Periodic Waves - College Physics

A wave travels along a rope with speed $v = \sqrt{T/\mu}$ where $T$ is tension and $\mu$ is linear mass density. If the tension is increased by 44% while keeping the rope material unchanged, the wave speed:
Answer: increases by 20% because wave speed is proportional to the square root of tension When you encounter wave speed problems, focus on how the given formula relates the variables. The wave speed equation $v = \sqrt{T/\mu}$ shows that speed depends on the square root of tension, not tension directly. Let's work through this step by step. If tension increases by 44%, the new tension becomes $T_{new} = T + 0.44T = 1.44T$ . Since the rope material is unchanged, $\mu$ stays constant. The new wave speed is: $v_{new} = \sqrt{\frac{1.44T}{\mu}} = \sqrt{1.44} \cdot \sqrt{\frac{T}{\mu}} = 1.2v$ Since $\sqrt{1.44} = 1.2$ , the new speed is 1.2 times the original speed, representing a 20% increase. Choice A incorrectly assumes wave speed is directly proportional to tension. If this were true, a 44% tension increase would yield a 44% speed increase, but the square root relationship prevents this. Choice B makes the right connection to square root proportionality but miscalculates the result. Perhaps this comes from incorrectly thinking that half of 44% equals 22%, but that's not how square roots work. Choice D dramatically overestimates the effect and mentions "wave propagation efficiency," which isn't relevant to this tension-speed relationship. The 88% figure has no mathematical basis in the given formula. Remember: when you see square root relationships in physics formulas, percentage changes in the variable under the radical always produce smaller percentage changes in the result. Always calculate $\sqrt{1 + \text{percentage change}}$ rather than guessing.
Two identical sinusoidal waves travel in the same direction on a string. Wave 1 has the form $y_1 = A\sin(kx - \omega t)$ and wave 2 has the form $y_2 = A\sin(kx - \omega t + \pi/3)$ . What is the amplitude of the resulting wave when they superpose?
Answer: $\sqrt{3}A$ When two waves superpose, you need to use the principle of superposition and trigonometric identities to find the resulting amplitude. This is a classic wave interference problem that tests your ability to add sinusoidal functions. To find the amplitude of the combined wave, add the two wave functions: $y_{total} = y_1 + y_2 = A\sin(kx - \omega t) + A\sin(kx - \omega t + \pi/3)$ Use the trigonometric identity for the sum of sines: $\sin(\alpha) + \sin(\beta) = 2\sin\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\alpha - \beta}{2}\right)$ Here, $\alpha = kx - \omega t$ and $\beta = kx - \omega t + \pi/3$ , so: - $\frac{\alpha + \beta}{2} = kx - \omega t + \pi/6$ - $\frac{\alpha - \beta}{2} = -\pi/6$ This gives: $y_{total} = 2A\sin(kx - \omega t + \pi/6)\cos(-\pi/6) = 2A\sin(kx - \omega t + \pi/6) \cdot \frac{\sqrt{3}}{2} = \sqrt{3}A\sin(kx - \omega t + \pi/6)$ The amplitude is $\sqrt{3}A$ , making D correct. Choice A ( $A$ ) would result from destructive interference, but these waves have a $\pi/3$ phase difference, not $\pi$ . Choice B ( $2A$ ) assumes perfect constructive interference (zero phase difference), ignoring the $\pi/3$ phase shift. Choice C ( $1.5A$ ) has no physical basis for this wave combination. Study tip: For wave superposition problems, always check the phase difference first. Use trigonometric identities systematically rather than guessing based on limiting cases.
A standing wave is formed on a string fixed at both ends. The string has length $L = 1.2$ m and the standing wave has exactly 3 antinodes between the fixed ends. If the wave speed on the string is 240 m/s, what is the frequency of the driving oscillation?
Answer: 300 Hz For a string fixed at both ends, nodes occur at the boundaries. With 3 antinodes between the fixed ends, the pattern is: node-antinode-node-antinode-node-antinode-node. This represents 3 half-wavelength segments, so $L = 3(\lambda/2)$ , giving $\lambda = 2L/3 = 2(1.2)/3 = 0.8$ m. Using $v = f\lambda$ : $f = v/\lambda = 240/0.8 = 300$ Hz.
Two periodic waves with the same frequency and amplitude travel in opposite directions on a string. At a point where they interfere constructively, the amplitude of oscillation is $A$ . What is the amplitude of each individual wave?
