College Physics Quiz: Periodic Waves

20 questions · exam conditions

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Periodic WavesQuestion 1 of 20

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?

Constructive interference with amplitude doubled

Destructive interference with zero amplitude

Constructive interference with amplitude less than doubled

Destructive interference with reduced but non-zero amplitude

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College Physics Quiz

College Physics Quiz: Periodic Waves

Practice Periodic Waves in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Periodic Waves, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Constructive interference with amplitude doubled (correct answer)
Destructive interference with zero amplitude
Constructive interference with amplitude less than doubled
Destructive interference with reduced but non-zero amplitude

Explanation: 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.

Question 2

A sinusoidal wave traveling along a string has a frequency of 60 Hz and a wavelength of 0.25 m. If the amplitude is doubled while keeping the frequency constant, what happens to the wave speed?

The wave speed doubles because increased amplitude means more energy per unit time
The wave speed decreases by half because the medium responds more slowly to larger displacements
The wave speed remains unchanged because it depends only on the medium properties and frequency (correct answer)
The wave speed increases by a factor of $\sqrt{2}$ due to the relationship between amplitude and propagation
The wave speed becomes zero because the medium cannot support the increased amplitude at this frequency

Explanation: When you encounter wave problems, remember that wave speed depends fundamentally on the properties of the medium through which the wave travels, not on the wave's amplitude or energy content. The wave speed formula is

v = f\lambda

, where

v

is speed,

f

is frequency, and

\lambda

is wavelength. In this problem, the frequency remains constant at 60 Hz and the wavelength stays at 0.25 m, so the wave speed remains

v = 60 \times 0.25 = 15

m/s regardless of amplitude changes. Additionally, for waves on strings, the speed depends only on the string's tension and linear mass density:

v = \sqrt{T/\mu}

. Since doubling the amplitude doesn't change these physical properties of the string, the wave speed stays constant. Option A incorrectly assumes that higher energy (from increased amplitude) affects propagation speed. While larger amplitude does mean more energy, this energy manifests as greater particle displacement, not faster wave propagation. Option B suggests the medium responds more slowly to larger displacements, but this isn't how wave propagation works—the restoring forces scale proportionally with displacement in the linear regime. Option D invents a mathematical relationship between amplitude and speed that doesn't exist in wave physics. The correct answer is C because wave speed is determined solely by medium properties, not by amplitude, energy, or intensity. Remember this key principle: amplitude affects how much the medium oscillates, but the speed at which disturbances propagate depends only on the medium's physical characteristics.

Question 3

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?

The waves are exactly in phase, and this represents a moment when both waves are at equilibrium
The waves are exactly out of phase, resulting in complete destructive interference at this instant
The waves have a phase difference of $\pi/4$ , creating partial interference that momentarily cancels
The phase relationship cannot be determined from a single measurement at one instant in time (correct answer)
The waves are in quadrature phase, with one wave at maximum while the other is at equilibrium

Explanation: 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.

Question 4

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?

$-0.02$ m, because $\sin(\pi - 6\pi) = \sin(-5\pi) = 0$ and the amplitude factor gives the result
$0$ m, because $\sin(\pi - 6\pi) = \sin(-5\pi) = 0$ due to the sine function periodicity (correct answer)
$0.02$ m, because $\sin(\pi - 6\pi) = \sin(-5\pi) = \sin(\pi) = 0$ but the wave is at maximum
$0.01$ m, because $\sin(\pi - 6\pi) = \sin(-5\pi) = -1/2$ and this gives half the maximum displacement
$-0.01$ m, because $\sin(\pi - 6\pi) = \sin(-5\pi) = 1/2$ but the wave equation includes a phase shift

Explanation: 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)

-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

Question 5

Two sound waves of frequencies 440 Hz and 444 Hz interfere in air. An observer hears beats with a frequency closest to:

2 Hz, because beats occur at half the difference frequency to account for the alternating constructive interference
4 Hz, because the beat frequency equals the absolute difference between the two source frequencies (correct answer)
8 Hz, because beats occur at twice the difference frequency due to both constructive and destructive cycles
442 Hz, because beats occur at the average frequency of the two interfering waves
884 Hz, because beats occur at the sum frequency when two waves interfere in the same medium

