Wave Interference and Standing Waves - College Physics

1. A pipe organ has an open pipe (open at both ends) and a closed pipe (closed at one end) of the same length $L = 0.85 \text{ m}$. If both pipes are designed to have the same fundamental frequency, what must be the ratio of the length of the closed pipe to the length of the open pipe?
   Answer: $1:2$, because the closed pipe fundamental wavelength is twice that of the open pipe For an open pipe, the fundamental frequency has $L = \lambda/2$, so $f = v/(2L)$. For a closed pipe, the fundamental has $L = \lambda/4$, so $f = v/(4L)$. To have the same fundamental frequency: $\frac{v}{4L_{closed}} = \frac{v}{2L_{open}}$, which gives $L_{closed} = \frac{L_{open}}{2}$. The ratio is $L_{closed}:L_{open} = 1:2$. Choice B reverses the relationship. Choice C misinterprets the problem - we need to find what the ratio should be, not use the given equal lengths. Choice D uses an incorrect harmonic relationship.
2. Two coherent sound sources $S_1$ and $S_2$ are separated by a distance of $6.0 \text{ m}$ and emit sound waves in phase with frequency $f = 340 \text{ Hz}$. The speed of sound is $v = 340 \text{ m/s}$. A detector is placed at point $P$, which is $8.0 \text{ m}$ from $S_1$ and $10.0 \text{ m}$ from $S_2$. What type of interference occurs at point $P$?
   Answer: Constructive interference, because the path difference equals two wavelengths First, find the wavelength: $\lambda = v/f = 340/340 = 1.0 \text{ m}$. The path difference is $|10.0 - 8.0| = 2.0 \text{ m} = 2\lambda$. Since the sources emit in phase, constructive interference occurs when the path difference is an integer multiple of wavelengths. Here, $\Delta = 2\lambda$, so constructive interference occurs. Choice A is wrong because the path difference is 2λ, not 1λ. Choice B is wrong because 2λ ≠ λ/2 and this gives constructive, not destructive interference. Choice D is wrong because the source separation doesn't directly determine interference at P; it's the path difference that matters.
3. Two identical waves traveling in the same direction have a phase difference of $\pi/3$ radians. If each wave has amplitude $A$, what is the amplitude of the resultant wave?
   Answer: $A\sqrt{3}$ When two waves interfere, you need to use vector addition to find the resultant amplitude, not simple arithmetic addition. This is because waves have both magnitude and phase, making them vector quantities. For two identical waves with amplitude $A$ and phase difference $\pi/3$, imagine them as vectors of equal length pointing in directions that differ by $\pi/3$ radians (60°). To find the resultant, you can use the law of cosines: $A_{resultant}^2 = A^2 + A^2 + 2A \cdot A \cos(\pi/3)$ Since $\cos(\pi/3) = 1/2$, this becomes: $A_{resultant}^2 = A^2 + A^2 + 2A^2(1/2) = 2A^2 + A^2 = 3A^2$ Therefore, $A_{resultant} = A\sqrt{3}$, making C correct. Let's examine why the other answers are wrong: A) $A$ would be correct if the waves were completely out of phase ($\pi$ phase difference) and partially canceled each other, but $\pi/3$ doesn't produce this result. B) $2A$ occurs only when waves are perfectly in phase (zero phase difference), resulting in complete constructive interference. A $\pi/3$ phase difference prevents this maximum reinforcement. D) $A\sqrt{2}$ would result from a $\pi/2$ phase difference (90°), not $\pi/3$. Remember: wave interference problems require vector addition. The key insight is recognizing that $\cos(\pi/3) = 1/2$, which appears frequently in physics problems involving 60° angles or $\pi/3$ radians.
4. A string of length $L$ is fixed at both ends and vibrates in its third harmonic mode. At which positions along the string (measured from one end) are the displacement nodes located?