Answer: $A/2$ For constructive interference between two waves of equal amplitude $A_0$ , the resulting amplitude is $A_{total} = A_0 + A_0 = 2A_0$ . Since we observe amplitude $A$ at the constructive interference point, we have $A = 2A_0$ , so each individual wave has amplitude $A_0 = A/2$ .
A wave pulse travels down a rope at 10 m/s. The rope then connects to a heavier rope where waves travel at 6 m/s. What fraction of the incident wave's amplitude is reflected back toward the source?
Answer: 1/4 At a boundary, the reflection coefficient for amplitude is $r = \frac{v_1 - v_2}{v_1 + v_2}$ , where $v_1$ and $v_2$ are the wave speeds in the first and second media. Here, $r = \frac{10 - 6}{10 + 6} = \frac{4}{16} = \frac{1}{4}$ . The magnitude of this coefficient gives the fraction of amplitude reflected, which is 1/4.
The intensity of a sound wave is proportional to the square of its amplitude. If the amplitude of a sound wave is reduced to one-third of its original value, the intensity becomes:
Answer: one-ninth of the original intensity, because intensity is proportional to the square of amplitude When you encounter questions about wave intensity and amplitude, remember that intensity describes the energy carried by a wave per unit area per unit time, while amplitude measures the maximum displacement from equilibrium. The key relationship here is that intensity is proportional to the square of amplitude: $I \propto A^2$ . This means if you change the amplitude by some factor, the intensity changes by the square of that factor. Let's work through this mathematically. If the original amplitude is $A_0$ and the new amplitude is $A_0/3$ , then: - Original intensity: $I_0 \propto A_0^2$ - New intensity: $I_{new} \propto (A_0/3)^2 = A_0^2/9$ - Therefore: $I_{new} = I_0/9$ The intensity becomes one-ninth of its original value, confirming answer C. Now let's examine why the other options are incorrect. Option A suggests a direct proportional relationship ( $I \propto A$ ), which ignores the crucial square relationship. Option B introduces frequency factors that aren't relevant to this amplitude-intensity relationship - while frequency affects wave energy, the question specifically isolates the amplitude effect. Option D mentions geometric spreading factors, which relate to how waves spread out over distance, but again, this question focuses solely on the amplitude-intensity relationship at a given point. Remember this pattern: whenever you see "proportional to the square" in physics, changing the input variable by a factor changes the output by that factor squared. This appears frequently with intensity-amplitude relationships, kinetic energy-velocity relationships, and gravitational force-distance relationships.
Two speakers separated by 4.0 m emit sound waves in phase with frequency 170 Hz. A listener at point P is 5.0 m from one speaker and 3.0 m from the other. If the speed of sound is 340 m/s, what type of interference occurs at point P?
Answer: Constructive interference with amplitude doubled First, find the wavelength: $\lambda = v/f = 340/170 = 2.0$ m. The path difference is $|5.0 - 3.0| = 2.0$ m. Since this equals exactly one wavelength ( $2.0 = 1 \times 2.0$ ), the waves arrive in phase, creating constructive interference. For two identical sources, constructive interference doubles the amplitude.
A sinusoidal wave has the form $y(x,t) = A\sin(kx - \omega t + \phi)$ . If at $t = 0$ and $x = 0$ the displacement is $y = A/2$ and the particle velocity $\partial y/\partial t$ is negative, what is the phase constant $\phi$ ?
Answer: $\pi/6$ When analyzing sinusoidal waves, you need to use both the displacement and velocity conditions to determine the phase constant. The wave equation $y(x,t) = A\sin(kx - \omega t + \phi)$ contains all the information about the wave's behavior. Start with the displacement condition. At $t = 0$ and $x = 0$ , you have $y(0,0) = A\sin(\phi) = A/2$ . This gives you $\sin(\phi) = 1/2$ , which means $\phi = \pi/6$ or $\phi = 5\pi/6$ . Now use the velocity condition to determine which value is correct. The particle velocity is $\frac{\partial y}{\partial t} = -A\omega\cos(kx - \omega t + \phi)$ . At $t = 0$ and $x = 0$ , this becomes $-A\omega\cos(\phi)$ . Since the velocity is negative and $-A\omega$ is negative (assuming positive $A$ and $\omega$ ), you need $\cos(\phi) > 0$ . For $\phi = \pi/6$ : $\cos(\pi/6) = \sqrt{3}/2 > 0$ ✓ For $\phi = 5\pi/6$ : $\cos(5\pi/6) = -\sqrt{3}/2 < 0$ ✗ Therefore, $\phi = \pi/6$ , which is answer D. The wrong answers represent common errors: A ( $5\pi/6$ ) satisfies the displacement condition but gives positive velocity. B ( $\pi/3$ ) and C ( $2\pi/3$ ) don't even satisfy the displacement condition since $\sin(\pi/3) = \sin(2\pi/3) = \sqrt{3}/2 \neq 1/2$ . Always check both displacement and velocity conditions systematically—one constrains the possible values, the other selects the correct one.
