Lecture Damped SHM and Wave Interference

Damped Simple Harmonic Motion (SHM)

Types of Damping (Example 1)

• Problem Statement: A spring with a constant of $75 ext{ N/m}$ has a $115 ext{ g}$ mass attached. The damping constant is $1 ext{ kg/s}$. Determine if the system is underdamped, critically damped, or overdamped.

• Given Values:

  • Spring constant: $k = 75 ext{ N/m}$ (or $75 ext{ kg} ext{ m/s}^2 ext{ m}$)

  • Mass: $m = 115 ext{ g} = 0.115 ext{ kg}$

  • Damping constant: $b = 1.0 ext{ kg/s}$

• Condition for Critical Damping:

  • A critically damped system has its damped angular frequency $\omega<em>D = 0$. This occurs when the term inside the square root is zero: $\omega</em>D = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}} = 0$

  • This implies: $\frac{k}{m} = \frac{b^2}{4m^2}$

  • Solving for the critical damping constant ($b<em>c$): $b</em>c^2 = 4m^2 \frac{k}{m} = 4mk$
    $b_c = 2\sqrt{mk}$

• Calculation of Critical Damping Constant:

  • $b_c = 2\sqrt{(0.115 \text{ kg})(75 \text{ N/m})} = 2\sqrt{8.625} \approx 2(2.937) \approx 5.87 \text{ kg/s}$

• Comparison and Conclusion:

  • Compare the given damping constant ($b = 1.0 \text{ kg/s}$) with the critical damping constant ($b_c \approx 5.87 \text{ kg/s}$).

  • Since b < b_c (i.e., 1.0 \text{ kg/s} < 5.87 \text{ kg/s}), the system is underdamped. This means the system will oscillate with decreasing amplitude.

Comparing Damped and Undamped Frequencies (Example 2)

• Problem Statement: A mass of $85 ext{ g}$ is undergoing underdamped SHM, attached to a vertical spring with a spring constant of $125 ext{ N/m}$. Its damping constant is $2.0 ext{ kg/s}$. Compare its damped frequency to its frequency if no damping existed.

• Given Values:

  • Mass: $m = 85 ext{ g} = 0.085 \text{ kg}$

  • Spring constant: $k = 125 \text{ N/m}$

  • Damping constant: $b = 2.0 \text{ kg/s}$

• Damped Angular Frequency ($\omega_D$):

  • Formula: $\omega_D = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}$

  • Calculation: $\omega_D = \sqrt{\frac{125 \text{ N/m}}{0.085 \text{ kg}} - \frac{(2.0 \text{ kg/s})^2}{4(0.085 \text{ kg})^2}}$

  • $\omega_D = \sqrt{1470.588 - \frac{4}{4(0.007225)}} = \sqrt{1470.588 - \frac{4}{0.0289}} = \sqrt{1470.588 - 138.408} = \sqrt{1332.18} \approx 36 \text{ rad/s}$

• Undamped (Natural) Angular Frequency ($\omega$):

  • Formula: $\omega = \sqrt{\frac{k}{m}}$

  • Calculation: $\omega = \sqrt{\frac{125 \text{ N/m}}{0.085 \text{ kg}}} = \sqrt{1470.588} \approx 38 \text{ rad/s}$

• Comparison:

  • The damped frequency ($36 \text{ rad/s}$) is slightly lower than the undamped frequency ($38 \text{ rad/s}$).

  • Observation: When the damping is small (lightly damped), the damped frequency is very close to the undamped natural frequency ($\omega_D \approx \omega$).

Amplitude Decay and Number of Cycles (Example 3)

• Problem Statement: A spring with a force constant $75 ext{ N/m}$ has a $115 ext{ g}$ mass attached. The system has a damping constant of $0.20 ext{ kg/s}$. If the mass is pulled down $8.0 ext{ cm}$ and released, how long will it take for its amplitude to be $2.0 ext{ cm}$? How many cycles will the system undergo during this time?

• Given Values:

  • Spring constant: $k = 75 \text{ N/m}$

  • Mass: $m = 115 \text{ g} = 0.115 \text{ kg}$

  • Damping constant: $b = 0.20 \text{ kg/s}$

  • Initial amplitude: $A_0 = 8.0 \text{ cm}$

  • Final amplitude: $A_t = 2.0 \text{ cm}$

• Amplitude Decay Equation:

  • The amplitude of an underdamped oscillator decreases exponentially with time:
    $y(t) = A_0 e^{-\frac{b}{2m}t}$

• Calculating Time for Amplitude Decay:

  • Substitute the given values:
    $2.0 \text{ cm} = (8.0 \text{ cm}) e^{-\frac{0.20 \text{ kg/s}}{2 \cdot 0.115 \text{ kg}}t}$

  • Divide by initial amplitude:
    $\frac{2.0}{8.0} = 0.25 = e^{-\frac{0.20}{0.230}t} = e^{-0.8696t}$

