleration (ẍ = -\frac{k}{m}x) with the mathematical acceleration (ẍ = -A\omega^2\cos(\omega t + \phi) = -\omega^2x):
  -\frac{k}{m}x = -\omega^2x (Since x = A\cos(\omega t + \phi))
  \omega^2 = \frac{k}{m} \Rightarrow \omega = \sqrt{\frac{k}{m}}

• Physical Implications: This equation shows the factors determining the angular frequency of a mass-spring system:

  • Spring Constant (k): A higher spring constant (stiffer spring) leads to a higher angular frequency, meaning faster oscillations.

  • Mass (m): A higher mass leads to a lower angular frequency, meaning slower oscillations.

• Period and Frequency: Angular frequency is also related to frequency (f) and period (T):
  \omega = 2\pi f = \frac{2\pi}{T}
  Therefore, the period of a mass-spring system is:
  T = 2\pi \sqrt{\frac{m}{k}}
  This implies that the period (T) is proportional to the square root of the mass \sqrt{m}, meaning as mass increases, the period increases but not linearly.

Concept Question: Properties of SHM on a Spring

• Consider an object on a spring oscillating: Which statement is correct?

  • A) At the spring's equilibrium position, the object's acceleration and velocity are maximum.

  • B) At the spring's maximum compression, the object's acceleration is zero.

  • C) At the spring's maximum stretch, the object's acceleration is maximum.

  • D) At spring's maximum compression, the object's acceleration and velocity are zero.

• Correct Answer: C

  • At maximum compression or extension (endpoints of motion):

    • Displacement (x) is maximum (=A).

    • Force (F = -kx) is greatest in magnitude.

    • Acceleration (a = -\frac{k}{m}x) is greatest in magnitude.

    • Velocity (v) is instantaneously zero as the object changes direction.

  • At equilibrium position (x=0):

    • Displacement (x) is zero.

    • Force (F = -kx) is zero.

    • Acceleration (a = -\frac{k}{m}x) is zero.

    • Velocity (v) is maximum.

Sample Problem: Velocity Calculation

• Problem: A 1.6\text{ kg} block on a horizontal frictionless surface is attached to a spring (k = 420\text{ N/m}). The block is pulled from equilibrium (x=0) to x = 0.080\text{ m} and released from rest. Find the velocity of the block at t=0.40\text{s}.

• Step 1: Find the position equation (x(t) = A\cos(\omega t + \phi)).

  • Amplitude (A): The block is pulled to 0.080\text{ m} from equilibrium, so A = 0.080\text{ m}.

  • Phase Constant (\phi): Released from rest at maximum positive displacement (x(0)=A).
    x(0) = A\cos(\phi) = A \Rightarrow \cos(\phi) = 1 \Rightarrow \phi = 0 radians.

  • Angular Frequency (\omega):
    \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{420\text{ N/m}}{1.6\text{ kg}}} = \sqrt{262.5}\text{ rad/s} \approx 16.20\text{ rad/s}.

  • Position equation: x(t) = (0.080\text{m})\cos(16.20t).

• Step 2: Find the velocity equation (v(t) = -A\omega\sin(\omega t + \phi)).
  v(t) = -(0.080\text{m})(16.20\text{ rad/s})\sin(16.20t). (Since \phi=0)

• Step 3: Calculate velocity at t=0.40\text{s}. v(0.40\text{s}) = -(0.080)(16.20)\sin(16.20 \times 0.40) radians v(0.40\text{s}) = -1.296\sin(6.48\text{ rad})

  • Since 6.48 rad is in the first quadrant (approximately 6.48 - 2\pi \approx 6.48 - 6.28 = 0.20 rad), \sin(6.48) is positive.
    v(0.40\text{s}) \approx -1.296 \times 0.198 \approx -0.257\text{ m/s}.
    This is closest to -0.1 m/s among the given options.

Sample Problem: Time for Specific Acceleration

• Problem: A 0.28\text{ kg} block on a frictionless surface is attached to a spring (k = 0.500\text{ kN/m} = 500\text{ N/m}). Pulled from equilibrium to +0.080\text{ m} and released from rest. When is the first time the block has an acceleration of 93\text{ m/s}^2?

• Method: Use Newton's second law (ma(t) = -kx(t)) and the position equation.

  • Position equation: $x(t) = A\cos(\omega t + \phi)$.

    • A = 0.080\text{m}.

    • Released from rest at x=A, so \phi = 0.