A sinusoidal wave has a period of 0.02 s and an amplitude of 3.0 cm. What is the maximum speed of a particle in the medium as the wave passes?
Answer: $9.4$ m/s, because the maximum particle speed is amplitude times angular frequency When you encounter wave problems involving particle motion, remember that you're analyzing how individual particles in the medium oscillate as the wave passes through—this is different from the wave's propagation speed. For a sinusoidal wave, particles undergo simple harmonic motion with displacement $y = A\sin(\omega t)$ , where $A$ is amplitude and $\omega$ is angular frequency. To find particle velocity, you take the time derivative: $v = \frac{dy}{dt} = A\omega\cos(\omega t)$ . The maximum particle speed occurs when $\cos(\omega t) = 1$ , giving $v_{max} = A\omega$ . First, calculate the angular frequency: $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.02} = 314.2$ rad/s. Then: $v_{max} = (0.03 \text{ m})(314.2 \text{ rad/s}) = 9.4$ m/s. This confirms answer C is correct. Answer A incorrectly suggests maximum speed occurs at maximum displacement, but particles momentarily stop at maximum displacement—maximum speed actually occurs at equilibrium position. Answer B uses frequency instead of angular frequency, missing the $2\pi$ factor in the conversion, giving $v = Af = (0.03)(50) = 1.5$ m/s. Answer D attempts the correct formula but arranges it as $\frac{2\pi A}{T}$ , which mathematically equals $A\omega$ but suggests conceptual confusion about the relationship. Remember this key pattern: maximum particle speed in wave motion always equals amplitude times angular frequency ( $A\omega$ ). This formula appears frequently in wave physics problems.
In a standing wave on a string, nodes are points where:
Answer: the displacement is always zero, and the particle velocity is always zero throughout the oscillation When analyzing standing waves, you need to understand what happens at specific points along the wave pattern. Standing waves form when two identical waves traveling in opposite directions interfere, creating stationary points called nodes and antinodes. At nodes, both the displacement and particle velocity are always zero throughout the entire oscillation cycle. This occurs because the two traveling waves are always perfectly out of phase at these locations, causing complete destructive interference. The string literally doesn't move at all at these points - neither up and down displacement nor any velocity motion occurs. Let's examine why the other options are incorrect. Option A suggests nodes have zero displacement but maximum alternating velocity - this actually describes antinodes, not nodes. At antinodes, displacement is zero only instantaneously as particles pass through equilibrium with maximum speed. Option B claims maximum displacement and zero velocity, which is physically impossible since maximum displacement occurs when particles momentarily stop before changing direction. Option D describes alternating maximum displacement with zero velocity, which again doesn't represent any real point in a standing wave pattern. The key distinction is that nodes are completely motionless points where the interference is perfectly destructive at all times. In contrast, antinodes are points of maximum motion where constructive interference occurs. Remember this pattern: in standing wave problems, "node" means "no motion" - zero displacement AND zero velocity always. This complete absence of motion is what distinguishes nodes from all other points along the standing wave.
Two periodic waves with the same amplitude and frequency are traveling in opposite directions along a rope. At a point where they meet, the displacement from equilibrium is measured to be zero at a particular instant. What can be concluded about the phase relationship between the waves at this location?
Answer: The phase relationship cannot be determined from a single measurement at one instant in time When two waves with identical amplitude and frequency travel in opposite directions, they create a standing wave pattern through superposition. The key insight here is that wave interference is a dynamic process that varies continuously in both space and time. A single measurement of zero displacement at one point and one instant provides insufficient information to determine the phase relationship. Zero displacement can occur in multiple scenarios: when both waves are momentarily at their equilibrium positions (in phase), when they have equal and opposite displacements that cancel out (out of phase), or even during intermediate phase relationships where the waves happen to sum to zero at that specific moment. Answer A incorrectly assumes that zero displacement automatically means both waves are at equilibrium simultaneously. While this is one possibility, it's not the only explanation for the observed zero displacement. Answer B makes the opposite error, concluding the waves must be out of phase. Though destructive interference could cause zero displacement, this is just one possible scenario among many. Answer C suggests a specific phase difference of $\pi/4$ without justification. Any phase relationship could potentially produce zero displacement at a particular instant, making this assumption unfounded. Answer D correctly recognizes that determining phase relationships requires multiple measurements across time or space. A complete picture of the wave pattern emerges only when you observe how the displacement changes over time at that point, or measure displacements at multiple locations simultaneously. Remember: wave interference problems often require temporal or spatial context beyond single-point, single-instant measurements to draw definitive conclusions about phase relationships.