  • Take the natural logarithm of both sides:
    $\ln(0.25) = -0.8696t$

  • $-1.38629 = -0.8696t$

  • Solve for $t$:
    $t = \frac{-1.38629}{-0.8696} \approx 1.594 \text{ s}$
    (Approximately $1.6 \text{ s}$ when rounded)

• Calculating Number of Cycles:

  • First, calculate the (undamped) period $T$ of oscillation (often used for simplified cycle count in damped systems):
    $T = 2\pi\sqrt{\frac{m}{k}}$

  • $T = 2\pi\sqrt{\frac{0.115 \text{ kg}}{75 \text{ N/m}}} = 2\pi\sqrt{0.001533} \approx 2\pi(0.03915) \approx 0.2460 \text{ s}$

  • Number of cycles ($n$) is the total time divided by the period:
    $n = \frac{t}{T} = \frac{1.594 \text{ s}}{0.2460 \text{ s}} \approx 6.479 \text{ cycles}$
    (Approximately $6.5 \text{ cycles}$ when rounded).

Travelling Waves

Introduction to Travelling Waves

• Nature: Connected particles undergoing Simple Harmonic Motion (SHM) result in the formation of travelling waves.

• General Equation for a Travelling Wave:

  • A wave travelling in the positive x-direction can be described by:
    $y(x,t) = A\sin(kx - \omega t + \phi)$

  • Components:

    • $A$: Amplitude (maximum displacement from equilibrium).

    • $k$: Wave number ($k = 2\pi/\lambda$), related to wavelength.

    • $\omega$: Angular frequency ($\omega = 2\pi f$), related to frequency and period.

    • $t$: Time.

    • $\phi$: Phase constant. This constant determines the state of the wave at $t=0$ and $x=0$. It is often not crucial for many wave problems.

Wave Properties from the Equation (Example 1)

• Problem Statement: Consider the wave described by the equation $y = 0.08 \sin(5\pi x - 4\pi t)$. What are its amplitude, speed, wavelength, frequency, and period?

• Given Equation: $y = 0.08 \sin(5\pi x - 4\pi t)$

• Comparison to General Form ($A\sin(kx - \omega t)$):

  • Amplitude: $A = 0.08 \text{ m}$ (or $8 \text{ cm}$)

  • Wave number: $k = 5\pi \text{ rad/m}$

  • Angular frequency: $\omega = 4\pi \text{ rad/s}$

• Derived Properties:

  • Wavelength ($\lambda$):
    $k = \frac{2\pi}{\lambda} \implies \lambda = \frac{2\pi}{k} = \frac{2\pi}{5\pi} = 0.40 \text{ m}$

  • Frequency ($f$):
    $\omega = 2\pi f \implies f = \frac{\omega}{2\pi} = \frac{4\pi}{2\pi} = 2.0 \text{ Hz}$

  • Period ($T$):
    $T = \frac{1}{f} = \frac{1}{2.0 \text{ Hz}} = 0.50 \text{ s}$

  • Speed ($v$):

    • Using frequency and wavelength: $v = f\lambda = (2.0 \text{ Hz})(0.40 \text{ m}) = 0.80 \text{ m/s}$

    • Using angular frequency and wave number: $v = \frac{\omega}{k} = \frac{4\pi}{5\pi} = 0.80 \text{ m/s}$

Ranking Waves by Wavelength (Example 2)

• Problem Statement: Rank the following waves according to their wavelengths (greatest to least):

  • I) $y = 2\sin(-3x + 6t)$

  • II) $y = 0.5\sin(5x - 2t)$

  • III) $y = 3\sin(2x + 4t)$

• Key Concept: Wavelength ($\lambda$) is inversely proportional to the wave number ($k$), i.e., $\lambda = \frac{2\pi}{k}$. Therefore, a smaller wave number corresponds to a larger wavelength.

  • We use the magnitude of the coefficient of $x$ for $k$. The sign indicates direction of travel, not wavelength.

• Wave Numbers:

  • I) $k = 3$

  • II) $k = 5$

  • III) $k = 2$

• Ranking (Greatest to Least Wavelength):

  • We need the order of $k$ from smallest to largest:

    • III ($k=2$) has the largest wavelength.

    • I ($k=3$) has the second largest wavelength.

    • II ($k=5$) has the smallest wavelength.

  • Therefore, the ranking is III > I > II.

Direction of Wave Travel (Example 3)

• Problem Statement: Which wave(s) are travelling in the negative x-direction?

  • I) $y = 2\sin(-3x + 6t)$

  • II) $y = 0.5\sin(5x - 2t)$

  • III) $y = 3\sin(2x + 4t)$

• Rule for Wave Direction:

  • If the signs of the $kx$ term and the $\omega t$ term are the same, the wave is moving in the negative x-direction. (e.g., $+kx + \omega t$ or $-kx - \omega t$)

  • If the signs of the $kx$ term and the $\omega t$ term are different, the wave is moving in the positive x-direction. (e.g., $+kx - \omega t$ or $-kx + \omega t$)

• Analysis of Waves:

  • I) $(-3x + 6t)$: Signs are different ($x$ term is negative, $t$ term is positive). -> Positive x-direction.