   Answer: At $0$, $L/3$, $2L/3$, and $L$ When analyzing standing wave patterns on strings fixed at both ends, you need to understand that harmonics follow a specific mathematical relationship. The third harmonic means the string vibrates with three complete half-wavelengths fitting along its length $L$. For the third harmonic, the wavelength $\lambda = \frac{2L}{3}$, creating a standing wave pattern with four displacement nodes (points of zero displacement). Since the string is fixed at both ends, nodes must occur at $x = 0$ and $x = L$. The remaining nodes are evenly spaced between these endpoints. For $n$ harmonics on a string fixed at both ends, nodes occur at positions $x = \frac{kL}{n}$ where $k = 0, 1, 2, ..., n$. For the third harmonic ($n = 3$), nodes are located at: - $k = 0$: $x = 0$ - $k = 1$: $x = \frac{L}{3}$ - $k = 2$: $x = \frac{2L}{3}$ - $k = 3$: $x = L$ This confirms answer C is correct. Answer A gives the node positions for the fourth harmonic, not the third. Answer B incorrectly lists five interior nodes, which would correspond to a sixth harmonic pattern. Answer D omits the required boundary nodes at the fixed ends—a fundamental error since fixed endpoints are always displacement nodes. Remember this pattern: for the $n$th harmonic on a string fixed at both ends, you'll have $(n+1)$ nodes total, including the two at the boundaries. The nodes divide the string into $n$ equal segments.
5. A standing wave on a string has the equation $y(x,t) = (0.06 \text{ m}) \sin(4\pi x) \cos(120\pi t)$, where $x$ is in meters and $t$ is in seconds. What is the wavelength of the component traveling waves that create this standing wave?
   Answer: $0.50$ m When you encounter a standing wave equation, you're looking at the superposition of two traveling waves moving in opposite directions. The key is recognizing the standard form and extracting the wave parameters. The given equation $y(x,t) = (0.06 \text{ m}) \sin(4\pi x) \cos(120\pi t)$ follows the standard standing wave pattern $y(x,t) = A \sin(kx) \cos(\omega t)$, where $k$ is the wave number and $\omega$ is the angular frequency. From the equation, you can identify $k = 4\pi \text{ rad/m}$. The wavelength $\lambda$ relates to the wave number by $k = \frac{2\pi}{\lambda}$. Solving for wavelength: $\lambda = \frac{2\pi}{k} = \frac{2\pi}{4\pi} = 0.50 \text{ m}$. This confirms answer choice (B). Looking at the incorrect options: (A) 0.25 m would correspond to $k = 8\pi$, which is double the actual wave number—a common error from confusing the relationship between $k$ and $\lambda$. (C) 1.0 m would give $k = 2\pi$, suggesting someone might have incorrectly used $k = \frac{\pi}{\lambda}$ instead of $k = \frac{2\pi}{\lambda}$. (D) 2.0 m corresponds to $k = \pi$, which could result from forgetting the factor of 2 in the wave number formula entirely. Study tip: Always memorize $k = \frac{2\pi}{\lambda}$ and $\omega = \frac{2\pi}{T}$ as your fundamental relationships. When you see a standing wave equation, immediately identify the coefficients of $x$ and $t$ to find $k$ and $\omega$.
6. A string fixed at both ends has length $L = 2.0 \text{ m}$ and supports standing waves. When the string vibrates in a particular mode, there are exactly 4 nodes (including the fixed ends) and 3 antinodes. If the wave speed on the string is $v = 200 \text{ m/s}$, what is the frequency of this vibration?
   Answer: $150 \text{ Hz}$, because this represents the third harmonic with wavelength $\lambda = 2L/3$ With 4 nodes including the ends, there are 3 internal nodes, making this the 3rd harmonic. For the nth harmonic of a string fixed at both ends: $L = n\frac{\lambda}{2}$. For $n = 3$: $\lambda = \frac{2L}{3} = \frac{2(2.0)}{3} = \frac{4}{3} \text{ m}$. The frequency is $f = \frac{v}{\lambda} = \frac{200}{4/3} = 150 \text{ Hz}$. Choice A uses wrong harmonic number and wavelength. Choice B incorrectly identifies this as 2nd harmonic. Choice D incorrectly calls this 4th harmonic and uses wrong wavelength.
7. Two sound waves of equal amplitude and frequency $f$ interfere. At a certain location, the waves have traveled distances of $3.4$ m and $4.1$ m respectively from their sources. If the speed of sound is $350$ m/s and the frequency is $500$ Hz, what type of interference occurs at this location?
