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 \omegaD = 0. This occurs when the term inside the square root is zero: \omegaD = \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 (bc): bc^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) = 0.5From 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:
ay(x,t) = \frac{\partial vy}{\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 (ay) is directly proportional to the negative of displacement (y) from equilibrium: ay = -\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(x2,t) - \Phi(x1,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)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)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 (k1 x - k2 x) is equal to n(2\pi) for integer n at x=1.0 \text{ m}. So we compare (k1 - k2) 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):
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}.
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{resultant} = D1(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 |D1|=3.0 \text{ m} and |D2|=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{SW}(x,t) = y1 + 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{SW}, k = 2\pi. From y1, k=2\pi. The second wave must have k=2\pi.
Angular Frequency (\omega): From y{SW}, \omega = \pi. From y1, \omega=\pi. The second wave must have \omega=\pi.
Direction: Since y1 is travelling in the positive x-direction ((kx - \omega t) form), the second wave y2 must travel in the negative x-direction ((kx + \omega t) form).
Constructing y2: Based on these parameters, the second wave must be:
y2(x,t) = 0.50\text{m} \sin(2\pi x + \pi t)Conclusion: This matches option E.
Note: The statement