  • II) $(5x - 2t)$: Signs are different ($x$ term is positive, $t$ term is negative). -> Positive x-direction.

  • III) $(2x + 4t)$: Signs are the same ($x$ term is positive, $t$ term is positive). -> Negative x-direction.

• Conclusion: Only wave III is travelling in the negative x-direction.

Greatest Wave Speed (Example 4)

• Problem Statement: Which wave has the greatest speed?

  • I) $y = 2\sin(-3x + 6t)$

  • II) $y = 0.5\sin(5x - 2t)$

  • III) $y = 3\sin(2x + 4t)$

• Key Concept: The speed of a wave ($v$) is given by $v = \frac{\omega}{k}$.

  • The signs in front of the $kx$ and $\omega t$ terms indicate direction, but for calculating speed, we use the magnitudes of angular frequency ($\omega$) and wave number ($k$), as speed cannot be negative.

• Analysis of Wave Speed ($v = \omega/k$):

  • I) $\omega = 6, k = 3 \implies v = \frac{6}{3} = 2$

  • II) $\omega = 2, k = 5 \implies v = \frac{2}{5} = 0.4$

  • III) $\omega = 4, k = 2 \implies v = \frac{4}{2} = 2$

• Conclusion: Both wave I and wave III have the greatest speed ($2$ units/s).

Equation from Wave Properties (Example 5)

• Problem Statement: A wave has a period of $0.5 \text{ s}$, an amplitude of $1.0 \text{ m}$, and is travelling in the negative x-direction at $4.0 \text{ m/s}$. Which of the following equations best describes the wave?

  • A) $y(x,t) = 1.0\text{m} \sin(\pi x - 2\pi t)$

  • B) $y(x,t) = 1.0\text{m} \sin(\pi x + 2\pi t)$

  • C) $y(x,t) = 1.0\text{m} \sin(2\pi x + 4\pi t)$

  • D) $y(x,t) = 1.0\text{m} \sin(\pi x - 4\pi t)$

  • E) $y(x,t) = 1.0\text{m} \sin(\pi x + 4\pi t)$

• Given Properties:

  • Amplitude: $A = 1.0 \text{ m}$

  • Period: $T = 0.5 \text{ s}$

  • Speed: $v = 4.0 \text{ m/s}$

  • Direction: Negative x-direction.

• Calculate Wave Parameters:

  • Angular Frequency ($\omega$):
    $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.5 \text{ s}} = 4\pi \text{ rad/s}$

  • Frequency ($f$): $f = 1/T = 1/0.5 = 2.0 \text{ Hz}$

  • Wavelength ($\lambda$):
    $\lambda = vT = (4.0 \text{ m/s})(0.5 \text{ s}) = 2.0 \text{ m}$

  • Wave Number ($k$):
    $k = \frac{2\pi}{\lambda} = \frac{2\pi}{2.0 \text{ m}} = \pi \text{ rad/m}$

• Formulating the Equation:

  • For a wave travelling in the negative x-direction, the general form is $y(x,t) = A\sin(kx + \omega t)$.

  • Substitute the calculated values: $y(x,t) = 1.0\text{m} \sin(\pi x + 4\pi t)$.

• Conclusion: This matches option E.

Equation from Graph - Snapshot in X (Example 6)

• Problem Statement: Which of the following equations could apply to the provided graph of $y(x)$ at a single moment in time (snapshot)?

  • A) $y(x,t) = 1.5 \text{ m} \sin(\pi x + 2\pi t)$

  • B) $y(x,t) = 1.5 \text{ m} \sin(\frac{\pi}{2} x - \pi t)$

  • C) $y(x,t) = 1.5 \text{ m} \sin(2\pi x - \pi t)$

  • D) $y(x,t) = 1.5 \text{ m} \sin(\pi x - 2\pi t)$

• Analysis from Graph:

  • Amplitude ($A$): The maximum displacement is $1.5 \text{ m}$. All options have this amplitude.

  • Wavelength ($\lambda$): Looking at the graph, a complete cycle occurs from $x=0$ to $x=1.0 \text{ m}$ (or from peak at $x=0.25$ to next peak at $x=1.25$). So, $\lambda = 1.0 \text{ m}$.

• Calculate Wave Number ($k$):

  • $k = \frac{2\pi}{\lambda} = \frac{2\pi}{1.0 \text{ m}} = 2\pi \text{ rad/m}$

• Comparing with Options:

  • We need an equation with $A = 1.5 \text{ m}$ and $k = 2\pi$. Only option C, $y(x,t) = 1.5 \text{ m} \sin(2\pi x - \pi t)$, satisfies the wave number condition (coefficient of $x$ is $2\pi$).