A transverse wave on a string is described by $y(x,t) = 0.02\sin(4\pi x - 60\pi t)$ where $x$ is in meters, $t$ is in seconds, and $y$ is in meters. At $t = 0.1$ s, what is the displacement at $x = 0.25$ m?
Answer: $0$ m, because $\sin(\pi - 6\pi) = \sin(-5\pi) = 0$ due to the sine function periodicity When you encounter a wave equation like this, you're working with the standard form $y(x,t) = A\sin(kx - \omega t)$ , where you need to substitute the given values of position and time to find the displacement. To find the displacement at $x = 0.25$ m and $t = 0.1$ s, substitute these values into the equation: $y(0.25, 0.1) = 0.02\sin(4\pi \cdot 0.25 - 60\pi \cdot 0.1)$ $= 0.02\sin(\pi - 6\pi)$ $= 0.02\sin(-5\pi)$ Since sine has a period of $2\pi$ , we can write $-5\pi = -2\pi - 2\pi - \pi = -\pi$ (after removing complete $2\pi$ cycles). Therefore $\sin(-5\pi) = \sin(-\pi) = 0$ . This gives us $y = 0.02 \times 0 = 0$ m. Looking at the wrong answers: Choice A incorrectly claims the result should be $-0.02$ m, but since $\sin(-5\pi) = 0$ , multiplying by the amplitude gives zero, not the negative amplitude. Choice C makes an error by suggesting the wave is at maximum when $\sin(-5\pi) = 0$ , which would occur when the sine equals ±1. Choice D incorrectly evaluates $\sin(-5\pi)$ as $-1/2$ , but $\sin(-5\pi) = 0$ , not $-1/2$ . Remember that when evaluating sine at multiples of $\pi$ , use the fact that $\sin(n\pi) = 0$ for any integer $n$ . Always double-check your trigonometric evaluations, especially when dealing with negative angles or large multiples of $\pi$ .
A sound wave in air has a frequency of 1000 Hz and travels at 340 m/s. If this wave enters water where the speed is 1500 m/s, what happens to the wavelength?
Answer: The wavelength increases to 1.5 m because the frequency remains constant but speed increases When a wave crosses from one medium to another, you need to understand which wave properties change and which stay constant. The fundamental relationship is $v = f\lambda$ , where speed (v), frequency (f), and wavelength (λ) are related. The key insight is that frequency remains constant when a wave enters a new medium - it's determined by the source, not the medium. However, wave speed changes based on the medium's properties. Since $v = f\lambda$ must always hold true, if frequency stays the same but speed changes, wavelength must adjust proportionally. Let's calculate: In air, $\lambda = \frac{v}{f} = \frac{340 \text{ m/s}}{1000 \text{ Hz}} = 0.34 \text{ m}$ . In water, $\lambda = \frac{1500 \text{ m/s}}{1000 \text{ Hz}} = 1.5 \text{ m}$ . The wavelength increases because the wave travels faster in water while maintaining the same frequency. Choice A incorrectly states the wave slows down in water (it actually speeds up) and gives the wrong wavelength value. Choice C wrongly suggests wavelength is intrinsic to the source - while frequency is intrinsic, wavelength depends on both source and medium. Choice D invents a non-existent "averaging" rule and provides an incorrect value. Remember this pattern: when waves enter a new medium, frequency stays constant (set by the source), speed changes (determined by medium properties), and wavelength adjusts accordingly through $v = f\lambda$ . Always identify what stays constant first, then solve for what changes.
A transverse wave pulse traveling along a rope has its maximum displacement at $x = 2.0$ m when $t = 0$ s. If the pulse travels at 5.0 m/s in the positive $x$ direction, at what position will the maximum displacement occur when $t = 1.5$ s?