    • \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{500\text{ N/m}}{0.28\text{ kg}}} = \sqrt{1785.71}\text{ rad/s} \approx 42.2577\text{ rad/s}.

    • x(t) = (0.080\text{m})\cos(42.2577t).

  • Acceleration equation: Substitute x(t) into ma(t) = -kx(t).
    ma(t) = -k (0.080\text{m})\cos(\omega t)
    a(t) = -\frac{k}{m} (0.080\text{m})\cos(\omega t)

  • Solve for t: Set a(t) = 93\text{ m/s}^2. 93 = -\frac{500}{0.28} (0.080)\cos(42.2577t) 93 = -1785.71 \times 0.080 \cos(42.2577t) 93 = -142.857\cos(42.2577t) \cos(42.2577t) = -\frac{93}{142.857} \approx -0.651 42.2577t = \cos^{-1}(-0.651)

    • The first positive angle for which cosine is -0.651 is approximately 2.27838 radians.
      42.2577t = 2.27838\text{ rad}
      t = \frac{2.27838}{42.2577} \approx 0.0539\text{ s}.
      Closest option: 0.054\text{ s}.

Spring Systems

Springs in Series or Parallel

• Springs in Parallel:

  • Configuration: Springs are arranged side-by-side, sharing the load, producing the same extension or compression when a force is applied. If a mass is connected directly to multiple springs, or if springs are arranged such that they both extend/compress when a mass moves (e.g., mass between two springs fixed to walls).

  • Effective Spring Constant (k{eff}): The total force from the springs is the sum of individual forces.
    F{total} = F1 + F2 = k1x + k2x = (k1 + k2)x
    k{eff} = k1 + k_2

• Springs in Series:

  • Configuration: Springs are connected end-to-end; the force through each spring is the same, but the total displacement is the sum of individual displacements.

  • Effective Spring Constant (k{eff}): The total extension is the sum of individual extensions.
    x{total} = x1 + x2
    Using Hooke's Law ($x = F/k$), and recognizing F is the same for both springs:
    \frac{F}{k{eff}} = \frac{F}{k1} + \frac{F}{k2} \frac{1}{k{eff}} = \frac{1}{k1} + \frac{1}{k2}

Concept Question: Effective Spring Constant

• Problem: Consider a mass attached to two springs (k1 = 20\text{ N/m}, k2 = 60\text{ N/m}) where the mass is between them and they are fixed to walls (e.g., horizontally). What is the effective spring constant?

• Solution: Even though it visually appears like a series arrangement, if the mass moves a distance x in either direction, one spring stretches by x and the other compresses by x. Both springs exert a force simultaneously opposing the motion, contributing to the total restoring force. This configuration acts exactly like two springs in parallel.
  k{eff} = k1 + k_2 = 20\text{ N/m} + 60\text{ N/m} = 80\text{ N/m}.

Concept Question: Angular Frequency with Series Springs

• Problem: A 1.2\text{ kg} mass is attached to two vertical springs connected in series. k1 = 200\text{ N/m}, k2 = 800\text{ N/m}. The mass is pulled down 5.0\text{ cm} and released. What is its angular frequency?

• Step 1: Calculate the effective spring constant for springs in series.
  \frac{1}{k{eff}} = \frac{1}{k1} + \frac{1}{k2} = \frac{1}{200\text{ N/m}} + \frac{1}{800\text{ N/m}} \frac{1}{k{eff}} = \frac{4}{800\text{ N/m}} + \frac{1}{800\text{ N/m}} = \frac{5}{800\text{ N/m}} = \frac{1}{160\text{ N/m}}
  k_{eff} = 160\text{ N/m}.

• Step 2: Calculate the angular frequency.
  \omega = \sqrt{\frac{k_{eff}}{m}} = \sqrt{\frac{160\text{ N/m}}{1.2\text{ kg}}} = \sqrt{133.33}\text{ rad/s} \approx 11.547\text{ rad/s}.
  Closest option: 12\text{ rad/s}.