  • The time-dependent part ($\omega t$) and direction cannot be determined from a single snapshot graph unless additional information is provided. However, the unique $k$ value allows selection.

• Conclusion: Option C best describes the wave shown in the graph.

Equation from Graph - Snapshot in Time and Detector Data (Example 7)

• Problem Statement: A harmonic wave with a speed of $0.50 \text{ m/s}$ in the positive x-direction passes by a detector. The detector records the amplitude of the wave and displays it in the graph (which shows $y(t)$ at a fixed $x$). The equation of the wave is…

  • A) $y(x,t) = 2.0\text{m} \sin(4\pi x - 2\pi t)$

  • B) $y(x,t) = 2.0\text{m} \sin(2\pi x - 2\pi t)$

  • C) $y(x,t) = 2.0\text{m} \sin(4\pi x - \pi t)$

  • D) $y(x,t) = 2.0\text{m} \sin(2\pi x - \pi t)$

  • E) $y(x,t) = 2.0\text{m} \sin(\frac{\pi}{2} x - 2\pi t)$

• Given Information:

  • Speed: $v = 0.50 \text{ m/s}$

  • Direction: Positive x-direction (implies $(kx - \omega t)$ form).

• Analysis from Graph ($y(t)$):

  • Amplitude ($A$): The maximum displacement is $2.0 \text{ m}$. All options have this amplitude.

  • Period ($T$): From the graph, one complete cycle occurs in $1.0 \text{ s}$ (e.g., from $t=0$ to $t=1.0 \text{ s}$ for the zero crossing, or peak at $t=0.25$ to next peak at $t=1.25$). So, $T = 1.0 \text{ s}$.

• Calculate Wave Parameters:

  • Angular Frequency ($\omega$):
    $\omega = \frac{2\pi}{T} = \frac{2\pi}{1.0 \text{ s}} = 2\pi \text{ rad/s}$
    This eliminates options C and D, which have $\omega = \pi$.

  • Frequency ($f$): $f = 1/T = 1/1.0 = 1.0 \text{ Hz}$

  • Wavelength ($\lambda$):
    $\lambda = \frac{v}{f} = \frac{0.50 \text{ m/s}}{1.0 \text{ Hz}} = 0.50 \text{ m}$

  • Wave Number ($k$):
    $k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.50 \text{ m}} = 4\pi \text{ rad/m}$

• Formulating the Equation:

  • For a wave travelling in the positive x-direction, the general form is $y(x,t) = A\sin(kx - \omega t)$.

  • Substitute the calculated values: $y(x,t) = 2.0\text{m} \sin(4\pi x - 2\pi t)$.

• Conclusion: This matches option A.

Equation with Phase Constant (Example 8)

• Problem Statement: Consider the graph of the travelling wave shown ($y(t)$ at $x=1.0 \text{ m}$), with an amplitude of $A=0.2 \text{ m}$. This data was observed at $x = 1.0 \text{ m}$. The wave has a speed of $2.0 \text{ m/s}$ and is travelling in the negative x-direction. Its equation is…

  • A) $A\sin(\pi x + 2\pi t + \frac{2\pi}{3})$

  • B) $A\sin(\pi x + 2\pi t - \frac{2\pi}{3})$

  • C) $A\sin(2\pi x + \pi t + \frac{2\pi}{3})$

  • D) $A\sin(2\pi x + \pi t - \frac{2\pi}{3})$

  • E) $A\sin(\pi x - 2\pi t - \frac{2\pi}{3})$

• Given Information:

  • Amplitude: $A = 0.2 \text{ m}$ (inferred from calculation steps, not directly stated in problem, but graph suggests a max amplitude to determine phase constant against)

  • Observation point: $x = 1.0 \text{ m}$

  • Speed: $v = 2.0 \text{ m/s}$

  • Direction: Negative x-direction.

• Analysis from Graph ($y(t)$ at $x=1.0 \text{ m}$):

  • Period ($T$): From the graph, one complete cycle is $1.0 \text{ s}$. So, $T = 1.0 \text{ s}$.

• Calculate Wave Parameters:

  • Angular Frequency ($\omega$):
    $\omega = \frac{2\pi}{T} = \frac{2\pi}{1.0 \text{ s}} = 2\pi \text{ rad/s}$
    This immediately rules out options C and D.

  • Wave Number ($k$):
    $v = \frac{\omega}{k} \implies k = \frac{\omega}{v} = \frac{2\pi \text{ rad/s}}{2.0 \text{ m/s}} = \pi \text{ rad/m}$

• Formulating the General Equation:

  • For a wave travelling in the negative x-direction, the general form including a phase constant is $y(x,t) = A\sin(kx + \omega t + \phi)$.