   Answer: Constructive, because the path difference is one wavelength When sound waves interfere, the key is determining whether they arrive in phase or out of phase at a given location. This depends on the path difference—how much farther one wave has traveled compared to the other. First, calculate the wavelength: $\lambda = \frac{v}{f} = \frac{350 \text{ m/s}}{500 \text{ Hz}} = 0.7 \text{ m}$ Next, find the path difference: $4.1 \text{ m} - 3.4 \text{ m} = 0.7 \text{ m}$ Since the path difference equals exactly one wavelength ($0.7 \text{ m}$), the waves arrive in phase and interfere constructively. When one wave travels exactly one full wavelength farther than another, it completes one full cycle and aligns perfectly with the other wave again. Looking at each choice: Answer A correctly identifies constructive interference and accurately states that the path difference is one wavelength. Answer B incorrectly claims destructive interference—this would only occur if the path difference were an odd multiple of half-wavelengths (like $0.35 \text{ m}$ or $1.05 \text{ m}$). Answer C correctly identifies constructive interference but wrongly states the path difference as two wavelengths, which would be $1.4 \text{ m}$. Answer D incorrectly claims destructive interference when the path difference is one wavelength—this actually produces constructive interference. Remember: constructive interference occurs when the path difference is a whole number of wavelengths ($0, \lambda, 2\lambda, 3\lambda..$), while destructive interference occurs when it's an odd multiple of half-wavelengths ($\frac{\lambda}{2}, \frac{3\lambda}{2}, \frac{5\lambda}{2}..$). Always calculate the wavelength first, then compare it to the path difference.
8. A guitar string of length $L = 0.65 \text{ m}$ is fixed at both ends and has a fundamental frequency of $330 \text{ Hz}$. When the guitarist presses the string down at the 5th fret (located at $L/4$ from one end), what is the new fundamental frequency of the vibrating portion?
   Answer: $440 \text{ Hz}$, because the effective length is $3L/4$ and this creates a perfect fourth interval When analyzing guitar string vibrations, you need to understand how the fundamental frequency relates to the effective vibrating length. For a string fixed at both ends, the fundamental frequency follows the relationship $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is tension, and $\mu$ is linear mass density. When you press the string at the 5th fret ($L/4$ from one end), the effective vibrating length becomes $L - L/4 = 3L/4$. Since tension and mass density remain constant, frequency is inversely proportional to length: $f_2 = f_1 \times \frac{L_1}{L_2}$. Calculating the new frequency: $f_2 = 330 \text{ Hz} \times \frac{L}{3L/4} = 330 \times \frac{4}{3} = 440 \text{ Hz}$. This creates a perfect fourth interval above the original note, which is exactly what the 5th fret is designed to produce in standard guitar tuning. Choice A incorrectly calculates $330 \times 3/4 = 247.5 \text{ Hz}$, suggesting frequency is directly (not inversely) proportional to length. Choice B assumes the length is halved rather than reduced to $3L/4$, leading to $330 \times 2 = 660 \text{ Hz}$. Choice C makes the same length error as B but also incorrectly states that frequency doubles when length halves (it should be inversely proportional). Remember: guitar frets are positioned to create specific musical intervals. The 5th fret always produces a perfect fourth (frequency ratio of 4:3), so if you know this musical relationship, you can quickly verify physics calculations.
9. A pipe open at both ends has a fundamental frequency of $340$ Hz. If one end is now closed, what will be the frequency of the new fundamental mode?
   Answer: $170$ Hz When you encounter problems about changing pipe configurations, you need to understand how boundary conditions affect standing wave patterns and fundamental frequencies. For a pipe open at both ends, both ends are displacement antinodes (pressure nodes), so the fundamental mode has one half-wavelength fitting inside the pipe length. The fundamental frequency is $f_1 = \frac{v}{2L}$, where $v$ is the speed of sound and $L$ is the pipe length. When one end is closed, you create a displacement node (pressure antinode) at the closed end while keeping a displacement antinode at the open end. This completely changes the standing wave pattern. Now the fundamental mode has only one quarter-wavelength fitting in the same pipe length: $f_1' = \frac{v}{4L}$. Since the original frequency was $340$ Hz $= \frac{v}{2L}$, we can find that $\frac{v}{4L} = \frac{1}{2} \times \frac{v}{2L} = \frac{340}{2} = 170$ Hz. Looking at the wrong answers: B) $340$ Hz assumes the frequency doesn't change, ignoring the boundary condition change. C) $680$ Hz incorrectly doubles the frequency, perhaps confusing this with a harmonic relationship. D) $85$ Hz divides by four instead of two, misunderstanding the wavelength relationship. The correct answer is A) $170$ Hz. Study tip: Remember that closing one end of an open pipe always halves the fundamental frequency because it doubles the effective wavelength from $\lambda/2$ to $\lambda/4$ fitting in the pipe.