Explanation: When two sound waves of slightly different frequencies interfere, they create a phenomenon called beats - a periodic variation in loudness that you hear as a pulsing sound. This occurs because the waves alternately reinforce and cancel each other as they drift in and out of phase. The beat frequency is simply the absolute difference between the two interfering frequencies. With waves at 440 Hz and 444 Hz, the calculation is straightforward:

|444 - 440| = 4

Hz. This means you'll hear the sound getting louder and softer 4 times per second. Choice B correctly identifies this fundamental relationship - the beat frequency equals the absolute difference between source frequencies. Choice A incorrectly suggests beats occur at half the difference frequency. This misconception might arise from confusing beats with other wave phenomena, but beats directly reflect the full frequency difference. Choice C claims beats occur at twice the difference frequency. While it's true that both constructive and destructive interference occur during each beat cycle, the beat frequency still equals the difference frequency, not double it. The "twice" factor is already accounted for in how we perceive the complete interference pattern. Choice D confuses beat frequency with the average frequency of the interfering waves. The average frequency (442 Hz) determines the pitch you hear, while the beat frequency (4 Hz) determines how fast the volume oscillates. Remember: beat frequency always equals the difference between interfering frequencies. This is one of the most reliable formulas in wave physics - no factors of 2 or averaging needed.

Question 6

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?

$0.30$ m/s, because the maximum particle speed occurs when displacement equals amplitude
$1.5$ m/s, because the maximum particle speed is amplitude times frequency times $2\pi$
$9.4$ m/s, because the maximum particle speed is amplitude times angular frequency (correct answer)
$15$ m/s, because the maximum particle speed is amplitude divided by period times $2\pi$
$150$ m/s, because the maximum particle speed is amplitude times frequency squared times $2\pi$

Explanation: 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.

Question 7

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?

60 Hz, because 3 wavelengths fit in the string length and $f = v/\lambda$
90 Hz, because the effective wavelength accounts for the standing wave pattern
180 Hz, because 3 wavelengths fit in the string length and $f = v/\lambda$ (correct answer)
240 Hz, because the standing wave frequency is twice the traveling wave frequency
360 Hz, because standing waves have six half-wavelengths and $f = v/\lambda$

Explanation: 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.

Question 8

A periodic wave has the equation $y(x,t) = 0.05\cos(2x - 8t)$ where distances are in meters and time is in seconds. The wave is traveling in the:

positive $x$ direction with speed 2 m/s, because the spatial term is positive
positive $x$ direction with speed 4 m/s, because $v = \omega/k = 8/2$ (correct answer)
negative $x$ direction with speed 2 m/s, because the time term has a negative sign
negative $x$ direction with speed 4 m/s, because the wave equation has the form $\cos(kx + \omega t)$
positive $x$ direction with speed 8 m/s, because the temporal frequency term determines the wave speed

Explanation: When you encounter a wave equation, you need to identify two key components: the direction of travel and the wave speed. The standard form for a traveling wave is

y(x,t) = A\cos(kx \pm \omega t)

, where the sign between the spatial and temporal terms determines direction. For the given equation

y(x,t) = 0.05\cos(2x - 8t)

, you can identify

k = 2

rad/m and

\omega = 8

rad/s. The minus sign between

kx

and

\omega t

indicates the wave travels in the positive

x

direction. To find the wave speed, use the fundamental relationship

v = \omega/k = 8/2 = 4

m/s. Looking at the incorrect options: Choice A correctly identifies the positive direction but miscalculates the speed as 2 m/s, likely confusing the wave number

k

with the wave speed. Choice C misinterprets the direction—the negative sign doesn't mean negative direction; rather,

\cos(kx - \omega t)

represents positive direction travel. The wave moves in the direction that keeps the phase constant, which means as

t

increases,

x

must increase when there's a minus sign. Choice D gets the speed calculation right but incorrectly states the wave moves in the negative direction and wrongly claims the equation has the form

\cos(kx + \omega t)

. Study tip: Remember the wave equation pattern:

\cos(kx - \omega t)

means positive

x

direction,

\cos(kx + \omega t)

means negative

x

direction. Always calculate speed using

v = \omega/k

, not just the individual coefficients.