Answer: $x = 9.5$ m, because the pulse travels from its initial position plus the distance covered When analyzing wave motion problems, you need to track how the wave's features move through space over time. The key insight is that every point on the wave pulse moves at the same speed as the wave itself. Here, the maximum displacement starts at $x = 2.0$ m at $t = 0$ s. Since the pulse travels at 5.0 m/s in the positive $x$ direction, you can find how far this maximum point travels using the basic equation: distance = speed × time. In 1.5 seconds, the maximum travels $5.0 \text{ m/s} \times 1.5 \text{ s} = 7.5$ m from its starting position. The new position is the original position plus the distance traveled: $2.0 \text{ m} + 7.5 \text{ m} = 9.5$ m. This makes choice C correct. Choice A incorrectly calculates the distance traveled as only 1.5 m, confusing the time value (1.5 s) with the distance. Choice B gives 7.5 m as the final position, but this is actually just the distance traveled—it ignores the initial starting position of 2.0 m. Choice D calculates the distance traveled correctly (7.5 m) but then adds this to an assumed starting position of 2.5 m, which isn't given in the problem. Remember this pattern: for wave motion problems, always identify what's moving (the wave feature), where it starts, and add the distance it travels (speed × time) to find its final position. Don't confuse distance traveled with final position.
A wave pulse traveling along a stretched string encounters a boundary where the string is attached to a much more massive string. Compared to the incident pulse, the reflected pulse will have:
Answer: reduced amplitude and inverted orientation, because the boundary acts like a fixed end When a wave pulse encounters a boundary between two different media, the behavior depends on the impedance difference between the materials. A string attached to a much more massive string creates a boundary that behaves essentially like a fixed end, since the massive string has much greater inertia and cannot move significantly. At this type of boundary, two important things happen: the pulse reflects with inverted phase (flipped orientation) and reduced amplitude. The inversion occurs because the boundary acts like a fixed end - when the incident pulse tries to displace the boundary upward, the massive string resists this motion, creating a reaction force that sends the pulse back downward. The amplitude reduction happens because some energy is transmitted into the massive string, even though it barely moves. Choice A is wrong because energy conservation doesn't prevent amplitude reduction or phase inversion - energy can be transmitted while the reflected portion changes. Choice C correctly identifies amplitude reduction due to energy transmission but incorrectly claims the orientation stays the same. Choice D correctly identifies the inversion due to the massive string's inability to move significantly, but wrongly suggests amplitude stays constant - this would only occur at a completely rigid boundary with zero energy transmission. The correct answer is B because both effects occur: reduced amplitude (energy is partially transmitted) and inverted orientation (the boundary acts like a fixed end). Remember this pattern: heavy-to-light boundaries invert the reflected wave, while light-to-heavy boundaries preserve orientation. The massive string case is effectively a light-to-fixed boundary, producing inversion.
Two wave pulses of equal amplitude travel toward each other along a rope. When they completely overlap, the amplitude of the resulting disturbance depends on:
Answer: the relative phase of the pulses, because this determines whether interference is constructive or destructive When two wave pulses meet on a rope, you're witnessing the principle of superposition - one of the most fundamental concepts in wave physics. The key insight is that waves don't actually "collide" like particles; instead, they pass through each other, and their amplitudes simply add together at every point. The correct answer is C because the relative phase relationship between the pulses determines how their amplitudes combine. If the pulses are in phase (peaks aligned with peaks), they undergo constructive interference, doubling the amplitude. If they're completely out of phase (peaks aligned with troughs), they undergo destructive interference, potentially canceling each other out entirely. Any phase difference between these extremes produces intermediate results. Option A is wrong because wave speed affects how quickly pulses approach each other, but not the amplitude when they overlap. Speed is determined by the medium properties, not the interference pattern. Option B incorrectly focuses on pulse width - while wider pulses overlap for longer durations, the maximum amplitude during overlap still depends on phase alignment, not width. Option D misunderstands the role of rope tension, which affects wave speed and the rope's ability to support waves, but doesn't change how amplitudes add during superposition. Remember this key principle: superposition depends only on how wave amplitudes align. Whether you're dealing with sound waves, light waves, or rope waves, constructive interference occurs when waves are in phase, destructive when they're out of phase. Focus on the phase relationship, not the medium properties or pulse characteristics.
A sound wave in air has a frequency of 440 Hz and travels at 343 m/s. If this wave enters water where the speed of sound is 1500 m/s, what happens to the period and wavelength?