10. A guitar string of length $0.60$ m is plucked and vibrates in its second harmonic. If the speed of waves on the string is $400$ m/s, what is the frequency of the sound produced?
    Answer: $667$ Hz When you encounter standing wave problems on guitar strings, you're dealing with harmonics and the relationship between wave speed, frequency, and wavelength. The key insight is understanding what "second harmonic" means for the wave pattern on the string. For a guitar string fixed at both ends, the second harmonic creates a standing wave pattern with one node in the middle and antinodes at both ends. This means the string length equals exactly one full wavelength: $L = \lambda$, so $\lambda = 0.60$ m. Using the wave equation $v = f\lambda$, we can solve for frequency: $f = \frac{v}{\lambda} = \frac{400 \text{ m/s}}{0.60 \text{ m}} = 667$ Hz. This confirms answer choice B is correct. Looking at the wrong answers: Choice A (333 Hz) would result if you mistakenly used $\lambda = 1.2$ m, perhaps thinking the second harmonic means half the fundamental wavelength. Choice C (200 Hz) comes from incorrectly assuming $\lambda = 2L = 1.2$ m, confusing the relationship between string length and wavelength for the second harmonic. Choice D (400 Hz) might tempt you if you confused the wave speed with the frequency, a common conceptual error. Remember this pattern: for the $n$th harmonic on a string fixed at both ends, the wavelength is $\lambda = \frac{2L}{n}$. For the second harmonic specifically, $\lambda = L$. Always identify which harmonic you're dealing with first, then apply the wave equation systematically.
11. Two loudspeakers separated by $4.0$ m emit sound waves in phase with wavelength $2.0$ m. A listener walks along a line parallel to the line connecting the speakers, at a distance of $6.0$ m from the midpoint between the speakers. At what distance from the point closest to the midpoint will the listener first encounter destructive interference?
    Answer: $1.5$ m When you encounter problems involving two coherent sound sources, you're dealing with wave interference patterns. The key is calculating the path difference between waves from each speaker to determine where constructive or destructive interference occurs. Let's set up coordinates with the midpoint between speakers at the origin. The speakers are at positions $(-2.0, 0)$ and $(2.0, 0)$, and the listener walks along the line $y = 6.0$ m. At position $(x, 6.0)$, the distances from each speaker are: - $d_1 = \sqrt{(x + 2.0)^2 + 6.0^2}$ - $d_2 = \sqrt{(x - 2.0)^2 + 6.0^2}$ For destructive interference, the path difference must equal an odd multiple of half-wavelengths: $|d_2 - d_1| = (2n-1)\frac{\lambda}{2}$. For the first occurrence, $n = 1$, so $|d_2 - d_1| = \frac{2.0}{2} = 1.0$ m. Assuming $x > 0$, we need $d_2 - d_1 = 1.0$. Expanding and solving: $\sqrt{(x-2)^2 + 36} - \sqrt{(x+2)^2 + 36} = 1.0$ Using algebraic manipulation (rationalizing by multiplying by the conjugate), this simplifies to: $\frac{-4x}{\sqrt{(x-2)^2 + 36} + \sqrt{(x+2)^2 + 36}} = 1.0$ Solving this equation yields $x = 1.5$ m, which is answer B. Answer A ($1.0$ m) incorrectly assumes the path difference equals the wavelength. Answer C ($2.0$ m) might come from confusing the speaker separation with the answer. Answer D ($2.5$ m) could result from calculation errors in the quadratic solution. Remember: destructive interference requires odd multiples of $\lambda/2$, while constructive interference needs even multiples of $\lambda/2$.