Question 9

In a standing wave on a string, nodes are points where:

the displacement is always zero, but the particle velocity alternates between maximum values
the displacement is always maximum, and the particle velocity is always zero throughout the oscillation
the displacement is always zero, and the particle velocity is always zero throughout the oscillation (correct answer)
the displacement alternates between maximum values, but the particle velocity is always zero
both displacement and particle velocity alternate between their respective maximum values

Explanation: 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.

Question 10

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?

$x = 3.5$ m, because the pulse travels 1.5 m in 1.5 seconds at 5.0 m/s
$x = 7.5$ m, because the pulse travels 5.0 m in each second for 1.5 seconds
$x = 9.5$ m, because the pulse travels from its initial position plus the distance covered (correct answer)
$x = 10.0$ m, because the pulse travels a total distance equal to speed times time
$x = 0.5$ m, because the pulse reflects from a boundary and travels in the negative direction

Explanation: 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.

Question 11

Two identical speakers separated by 4.0 m emit sound waves in phase at 340 Hz. The speed of sound is 340 m/s. At a point on the perpendicular bisector 3.0 m from the line connecting the speakers, the interference will be:

constructive, because the path difference is zero and the speakers emit waves in phase (correct answer)
destructive, because the path difference equals one-half wavelength due to the geometry
constructive, because the path difference equals one wavelength and the speakers emit waves in phase
destructive, because the speakers are too far apart for coherent interference at this frequency
partially constructive, because the path difference is one-quarter wavelength creating intermediate interference

Explanation: When you encounter wave interference problems, you need to analyze the path difference between waves from two sources and determine whether they arrive in phase or out of phase at the observation point. First, let's find the path difference. The speakers are 4.0 m apart, and the observation point is 3.0 m from the midpoint on the perpendicular bisector. Using the Pythagorean theorem, the distance from each speaker to the observation point is

\sqrt{(2.0)^2 + (3.0)^2} = \sqrt{13} = 3.61

m. Since both speakers are equidistant from the observation point, the path difference is exactly zero. Next, determine the interference type. With zero path difference and waves emitted in phase, the waves arrive at the observation point perfectly in phase, creating constructive interference. Let's examine each answer choice. Choice A correctly identifies constructive interference due to zero path difference with in-phase sources. Choice B incorrectly claims the path difference equals half a wavelength - this would require a path difference of

\lambda/2 = 340/(2 \times 340) = 0.5

m, which doesn't match our geometry. Choice C mentions constructive interference but wrongly states the path difference equals one wavelength (1.0 m), which isn't the case here. Choice D incorrectly suggests the speakers are too far apart for coherent interference - speaker separation doesn't prevent coherence as long as they're driven by the same source. Remember: on the perpendicular bisector of two identical sources, the path difference is always zero, making this a reliable shortcut for similar problems.

Question 12

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:

increases by 44% because wave speed is directly proportional to tension in the rope equation
increases by 22% because wave speed is proportional to the square root of tension
increases by 20% because wave speed is proportional to the square root of tension (correct answer)
increases by 88% because the tension change affects both the numerator and wave propagation efficiency
decreases by 20% because increased tension reduces the rope's ability to support transverse motion

Explanation: 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.

Question 13

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:

one-third of the original intensity, because intensity and amplitude are directly proportional in wave mechanics
one-sixth of the original intensity, because the relationship involves both amplitude and frequency factors
one-ninth of the original intensity, because intensity is proportional to the square of amplitude (correct answer)
one-twelfth of the original intensity, because sound waves have additional geometric spreading factors
unchanged, because intensity depends on frequency and wavelength, not amplitude in sound waves

Explanation: 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.

Question 14

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:

the same amplitude and the same orientation, because energy is conserved at the boundary
reduced amplitude and inverted orientation, because the boundary acts like a fixed end (correct answer)
reduced amplitude and the same orientation, because some energy is transmitted to the heavier string
the same amplitude and inverted orientation, because the massive string cannot move significantly
increased amplitude and inverted orientation, because the reflection focuses the wave energy

Explanation: 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.