Springs and Energy

• Total Energy in SHM: The total mechanical energy (E) of a mass-spring system in SHM is conserved and is the sum of kinetic energy (KE) and potential (elastic) energy (PE{elastic}). E = KE + PE{elastic} = \frac{1}{2}mv^2 + \frac{1}{2}kx^2

• Energy at Extremes: At the maximum displacement (amplitude A), velocity is zero, so all energy is elastic potential energy:
  E = \frac{1}{2}kA^2

• Energy at Equilibrium: At the equilibrium position (x=0), potential energy is zero, so all energy is kinetic energy (maximum kinetic energy):
  E = \frac{1}{2}mv_{max}^2

Sample Problem: Speed Using Energy Conservation

• Problem: A 2.0\text{ kg} mass is hung from a spring, stretching it 15\text{ cm} (equilibrium position for oscillation). The spring is then pulled down a further 12\text{ cm} and released (amplitude of oscillation A = 12\text{ cm} = 0.12\text{ m}). What is the speed of the mass when it passes through the point where the spring was stretched by 15\text{ cm} (i.e., the equilibrium position of the oscillation)?

• Step 1: Find the spring constant (k).
  When the mass hangs in static equilibrium, the gravitational force balances the spring force at the initial stretch of x0 = 0.15\text{ m}. kx0 = mg \Rightarrow k = \frac{mg}{x_0} = \frac{(2.0\text{ kg})(9.8\text{ m/s}^2)}{0.15\text{ m}} \approx 130.67\text{ N/m}.

• Step 2: Use conservation of energy to find maximum speed (v{max}).
  When released from rest at amplitude A, all energy is potential. At the equilibrium position (x=0), all energy is kinetic.
  \frac{1}{2}kA^2 = \frac{1}{2}mv{max}^2
  v{max}^2 = \frac{kA^2}{m} \Rightarrow v{max} = A\sqrt{\frac{k}{m}}
  v_{max} = (0.12\text{ m})\sqrt{\frac{130.67\text{ N/m}}{2.0\text{ kg}}} = (0.12\text{ m})\sqrt{65.335}\text{ rad/s} \approx (0.12)(8.083)\text{ m/s} \approx 0.97\text{ m/s}.

Sample Problem: Time When KE = PE

• Problem: A SHM oscillator consists of a 1.00 \times 10^2\text{ g} (0.100\text{ kg}) mass attached to a spring (k = 1.00 \times 10^2\text{ N/m}). The oscillator is displaced 20.0\text{ cm} (0.20\text{ m}) from equilibrium and released. How long after release will its kinetic energy equal its elastic energy?

• Step 1: Determine the position (x) when KE = PE.
  Total energy E = \frac{1}{2}kA^2. If KE = PE, then each must be half of the total energy: KE = PE = \frac{1}{2}E = \frac{1}{4}kA^2.
  Set potential energy equal to a quarter of the total energy:
  \frac{1}{2}kx^2 = \frac{1}{4}kA^2
  x^2 = \frac{1}{2}A^2 \Rightarrow x = \pm \frac{A}{\sqrt{2}}.

• Step 2: Find the time (t) using the position function.

  • Amplitude (A): 0.20\text{ m}.

  • Phase Constant (\phi): Released from rest at maximum displacement (x(0)=A), so \phi=0.

  • Position function: x(t) = A\cos(\omega t).

  • Substitute x: \frac{A}{\sqrt{2}} = A\cos(\omega t) \Rightarrow \cos(\omega t) = \frac{1}{\sqrt{2}}.

    • The first positive angle whose cosine is \frac{1}{\sqrt{2}} is \frac{\pi}{4} radians.
      \omega t = \frac{\pi}{4}

  • Angular Frequency (\omega):
    \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100\text{ N/m}}{0.100\text{ kg}}} = \sqrt{1000}\text{ rad/s} \approx 31.62\text{ rad/s}.

  • Solve for t:
    t = \frac{\pi/4}{\omega} = \frac{\pi/4}{\sqrt{1000}} \approx \frac{0.7854}{31.62} \approx 0.0248\text{ s}.

Concept Question: Speed at Equilibrium for Different Masses

• Problem: Two identical springs. One mass M, the other 2M. Both pulled down by identical amounts (amplitude A) and released. Do both oscillators have the same speed at equilibrium? (True/False)

• Solution: False

  • Maximum speed at equilibrium is given by v{max} = A\omega or using energy, \frac{1}{2}kA^2 = \frac{1}{2}mv{max}^2 \Rightarrow v_{max} = A\sqrt{\frac{k}{m}}.

  • Since the masses are different (M vs. 2M), their angular frequencies (\omega = \sqrt{k/m}) will be different. The larger mass will have a smaller angular frequency.

  • Therefore, the larger mass will have a smaller maximum speed at equilibrium (v{max,2M} = A\sqrt{k/(2M)} = v{max,M}/\sqrt{2}).