12. Two waves traveling in opposite directions on a string have equations $y_1 = 0.03 \sin(2x - 6t)$ and $y_2 = 0.03 \sin(2x + 6t)$, where $x$ and $y$ are in meters and $t$ is in seconds. At $t = 0$, what is the displacement of the string at $x = \pi/4$ m?
    Answer: $0.06$ m When you encounter two waves traveling in opposite directions, you're dealing with wave superposition. The key principle is that when waves overlap, their displacements add algebraically at each point. To find the total displacement at $x = \pi/4$ m and $t = 0$, you need to evaluate both wave equations at these values and sum them. For the first wave: $y_1 = 0.03 \sin(2x - 6t) = 0.03 \sin(2 \cdot \pi/4 - 6 \cdot 0) = 0.03 \sin(\pi/2) = 0.03(1) = 0.03$ m For the second wave: $y_2 = 0.03 \sin(2x + 6t) = 0.03 \sin(2 \cdot \pi/4 + 6 \cdot 0) = 0.03 \sin(\pi/2) = 0.03(1) = 0.03$ m The total displacement is: $y_{total} = y_1 + y_2 = 0.03 + 0.03 = 0.06$ m Answer A (0.00 m) would occur if the waves were completely out of phase, causing destructive interference. Answer B (0.03 m) represents the displacement of just one wave, ignoring superposition entirely. Answer D ($0.03\sqrt{2}$ m) might tempt you if you incorrectly think the waves add as vectors rather than scalars, but wave displacements add algebraically along the same axis. Remember: when analyzing wave superposition problems, always evaluate each wave function separately at the given coordinates, then add the results. The mathematics is straightforward once you apply the superposition principle correctly.
13. A standing wave on a string has 4 antinodes. If the string length is $1.2$ m and is fixed at both ends, what is the wavelength of the traveling waves that form this standing wave?
    Answer: $0.6$ m When you encounter standing wave problems with fixed ends, you need to visualize how the wave pattern relates to the string length. A standing wave forms when traveling waves reflect back and forth, creating stationary points (nodes) and maximum oscillation points (antinodes). For a string fixed at both ends, the ends must always be nodes since they can't move. With 4 antinodes, you have the pattern: node-antinode-node-antinode-node-antinode-node-antinode-node. This creates 4 complete "humps" or segments along the string. Each segment between consecutive nodes represents half a wavelength ($\lambda/2$). Since you have 4 antinodes, you have 4 segments of $\lambda/2$ each. The total string length equals: $1.2 \text{ m} = 4 \times \frac{\lambda}{2} = 2\lambda$ Solving for wavelength: $\lambda = \frac{1.2 \text{ m}}{2} = 0.6 \text{ m}$ Answer choice A ($0.3$ m) represents $\lambda/2$—the length of one segment, not the full wavelength. Answer choice C ($1.2$ m) incorrectly assumes the string length equals one wavelength, which would only occur with 2 antinodes. Answer choice D ($2.4$ m) doubles the string length, perhaps from confusion about the relationship between standing wave length and wavelength. The correct answer is B ($0.6$ m). Study tip: For standing waves with fixed ends, remember that the number of antinodes equals the number of half-wavelengths that fit in the string. Use the formula: string length = (number of antinodes) × ($\lambda/2$).
14. Two waves of equal frequency and amplitude travel toward each other on a string. When they overlap completely, they create a standing wave with nodes spaced $1.5$ m apart. What was the wavelength of each original traveling wave?
    Answer: $3.0$ m When two waves with equal frequency and amplitude travel in opposite directions and overlap, they create a standing wave pattern through constructive and destructive interference. Understanding the relationship between the original wavelength and the resulting node spacing is crucial for solving these problems. In a standing wave, nodes are points that remain stationary (zero amplitude), while antinodes oscillate with maximum amplitude. The key insight is that consecutive nodes are separated by exactly half the wavelength of the original traveling waves. This occurs because destructive interference (creating nodes) happens when waves are 180° out of phase, which corresponds to half a wavelength difference. Since the nodes are spaced $1.5$ m apart, and this distance equals half the original wavelength, we can write: $\frac{\lambda}{2} = 1.5$ m. Solving for the wavelength: $\lambda = 3.0$ m. This confirms answer B is correct. Let's examine the wrong answers: A) $1.5$ m incorrectly assumes the node spacing equals the full wavelength rather than half the wavelength. C) $0.75$ m represents one-quarter wavelength, which would be the distance from a node to an adjacent antinode. D) $6.0$ m doubles the correct answer, perhaps from confusing the relationship between wavelength and node spacing. Remember this pattern: in standing wave problems, always check whether the given distance represents the spacing between nodes (half wavelength), between antinodes (also half wavelength), or between a node and antinode (quarter wavelength). Node-to-node spacing is the most commonly tested relationship.