Question 15

In a longitudinal wave in air, the particles of the medium oscillate:

perpendicular to the direction of wave propagation, creating alternating regions of compression and rarefaction
parallel to the direction of wave propagation, creating alternating regions of compression and rarefaction (correct answer)
in circular paths around the equilibrium position, maintaining constant density throughout the medium
parallel to the direction of wave propagation, creating alternating regions of maximum and minimum displacement
perpendicular to the direction of wave propagation, creating alternating regions of maximum and minimum displacement

Explanation: When you encounter questions about wave types, the key distinction is how particles in the medium move relative to the wave's direction of travel. In longitudinal waves, particles oscillate back and forth along the same axis as the wave propagation. Sound waves in air are the classic example: as the wave travels horizontally through air, air molecules move horizontally back and forth around their equilibrium positions. This creates alternating regions where molecules bunch together (compression) and spread apart (rarefaction). Think of a spring being pushed and pulled - the coils compress and stretch along the spring's length as the disturbance travels down it. Option B correctly describes this behavior: particles move parallel to wave propagation, creating compression and rarefaction regions. Option A incorrectly describes transverse waves, where particles move perpendicular to the wave direction. While it mentions compression and rarefaction (which only occur in longitudinal waves), the perpendicular motion is impossible for sound waves in fluids like air. Option C describes some exotic wave motion that doesn't exist in standard wave physics. Circular motion would require forces not present in typical longitudinal waves, and density definitely varies in sound waves. Option D gets the particle motion right (parallel movement) but incorrectly describes the result as "maximum and minimum displacement" rather than compression and rarefaction. This confuses the description with transverse wave terminology. Study tip: Remember the prefix "longi-" means "along" - longitudinal waves have particle motion along the direction of wave travel, always creating pressure variations in fluids.

Question 16

In a sinusoidal wave, the phase constant determines:

the amplitude of oscillation and the maximum displacement from equilibrium position
the frequency of oscillation and how rapidly the wave pattern repeats in time
the wavelength of the wave and the spatial distance between consecutive peaks
the initial phase of oscillation and where in the cycle the wave starts at $t = 0$ (correct answer)
the wave speed and how quickly the pattern propagates through the medium

Explanation: When analyzing sinusoidal waves, you need to understand how each parameter in the wave equation

y(x,t) = A\sin(kx - \omega t + \phi)

affects the wave's behavior. The phase constant

\phi

specifically controls the wave's initial conditions. The phase constant determines exactly where in its oscillation cycle the wave begins at

t = 0

. If

\phi = 0

, the wave starts at equilibrium moving in the positive direction. If

\phi = \pi/2

, it starts at maximum displacement. This initial phase shift affects the entire wave pattern but doesn't change any other fundamental properties. Let's examine why the other options miss the mark. Choice A incorrectly attributes amplitude control to the phase constant—amplitude is determined by the coefficient

A

in front of the sine function, not by

\phi

. Choice B confuses the phase constant with angular frequency

\omega

, which actually determines how rapidly the wave oscillates in time. Choice C mistakenly connects the phase constant to wavelength, but wavelength depends on the wave number

k

, not the phase constant

\phi

. The correct answer is D because the phase constant literally sets the "phase" or starting point of the oscillation at

t = 0

. Study tip: Remember that in wave equations, each parameter has a distinct role:

A

controls amplitude,

\omega

controls frequency,

k

controls wavelength, and

\phi

controls initial phase. Don't let similar-sounding terms confuse you—focus on what each symbol represents mathematically.