  • Substituting values for $A, k, \omega$: $y(x,t) = A\sin(\pi x + 2\pi t + \phi)$.
    This eliminates option E because of the direction ($-2\pi t$). This leaves A and B. The difference is the sign of the phase.

• Determining the Phase Constant ($\phi):

  • From the graph, at $x=1.0 \text{ m}$ and $t=0$, the displacement is $y = 0.1 \text{ m}$.

  • Substitute these values into the equation:
    $0.1 = A\sin(\pi (1.0) + 2\pi (0) + \phi)$
    $0.1 = 0.2\sin(\pi + \phi)$
    $\sin(\pi + \phi) = 0.5$

  • From the values in the slide's solution (Page 29), it uses a cosine function for determining the phase, i.e., $0.1 = 0.2\cos(\pi + \phi)$, which yields $\cos(\pi + \phi) = 0.5$. This implies $\pi + \phi = \frac{\pi}{3}$ (or $-\frac{\pi}{3}$ to consider other quadrants).

  • Solving for $\phi$ given $\pi + \phi = \frac{\pi}{3}$, we get $\phi = \frac{\pi}{3} - \pi = -\frac{2\pi}{3}$.

  • Note: There's an inconsistency between the cosine-based calculation for phase and the sine options. Assuming the phase constant value is derived as per the slide's calculation, using the cosine approach: the phase of $-\frac{2\pi}{3}$ means the wave equation should incorporate this term. If we assume the given options are correct and the calculation intends to match one, we proceed with $\phi = -\frac{2\pi}{3}$.

• Conclusion: The best match, following the parameters and the phase constant calculation method presented in the slide (despite the potential sine/cosine discrepancy for phase calculation), is option B.
  $y(x,t) = A\sin(\pi x + 2\pi t - \frac{2\pi}{3})$

Particle Motion in Travelling Waves

Analyzing Particle Motion (Example 1)

• Problem Statement: Consider the wave $y(x,t) = 2 \sin(3\pi x - \pi t)$ at $x = 0.50 \text{ m}$. When is the first time (after $t=0$) that the particles in the medium have:

  • A) Zero velocity?

  • B) A maximum positive velocity?

  • C) A maximum negative velocity?

  • D) Maximum positive acceleration?

  • E) Maximum negative acceleration?

• Particle Displacement at $x = 0.50 \text{ m}$:

  • Substitute $x = 0.50 \text{ m}$ into the wave equation:
    $y(0.5,t) = 2.0\text{m} \sin(3\pi (0.5) - \pi t) = 2.0\text{m} \sin(\frac{3\pi}{2} - \pi t)$

• Particle Velocity ($v_y$):

  • The particle velocity is the time derivative of the displacement:
    $v_y(x,t) = \frac{\partial y}{\partial t} = \frac{\partial}{\partial t} (A\sin(kx - \omega t)) = -A\omega\cos(kx - \omega t)$

  • At $x=0.50 \text{ m}$: $v_y(0.5,t) = - (2.0 \text{ m})(\pi \text{ rad/s}) \cos(\frac{3\pi}{2} - \pi t) = -2\pi \cos(\frac{3\pi}{2} - \pi t)$

• Particle Acceleration ($a_y$):

  • The particle acceleration is the time derivative of the velocity:
    $a<em>y(x,t) = \frac{\partial v</em>y}{\partial t} = \frac{\partial}{\partial t} (-A\omega\cos(kx - \omega t)) = -A\omega^2\sin(kx - \omega t)$

  • At $x = 0.50 \text{ m}$: $a_y(0.5,t) = -(2.0 \text{ m})(\pi \text{ rad/s})^2 \sin(\frac{3\pi}{2} - \pi t) = -2\pi^2 \sin(\frac{3\pi}{2} - \pi t)$

• Analysis of Particle Motion:

  • A) Zero Velocity: Velocity is zero when the particle is at its maximum displacement (turning points), i.e., $y(0.5,t) = \pm A$. This means $\cos(\frac{3\pi}{2} - \pi t) = 0$.

    • This occurs when $(\frac{3\pi}{2} - \pi t)$ equals $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, …$ or $-\frac{\pi}{2}, …$

    • $\frac{3\pi}{2} - \pi t = \frac{\pi}{2} \implies \pi t = \pi \implies t = 1.0 \text{ s}$

    • $\frac{3\pi}{2} - \pi t = \frac{3\pi}{2} \implies \pi t = 0 \implies t = 0 \text{ s}$

    • $\frac{3\pi}{2} - \pi t = -\frac{\pi}{2} \implies \pi t = 2\pi \implies t = 2.0 \text{ s}$

    • The first time after $t=0$ is $t = 1.0 \text{ s}$.

  • B) Maximum Positive Velocity: Velocity is max positive when particle is at equilibrium and moving upwards. For $v_y = -2\pi \cos(\frac{3\pi}{2} - \pi t)$, max positive occurs when $\cos(\frac{3\pi}{2} - \pi t) = -1$.