15. A rope is fixed at one end and free at the other. When it vibrates in its fundamental mode, the length of the rope is $1.5$ m. What is the wavelength of the wave on the rope?
    Answer: $6.0$ m When you encounter standing wave problems, you need to understand the relationship between the physical length of the medium and the wavelength of the wave pattern formed. For a rope fixed at one end and free at the other, the fundamental mode creates a specific standing wave pattern. The fixed end must be a node (zero displacement), while the free end must be an antinode (maximum displacement). In the fundamental mode, this boundary condition creates exactly one-quarter of a complete wavelength along the rope's length. Since the rope length equals one-quarter wavelength: $L = \frac{\lambda}{4}$ Given $L = 1.5$ m, we can solve for wavelength: $\lambda = 4L = 4(1.5) = 6.0$ m Choice A ($1.5$ m) incorrectly assumes the rope length equals one full wavelength, which would apply to a rope fixed at both ends in its fundamental mode. Choice B ($3.0$ m) mistakenly treats the rope length as half a wavelength, another common boundary condition mix-up. Choice D ($0.75$ m) suggests the rope contains multiple wavelengths, which would represent a higher harmonic, not the fundamental mode. Study tip: Memorize the fundamental relationships for different boundary conditions. Fixed-free systems always have $L = \frac{\lambda}{4}$ in the fundamental mode, while fixed-fixed systems have $L = \frac{\lambda}{2}$. Draw the standing wave pattern to visualize where nodes and antinodes must occur based on the physical constraints.
16. Two waves traveling in opposite directions on a string have equations $y_1 = 0.05\sin(4x - 20t)$ and $y_2 = 0.05\sin(4x + 20t)$, where $x$ and $y$ are in meters and $t$ is in seconds. At what positions along the string do nodes (points of zero amplitude) occur in the resulting standing wave?
    Answer: $x = \frac{(2n+1)\pi}{8}$ where $n$ is any integer, because nodes occur at odd multiples of $\lambda/4$ The superposition gives: $y = y_1 + y_2 = 0.05[\sin(4x - 20t) + \sin(4x + 20t)] = 0.1\sin(4x)\cos(20t)$. Nodes occur where the amplitude $0.1\sin(4x) = 0$, so $4x = n\pi$ where $n$ is any integer, giving $x = \frac{n\pi}{4}$. However, this includes both nodes and antinodes. For just nodes (zero amplitude), we need $\sin(4x) = 0$ at positions that aren't antinodes. Since $k = 4$, we have $\lambda = 2\pi/k = \pi/2$. Nodes occur at $x = \frac{(2n+1)\pi}{8}$ (odd multiples of $\lambda/4$). Choice A has wrong spacing. Choice B includes antinodes. Choice C includes both nodes and antinodes.
17. A string fixed at both ends vibrates in a standing wave pattern. The distance between two consecutive nodes is $0.8$ m. If the wave speed on the string is $320$ m/s, what is the frequency of vibration?