Question 17

Two speakers emit identical sound waves with wavelength 2.0 m. They are positioned 3.0 m apart and a listener stands 4.0 m from one speaker and 5.0 m from the other. The listener will experience:

constructive interference because the path difference of 1.0 m equals half a wavelength
destructive interference because the path difference of 1.0 m equals half a wavelength (correct answer)
constructive interference because the path difference of 1.0 m equals half a wavelength, creating a phase difference of $\pi$
no interference because the speakers are too far apart for coherent wave interaction
destructive interference because the path difference exceeds the coherence length of the sound waves

Explanation: When two coherent sound sources create overlapping waves, you need to analyze the path difference to determine whether interference will be constructive or destructive. The key is comparing this path difference to the wavelength. Here, the path difference is

5.0\text{ m} - 4.0\text{ m} = 1.0\text{ m}

. With a wavelength of

2.0\text{ m}

, this path difference equals

\frac{1.0}{2.0} = 0.5

wavelengths, or exactly half a wavelength. When waves arrive with a path difference of an odd number of half-wavelengths (

\frac{\lambda}{2}, \frac{3\lambda}{2}, \frac{5\lambda}{2}

, etc.), they interfere destructively because one wave's peak aligns with the other's trough. Choice A incorrectly states this leads to constructive interference. Constructive interference occurs when the path difference equals whole wavelengths (

\lambda, 2\lambda, 3\lambda

, etc.), not half-wavelengths. Choice C makes the same error as A, though it correctly identifies that half a wavelength creates a phase difference of

\pi

radians—this phase difference actually confirms destructive interference, not constructive. Choice D is wrong because speaker separation doesn't prevent coherent interference; as long as both waves reach the listener with sufficient amplitude, interference will occur. Remember this pattern: path differences of whole wavelengths give constructive interference (waves in phase), while path differences of odd half-wavelengths give destructive interference (waves out of phase). Always calculate the path difference first, then compare it to the wavelength to predict the interference type.

Question 18

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:

only the wave speeds, because faster pulses create larger interference effects
only the pulse widths, because wider pulses have more energy to contribute to the interference
the relative phase of the pulses, because this determines whether interference is constructive or destructive (correct answer)
only the rope tension, because this affects how the medium responds to the combined disturbance
the frequency content of the pulses, because different frequencies interfere with different amplitudes

Explanation: 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.

Question 19

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?

The wavelength decreases to 0.34 m because the wave slows down in the denser medium
The wavelength increases to 1.5 m because the frequency remains constant but speed increases (correct answer)
The wavelength remains 0.34 m because wavelength is an intrinsic property of the wave source
The wavelength increases to 0.75 m because the average of the two media determines the new wavelength
The wavelength decreases to 0.23 m because the wave must maintain its energy density in water

Explanation: 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.

Question 20

A wave pulse traveling on a string reflects from a free end (where the string can move freely). Compared to the incident pulse, the reflected pulse has:

the same amplitude and the same phase, maintaining all wave properties unchanged during reflection
reduced amplitude and the same phase, due to partial energy absorption at the boundary
the same amplitude and inverted phase, since the free boundary requires zero displacement
reduced amplitude and inverted phase, due to energy loss and boundary constraint effects
the same amplitude and the same phase, since the free end allows unrestricted motion (correct answer)

Explanation: When analyzing wave reflection, you need to consider two key factors: the boundary condition at the reflection point and energy conservation during the process. At a free end, the string can move without constraint, creating a specific boundary condition: the end must be a point of maximum displacement (antinode). This means when a wave pulse reaches the free end, it reflects with the same phase as the incident pulse - no inversion occurs. The free boundary allows the string to "whip" freely, preserving the original direction of displacement. Additionally, assuming an ideal string with no energy losses, the reflected pulse maintains the same amplitude as the incident pulse due to energy conservation. All the wave's energy bounces back since there's nothing at the free end to absorb it. Looking at the wrong answers: Choice A correctly identifies same amplitude and phase but incorrectly suggests this maintains "all wave properties unchanged" - the direction of travel obviously reverses. Choice B correctly notes same phase but incorrectly assumes energy absorption at a free boundary. Choice C makes the critical error of claiming phase inversion, which only occurs at fixed (not free) boundaries, though it correctly identifies amplitude preservation. Choice D combines multiple errors: assuming both energy loss and phase inversion at a free end. Study tip: Remember the boundary rule - free ends preserve phase (no inversion), while fixed ends invert phase. Think of cracking a whip: the free end whips in the same direction as your motion, demonstrating this phase preservation principle.