Pendulums

• Small Angle Approximation: The equations of motion for a simple pendulum to behave as a Simple Harmonic Oscillator only apply when the displacement angle (\theta) from the vertical is small (typically \theta \le 6^\circ or \frac{\pi}{30} radians).

  • In the small angle approximation, \sin\theta \approx \tan\theta \approx \theta (where \theta is in radians).

  • This approximation does not apply to the cosine function; for small \theta, \cos\theta \approx 1 - \frac{\theta^2}{2}.

• Energy Equations: Energy conservation principles for a pendulum apply regardless of the displacement angle.

Concept Question: Small Angle Approximation Comparison

• Problem: Which of the following statements about small angle approximation is NOT correct?

  • A) \sin(\frac{\pi}{30}) \approx \tan(\frac{\pi}{30})

  • B) \sin(\frac{\pi}{90}) \approx \frac{\pi}{90}

  • C) \cos(\frac{\pi}{180}) \approx \frac{\pi}{180}

  • D) \tan(\frac{\pi}{180}) \approx \frac{\pi}{180}

• Correction: C is false. The small angle approximation \sin\theta \approx \tan\theta \approx \theta (in radians) only applies to sine and tangent functions, not cosine. For small angles, \cos\theta \approx 1.

Concept Question: Pendulum Frequency vs. Mass

• Problem: Which graph represents the relationship between a pendulum's frequency and the mass attached to it?

• Solution: The frequency (f) and period (T) of a simple pendulum depend only on two factors:

  1. Its length (L).

  2. The gravitational field intensity (g).

  • The formula for the period is T = 2\pi\sqrt{\frac{L}{g}}, and for frequency is f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}.

  • There is no dependency on the mass of the bob in the ideal simple pendulum model.

  • Therefore, a graph showing constant frequency regardless of mass is the correct representation.

Sample Problem: Pendulum Period on Different Planets

• Problem: A pendulum has a period of 1\text{s} on Earth (TE = 1\text{s}). If taken to the Moon where gravity is 1/6th as strong (gM = gE/6), what would be its period (TM)?

• Solution: Use the pendulum period formula T = 2\pi\sqrt{\frac{L}{g}}.

  • Ratio of periods:
    \frac{TM}{TE} = \frac{2\pi\sqrt{L/gM}}{2\pi\sqrt{L/gE}} = \sqrt{\frac{gE}{gM}}

  • Substitute gM = gE/6:
    \frac{TM}{TE} = \sqrt{\frac{gE}{gE/6}} = \sqrt{6}

  • Solve for TM: TM = T_E \sqrt{6} = (1\text{s})\sqrt{6} \approx 2.45\text{ s}.

Sample Problem: Pendulum Motion Calculation

• Problem: A bowling ball pendulum has a length of 5.0\text{ m}. It is initially displaced by \frac{\pi}{30} radians from the vertical and released (g=9.8\text{ m/s}^2).

  • A) How long after it is released does it reach its lowest point?

  • B) How long after it is released does it make an angle of \frac{\pi}{60} with the vertical?

• Part A: Time to Lowest Point

  • The lowest point is the equilibrium position, which is one-quarter of a full period (T/4).

  • Calculate the period:
    T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{5.0\text{ m}}{9.8\text{ m/s}^2}} = 2\pi\sqrt{0.5102} \approx 2\pi(0.714) \approx 4.486\text{ s}.

  • Time to lowest point:
    t = \frac{T}{4} = \frac{4.486\text{ s}}{4} \approx 1.12\text{ s}.

• Part B: Time to Specific Angle

  • Angular Position Function: For small angles, the angular displacement \theta(t) follows SHM:
    \theta(t) = \theta_{max}\cos(\omega t + \phi)

  • Initial Conditions: Released from rest at \theta_{max} = \frac{\pi}{30} (positive displacement), so \phi=0.

  • Angular Frequency (\omega):
    \omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.8\text{ m/s}^2}{5.0\text{ m}}} = \sqrt{1.96}\text{ rad/s} = 1.4\text{ rad/s}.

  • Equation: \theta(t) = (\frac{\pi}{30})\cos(1.4t).

  • Solve for t when \theta(t) = \frac{\pi}{60}: \frac{\pi}{60} = (\frac{\pi}{30})\cos(1.4t) \cos(1.4t) = \frac{(\pi/60)}{(\pi/30)} = \frac{1}{2} 1.4t = \cos^{-1}(\frac{1}{2})

    • The first positive angle for which cosine is 1/2 is \frac{\pi}{3} radians.
      1.4t = \frac{\pi}{3}\text{ rad}
      t = \frac{\pi/3}{1.4} \approx \frac{1.047}{1.4} \approx 0.748\text{ s}.