    Answer: $200$ Hz Standing wave problems require understanding the relationship between wavelength, wave speed, and frequency, plus the specific geometry of nodes and antinodes. In a standing wave on a string fixed at both ends, nodes are points of zero displacement that remain stationary. The distance between consecutive nodes is always half a wavelength ($\lambda/2$). Since consecutive nodes are $0.8$ m apart, we have: $\frac{\lambda}{2} = 0.8 \text{ m}$ Therefore: $\lambda = 1.6 \text{ m}$ Using the wave equation $v = f\lambda$, where $v = 320$ m/s: $f = \frac{v}{\lambda} = \frac{320}{1.6} = 200 \text{ Hz}$ Looking at the wrong answers: Choice B ($400$ Hz) results from incorrectly assuming the distance between consecutive nodes equals the full wavelength rather than half the wavelength. Choice C ($160$ Hz) comes from using $\lambda = 2.0$ m, which would be the distance between alternate nodes (a full wavelength apart). Choice D ($100$ Hz) represents using $\lambda = 3.2$ m, perhaps from doubling the wavelength incorrectly. The correct answer is A) $200$ Hz. Study tip: Always remember that consecutive nodes (or consecutive antinodes) in standing waves are separated by $\lambda/2$, not $\lambda$. This is one of the most common sources of error in standing wave problems. Draw a quick sketch of the wave pattern to visualize the node spacing if you're unsure.
18. A pipe closed at one end resonates at its fundamental frequency when the air column length is $0.25$ m. What is the wavelength of the sound wave in air?
    Answer: $1.0$ m When you encounter resonance in pipes, you need to understand the relationship between pipe length and wavelength based on the pipe's configuration. A pipe closed at one end creates a specific standing wave pattern that's different from an open pipe. In a pipe closed at one end, the closed end must be a displacement node (where air particles can't move), while the open end must be a displacement antinode (where air particles move freely). At the fundamental frequency, this creates the simplest standing wave pattern: exactly one-quarter of a complete wavelength fits inside the pipe length. This means: $L = \frac{\lambda}{4}$ Where $L$ is the pipe length and $\lambda$ is the wavelength. Given that $L = 0.25$ m, we can solve for wavelength: $0.25 = \frac{\lambda}{4}$ $\lambda = 4 \times 0.25 = 1.0$ m Looking at the wrong answers: Choice A ($0.25$ m) incorrectly assumes the pipe length equals the wavelength, which would be true for an open pipe at fundamental frequency. Choice B ($0.50$ m) represents half a wavelength, which might come from confusing this with other resonance conditions. Choice D ($0.125$ m) appears to come from dividing the pipe length by 2 instead of multiplying by 4, reversing the correct relationship. Remember this key pattern: for pipes closed at one end, the fundamental resonance occurs when the pipe length equals one-quarter wavelength. This is different from open pipes, where the fundamental occurs at half wavelength. Always identify the pipe type first.
19. In Young's double-slit experiment, monochromatic light of wavelength $\lambda = 500 \text{ nm}$ passes through two slits separated by $d = 0.20 \text{ mm}$. The interference pattern is observed on a screen located $L = 2.0 \text{ m}$ from the slits. If the experiment is repeated underwater where the refractive index is $n = 1.33$, how does the fringe spacing change?
    Answer: The fringe spacing decreases by a factor of $1.33$ because the wavelength decreases in water The fringe spacing in air is $\Delta y = \frac{\lambda L}{d}$. In water, the wavelength becomes $\lambda' = \frac{\lambda}{n} = \frac{500}{1.33} \text{ nm}$. The new fringe spacing is $\Delta y' = \frac{\lambda' L}{d} = \frac{\lambda L}{nd} = \frac{\Delta y}{n}$. So the fringe spacing decreases by a factor of $n = 1.33$. Choice A incorrectly states wavelength increases in water. Choice C ignores the wavelength change in the medium. Choice D uses an incorrect factor ($1.33^2 ≈ 1.77$) and misunderstands the physics.
20. A standing wave is established on a string of length $L = 1.2 \text{ m}$ that is fixed at both ends. The string vibrates in its third harmonic with a frequency of $180 \text{ Hz}$. What is the speed of waves on this string?
    Answer: $144 \text{ m/s}$, because the wavelength equals two-thirds the string length For a string fixed at both ends, the third harmonic has 3 half-wavelengths fitting in the length: $L = 3(\lambda/2)$, so $\lambda = 2L/3 = 2(1.2)/3 = 0.8 \text{ m}$. The wave speed is $v = f\lambda = 180 × 0.8 = 144 \text{ m/s}$. Choice A incorrectly uses $L/3$ for wavelength. Choice C incorrectly assumes the wavelength equals $L/3$. Choice D incorrectly uses $\lambda = 2L$, which would apply to a different boundary condition scenario.

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