    • This requires $(\frac{3\pi}{2} - \pi t) = \pi \implies \pi t = \frac{\pi}{2} \implies t = 0.5 \text{ s}$

  • C) Maximum Negative Velocity: Velocity is max negative when particle is at equilibrium and moving downwards. For $v_y = -2\pi \cos(\frac{3\pi}{2} - \pi t)$, max negative occurs when $\cos(\frac{3\pi}{2} - \pi t) = 1$.

    • This requires $(\frac{3\pi}{2} - \pi t) = 0 \implies \pi t = \frac{3\pi}{2} \implies t = 1.5 \text{ s}$

  • D) Maximum Positive Acceleration: Acceleration is max positive when displacement is most negative (at a trough), i.e., $y(0.5,t) = -A$. For $a_y = -2\pi^2 \sin(\frac{3\pi}{2} - \pi t)$, max positive acceleration occurs when $\sin(\frac{3\pi}{2} - \pi t) = -1$.

    • This requires $(\frac{3\pi}{2} - \pi t) = \frac{3\pi}{2} \implies \pi t = 0 \implies t = 0 \text{ s}$

    • or $(\frac{3\pi}{2} - \pi t) = -\frac{\pi}{2} \implies \pi t = 2\pi \implies t = 2.0 \text{ s}$

    • The first time after $t=0$ is $t = 2.0 \text{ s}$.

  • E) Maximum Negative Acceleration: Acceleration is max negative when displacement is most positive (at a peak), i.e., $y(0.5,t) = A$. For $a_y = -2\pi^2 \sin(\frac{3\pi}{2} - \pi t)$, max negative acceleration occurs when $\sin(\frac{3\pi}{2} - \pi t) = 1$.

    • This requires $(\frac{3\pi}{2} - \pi t) = \frac{\pi}{2} \implies \pi t = \pi \implies t = 1.0 \text{ s}$

Interpreting Graphs - Particle Velocity

• Problem Statement: Given a snapshot graph of a travelling wave $y(x)$, which points have particle velocities in the negative direction?

  • Key Concept (as per slide): For a graph of wave displacement $y$ versus position $x$ ($y(x)$), the local slope of the wave at a particle's location can indicate the particle's vertical velocity. Specifically, a negative slope of the $y(x)$ curve is associated with a negative particle velocity (downward movement), assuming the wave propagates, for example, to the right.

• Analysis of Points (A-H):

  • Points A, D, G are at crests or troughs (where $y(x)$ slope is zero), meaning particle velocity is instantaneously zero.

  • Looking at the graph, points B, C, and H are located where the slope of the $y(x)$ curve is negative (i.e., the wave is descending).

• Conclusion: Points B, C, G have velocities in the negative direction based on the 'negative slope = negative velocity' interpretation.

Interpreting Graphs - Greatest Positive Acceleration

• Problem Statement: At which point does the particle indicated have the greatest positive acceleration?

• Key Concept: For Simple Harmonic Motion (which particles in a travelling wave undergo), acceleration ($a<em>y$) is directly proportional to the negative of displacement ($y$) from equilibrium: $a</em>y = -\omega^2 y$.

  • Therefore, the acceleration is maximum positive when the displacement ($y$) is maximum negative (i.e., at a trough).

• Analysis of Points (A, D, G):

  • Point A is at a crest ($y$ is maximum positive), so acceleration is maximum negative.

  • Point D is at a trough ($y$ is maximum negative), so acceleration is maximum positive.

  • Point G is at a crest ($y$ is maximum positive), so acceleration is maximum negative.

• Conclusion: Point D has the greatest positive acceleration. This is because at a trough, the particle is momentarily at rest before moving upwards, implying a strong upward (positive) acceleration.

Wave Interference

Types of Interference

• Definition: Wave interference occurs when two waves meet and combine to form a resultant wave.

• Constructive Interference:

  • Occurs when the resultant amplitude is larger than the amplitudes of the individual waves.

  • This happens when the waves meet in phase (i.e., their phase difference is an even multiple of $\pi$: $0, \pm 2\pi, \pm 4\pi, …$).

• Destructive Interference:

  • Occurs when the resultant amplitude is smaller than the amplitudes of the individual waves.

  • This happens when the waves meet out of phase (i.e., their phase difference is an odd multiple of $\pi$: $\pm\pi, \pm 3\pi, \pm 5\pi, …$).

Phase Difference Between Points on a Wave (Example 1)

• Problem Statement: Consider the wave $D(x,t) = 5\sin(2\pi x - 3t)$. What is the phase difference between points on this wave at $x = 1.5 \text{ m}$ and $x = 2.0 \text{ m}$?

• Given Wave Equation: $D(x,t) = 5\sin(2\pi x - 3t)$

• Phase Function: The phase of the wave is $\Phi(x,t) = 2\pi x - 3t$.