Pendulums and Energy

Sample Problem: Release Angle from Speed at Lowest Point

• Problem: A 1.5\text{ m} long pendulum has a speed of 3.83\text{m/s} at its lowest point. From what angle (with respect to the vertical) was it released? (g=9.80\text{ m/s}^2).

• Solution: Use conservation of mechanical energy. Total energy at release (all potential) equals total energy at the lowest point (all kinetic).

  • Potential Energy at Release: {PE}{initial} = mgh0. The height h0 is given by L(1 - \cos\theta0), where \theta_0 is the release angle.

  • Kinetic Energy at Lowest Point: {KE}{final} = \frac{1}{2}mv{max}^2.

  • Conservation: {PE}{initial} = {KE}{final}
    mgL(1 - \cos\theta0) = \frac{1}{2}mv{max}^2
    gL(1 - \cos\theta0) = \frac{1}{2}v{max}^2
    1 - \cos\theta0 = \frac{v{max}^2}{2gL}
    \cos\theta0 = 1 - \frac{v{max}^2}{2gL}

  • Substitute values:
    \cos\theta0 = 1 - \frac{(3.83\text{ m/s})^2}{2(9.80\text{ m/s}^2)(1.5\text{ m})} \cos\theta0 = 1 - \frac{14.6689}{29.4} \approx 1 - 0.4989 \approx 0.5011
    \theta_0 = \cos^{-1}(0.5011) \approx 59.9^[\circ] \approx 60.0^[\circ].

Concept Question: Ranking Pendulums by Maximum Speed

• Problem: Rank three pendulums (I: L=2.00m, 60deg; II: L=1.00m, 90deg; III: L=3.00m, 45deg) according to their maximum speed. Mass is given but is irrelevant.

• Solution: Use the energy conservation derived formula for v{max}^2, which is v{max}^2 = 2gL(1 - \cos\theta_0).

  • Pendulum I: L=2.00\text{m}, \theta0 = 60^\circ. vI^2 = 2g(2.00)(1 - \cos 60^\circ) = 4g(1 - 0.5) = 2g.

  • Pendulum II: L=1.00\text{m}, \theta0 = 90^\circ. v{II}^2 = 2g(1.00)(1 - \cos 90^\circ) = 2g(1 - 0) = 2g.

  • Pendulum III: L=3.00\text{m}, \theta0 = 45^\circ. v{III}^2 = 2g(3.00)(1 - \cos 45^\circ) = 6g(1 - 0.707) = 6g(0.293) \approx 1.758g.

• Ranking: Since vI^2 = v{II}^2 = 2g and v{III}^2 \approx 1.758g, we have vI = v{II} > v{III}.

Sample Problem: Released Angle from Current Speed and Angle

• Problem: A 2.0\text{ m} long pendulum has a mass traveling at 2.0\text{ m/s} when the pendulum makes an angle of 70^\circ with respect to the ceiling. At what angle (with respect to the ceiling) was the mass released from rest to reach this speed?

• Note on Angles: The standard pendulum energy equations use angles with respect to the vertical. An angle of 70^\circ with respect to the ceiling means 90^\circ - 70^\circ = 20^\circ with respect to the vertical (let this be \theta).

• Solution: Use the energy conservation formula relating current speed and angle to the released angle: \cos\theta_0 = \cos\theta - \frac{v^2}{2gL}

  • Current speed v = 2.0\text{ m/s}.

  • Current angle with vertical \theta = 20^\circ.

  • Length L = 2.0\text{ m}.

  • Gravity g = 9.8\text{ m/s}^2.

  • Substitute values:
    \cos\theta0 = \cos(20^\circ) - \frac{(2.0\text{ m/s})^2}{2(9.8\text{ m/s}^2)(2.0\text{ m})} \cos\theta0 = 0.9397 - \frac{4.0}{39.2}
    \cos\theta0 = 0.9397 - 0.1020 \approx 0.8377 \theta0 = \cos^{-1}(0.8377) \approx 33.1^\circ (This is the angle with respect to the vertical).

• Convert back to Angle with Ceiling: The question asks for the angle with respect to the ceiling.
  Angle with ceiling = 90^\circ - \theta_0 = 90^\circ - 33.1^\circ = 56.9^\circ.