Answer: Period remains constant; wavelength increases by factor 4.4 When a wave crosses a boundary, frequency (and thus period $T = 1/f$ ) remains constant - this is determined by the source. The wavelength changes according to $\lambda = v/f$ . In air: $\lambda_{air} = 343/440 = 0.78$ m. In water: $\lambda_{water} = 1500/440 = 3.41$ m. The ratio is $1500/343 ≈ 4.4$ , so wavelength increases by this factor while period stays the same.
A sinusoidal wave travels along a string with frequency $f = 60$ Hz and wavelength $\lambda = 0.5$ m. If the amplitude is doubled while keeping the frequency constant, and the tension in the string is simultaneously increased by a factor of 4, what is the new wavelength?
Answer: 1.0 m The wave speed on a string is $v = \sqrt{T/\mu}$ , where $T$ is tension and $\mu$ is linear mass density. Initially, $v_1 = f\lambda_1 = 60 \times 0.5 = 30$ m/s. When tension increases by factor 4, the new speed is $v_2 = 2v_1 = 60$ m/s (since $\sqrt{4} = 2$ ). With frequency unchanged at 60 Hz, the new wavelength is $\lambda_2 = v_2/f = 60/60 = 1.0$ m. The amplitude change doesn't affect wavelength.
A standing wave pattern on a 2.0 m string shows 3 complete wavelengths. If the wave speed is 120 m/s, what is the frequency?
Answer: 180 Hz, because 3 wavelengths fit in the string length and $f = v/\lambda$ When you encounter standing wave problems, focus on the fundamental relationship between wave properties: wave speed, wavelength, and frequency are connected by $v = f\lambda$ . Here, you need to find the wavelength first. If 3 complete wavelengths fit in a 2.0 m string, then each wavelength is $\lambda = \frac{2.0 \text{ m}}{3} = 0.667 \text{ m}$ . With wave speed $v = 120 \text{ m/s}$ , you can calculate frequency using $f = \frac{v}{\lambda} = \frac{120}{0.667} = 180 \text{ Hz}$ . Choice A calculates the wavelength correctly but makes an arithmetic error in the final division, getting 60 Hz instead of 180 Hz. This shows the importance of double-checking your math. Choice B introduces the concept of an "effective wavelength" for standing waves, which is a misconception. Standing waves don't change the fundamental wavelength - they're simply the interference pattern of two traveling waves. The wavelength remains the same as it would for a traveling wave. Choice D incorrectly assumes that standing wave frequency is twice the traveling wave frequency. This confuses standing waves with some other wave phenomena. Standing waves have the same frequency as the traveling waves that create them through interference. The key insight is that standing wave patterns directly show you the wavelength - you can literally count how many complete waves fit in the given length. Once you have wavelength and wave speed, frequency follows directly from $f = v/\lambda$ . Always verify your arithmetic, as calculation errors are common in wave problems.
A standing wave is established on a string fixed at both ends. If the string length is 1.2 m and the wave speed is 240 m/s, what is the frequency of the third harmonic?
Answer: 300 Hz, because the third harmonic has wavelength $\lambda = 2L/3$ and $f = v/\lambda$ Standing waves on strings fixed at both ends follow a predictable harmonic pattern that's crucial for understanding resonance. When you see questions about harmonics on fixed strings, remember that the string length must accommodate complete wave patterns with nodes (zero displacement) at both ends. For a string fixed at both ends, the wavelength of the nth harmonic follows the formula $\lambda_n = \frac{2L}{n}$ , where L is the string length. For the third harmonic (n = 3), this gives us $\lambda_3 = \frac{2(1.2)}{3} = 0.8 \text{ m}$ . Using the wave equation $f = \frac{v}{\lambda}$ , we get $f_3 = \frac{240}{0.8} = 300 \text{ Hz}$ . Let's examine why the other answers miss the mark. Choice A correctly identifies the wavelength as $2L/3$ but miscalculates the frequency, getting 100 Hz instead of 300 Hz—likely from an arithmetic error. Choice B uses the wrong wavelength formula ( $L/3$ instead of $2L/3$ ), which would apply to a different wave scenario, leading to 200 Hz. Choice D incorrectly uses $\lambda = L/2$ , which is actually the wavelength for the second harmonic, not the third, resulting in 360 Hz. The correct answer is C: 300 Hz with the proper wavelength calculation. Study tip: Memorize the harmonic formula $\lambda_n = \frac{2L}{n}$ for strings fixed at both ends. The "2L" comes from fitting n half-wavelengths into the string length. Always double-check your arithmetic when applying $f = v/\lambda$ —calculation errors are common traps on physics exams.

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