• Phase at Different Points:

  • At $x_1 = 1.5 \text{ m}$: $\Phi(1.5,t) = 2\pi (1.5) - 3t = 3\pi - 3t$

  • At $x_2 = 2.0 \text{ m}$: $\Phi(2.0,t) = 2\pi (2.0) - 3t = 4\pi - 3t$

• Phase Difference ($\Delta\Phi$):

  • $\Delta\Phi = \Phi(x<em>2,t) - \Phi(x</em>1,t) = (4\pi - 3t) - (3\pi - 3t) = 4\pi - 3t - 3\pi + 3t = \pi$

• Conclusion: The phase difference between these two points is $\pi$ radians (or $180^ ext{o}$). Since this is an odd multiple of $\pi$, the points are out of phase with each other.

Identifying In-Phase Waves (Example 2)

• Problem Statement: Which of the following pairs of waves would be in phase at $x = 1.0 \text{ m}$?

  • A) D(x,t) = 3.2\text{m} \sin(5\pi x - \omega t)$,
    D(x,t) = 3.2\text{m} \sin(4\pi x - \omega t)$</p></li><li><p>B)$D(x,t) = 2.7\text{m} \sin(\frac{5}{2}\pi x - \omega t)$,
    $D(x,t) = 4.2\text{m} \sin(\frac{1}{2}\pi x - \omega t)$

  • C) D(x,t) = 1.1\text{m} \sin(3\pi x - \omega t)$,
    D(x,t) = 3.2\text{m} \sin(\pi x - \omega t)$</p></li><li><p>D)$D(x,t) = 1.7\text{m} \sin(\frac{3}{2}\pi x - \omega t)$,
    $D(x,t) = 3.6\text{m} \sin(\frac{1}{2}\pi x - \omega t)$

• Key Concept: Two waves are in phase at a specific point if their phases (excluding the common time-dependent term $\omega t$) are the same or differ by an integer multiple of $2\pi$ at that point. We need to check if $(k<em>1 x - k</em>2 x)$ is equal to $n(2\pi)$ for integer $n$ at $x=1.0 \text{ m}$. So we compare $(k<em>1 - k</em>2)$ with $n(2\pi)$. (Note that amplitude difference doesn't prevent being in phase, it just affects constructive/destructive levels).

• Analysis at $x = 1.0 \text{ m}$:

  • A) $(5\pi - 4\pi)(1.0) = \pi$. Not a multiple of $2\pi$.

  • B) $(\frac{5}{2}\pi - \frac{1}{2}\pi)(1.0) = \frac{4}{2}\pi = 2\pi$. This is $1 \cdot (2\pi)$, so they are in phase.

  • C) $(3\pi - \pi)(1.0) = 2\pi$. This is $1 \cdot (2\pi)$, so they are in phase.

  • D) $(\frac{3}{2}\pi - \frac{1}{2}\pi)(1.0) = \pi$. Not a multiple of $2\pi$.

• Conclusion: Pairs B and C are in phase at $x = 1.0 \text{ m}$. Therefore, two of the above pairs are in phase.

Standing Waves and Reflection

Wave Reflection

• Hard Reflection (Fixed End):

  • Occurs when a wave reflects from a fixed boundary (e.g., a string tied to a wall).

  • The reflected wave is inverted (undergoes a phase change of $\pi$ or $180^ ext{o}$).

• Soft Reflection (Free End):

  • Occurs when a wave reflects from a free boundary (e.g., a string attached to a light ring that can slide freely on a rod).

  • The reflected wave is not inverted (no phase change).

Reflected Wave Equation (Example 1)

• Problem Statement: The wave described by $D(x,t) = -2.0\text{m} \sin(\pi x + 2\pi t)$ reflects from a fixed end. Which of the following best represents the reflected wave?

  • A) $D(x,t) = -2.0\text{m} \sin(\pi x + 2\pi t)$

  • B) $D(x,t) = 2.0\text{m} \sin(\pi x + 2\pi t)$

  • C) $D(x,t) = -2.0\text{m} \sin(\pi x - 2\pi t)$

  • D) $D(x,t) = 2.0\text{m} \sin(\pi x - 2\pi t)$

• Given Incident Wave: $D_i(x,t) = -2.0\text{m} \sin(\pi x + 2\pi t)$

• Analysis of Reflection from a Fixed End (Hard Reflection):

  1. Inversion of Amplitude: A hard reflection causes the wave to invert. This means the amplitude sign changes. The initial amplitude is $-2.0 \text{ m}$, so the reflected amplitude will be $+2.0 \text{ m}$.

  2. Reversal of Direction: The incident wave has the form $(kx + \omega t)$, indicating it travels in the negative x-direction. The reflected wave must travel in the opposite (positive x) direction. This means the phase term will change from $(\pi x + 2\pi t)$ to $(\pi x - 2\pi t)$.

• Formulating the Reflected Wave Equation:

  • Combining these two changes: $D_r(x,t) = +2.0\text{m} \sin(\pi x - 2\pi t)$.

• Conclusion: This matches option D.

Resultant Amplitude and Type of Interference (Example 2)

• Problem Statement: The following two waves meet at $t = 0.5 \text{ s}$ and at $x = 1.0 \text{ m}$. What is the resulting amplitude, and is this an example of constructive or destructive interference?

  • $D_1(x,t) = -3.0\text{m} \sin(2\pi x - \pi t)$

  • $D_2(x,t) = 2.0\text{m} \sin(\pi x + \pi t)$

• Calculate Displacement of $D_1$ at $x=1.0 \text{ m}, t=0.5 \text{ s}$:

  • $D_1(1.0, 0.5) = -3.0 \sin(2\pi (1.0) - \pi (0.5)) = -3.0 \sin(2\pi - 0.5\pi) = -3.0 \sin(1.5\pi)$

  • Since $\sin(1.5\pi) = -1$, $D_1(1.0, 0.5) = -3.0(-1) = 3.0 \text{ m}$.

• Calculate Displacement of $D_2$ at $x=1.0 \text{ m}, t=0.5 \text{ s}$:

  • $D_2(1.0, 0.5) = 2.0 \sin(\pi (1.0) + \pi (0.5)) = 2.0 \sin(1.0\pi + 0.5\pi) = 2.0 \sin(1.5\pi)$

  • Since $\sin(1.5\pi) = -1$, $D_2(1.0, 0.5) = 2.0(-1) = -2.0 \text{ m}$.

• Resultant Amplitude:

  • The resultant displacement at this point in time and space is the superposition of the individual displacements:
    $D<em>{resultant} = D</em>1(1.0, 0.5) + D_2(1.0, 0.5) = 3.0 \text{ m} + (-2.0 \text{ m}) = 1.0 \text{ m}$

• Type of Interference:

  • The individual wave amplitudes are $|D<em>1|=3.0 \text{ m}$ and $|D</em>2|=2.0 \text{ m}$. The resultant amplitude is $1.0 \text{ m}$.

  • Since the resulting amplitude ($1.0 \text{ m}$) is smaller than the amplitude of either individual wave, this is an example of destructive interference.

Composing Standing Waves (Example 3)

• Problem Statement: A standing wave is described by the equation $y(x,t) = 1.0\text{m} \sin(2\pi x) \cos(\pi t)$. If one of the waves producing it was $y_1(x,t) = 0.50\text{m} \sin(2\pi x - \pi t)$, then the other wave is given by:

  • A) $y_2(x,t) = -0.50\text{m} \sin(2\pi x - \pi t)$

  • B) $y_2(x,t) = 1.00\text{m} \sin(2\pi x + \pi t)$

  • C) $y_2(x,t) = -0.50\text{m} \sin(2\pi x + \pi t)$

  • D) $y_2(x,t) = -1.00\text{m} \sin(2\pi x - \pi t)$

  • E) $y_2(x,t) = 0.50\text{m} \sin(2\pi x + \pi t)$

• General Form of Standing Wave: A standing wave is often formed by the superposition of two identical travelling waves moving in opposite directions:

  • If $y_1(x,t) = A\sin(kx - \omega t)$

  • And $y_2(x,t) = A\sin(kx + \omega t)$

  • Then the standing wave $y<em>{SW}(x,t) = y</em>1 + y_2 = 2A\sin(kx)\cos(\omega t)$.

• Given Standing Wave and One Component Wave:

  • $y_{SW}(x,t) = 1.0\text{m} \sin(2\pi x) \cos(\pi t)$

  • $y_1(x,t) = 0.50\text{m} \sin(2\pi x - \pi t)$

• Deriving Parameters of the Second Wave ($y_2$):

  • Amplitude: By comparing $y_{SW}$ to $2A\sin(kx)\cos(\omega t)$, we see that $2A = 1.0\text{m}$, so $A = 0.50\text{m}$. The second wave must have the same amplitude: $A = 0.50\text{m}$.

  • Wave Number ($k$): From $y<em>{SW}$, $k = 2\pi$. From $y</em>1$, $k=2\pi$. The second wave must have $k=2\pi$.

  • Angular Frequency ($\omega$): From $y<em>{SW}$, $\omega = \pi$. From $y</em>1$, $\omega=\pi$. The second wave must have $\omega=\pi$.

  • Direction: Since $y<em>1$ is travelling in the positive x-direction ($(kx - \omega t)$ form), the second wave $y</em>2$ must travel in the negative x-direction ($(kx + \omega t)$ form).

• Constructing $y<em>2$: Based on these parameters, the second wave must be:
  $y</em>2(x,t) = 0.50\text{m} \sin(2\pi x + \pi t)$

• Conclusion: This matches option E.

  • Note: The statement