College Physics Quiz: Properties Of Wave Pulses And Waves

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Properties Of Wave Pulses And WavesQuestion 1 of 20

A wave pulse travels along a string with speed $v = 20 \text{ m/s}$ . If the tension in the string is quadrupled while keeping the linear mass density constant, what is the new wave speed?

10 m/s

20 m/s

40 m/s

80 m/s

160 m/s

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College Physics Quiz

College Physics Quiz: Properties Of Wave Pulses And Waves

Practice Properties Of Wave Pulses And 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 Properties Of Wave Pulses And 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

A wave pulse travels along a string with speed $v = 20 \text{ m/s}$ . If the tension in the string is quadrupled while keeping the linear mass density constant, what is the new wave speed?

10 m/s
20 m/s
40 m/s (correct answer)
80 m/s
160 m/s

Explanation: When you encounter wave speed problems, remember that the speed of a wave on a string depends on the string's physical properties according to the formula

v = \sqrt{\frac{T}{\mu}}

, where

T

is tension and

\mu

is linear mass density (mass per unit length). Starting with the initial speed of 20 m/s, when tension is quadrupled while linear mass density remains constant, you can set up a ratio. If the new tension is

T_{new} = 4T_{old}

, then:

\frac{v_{new}}{v_{old}} = \sqrt{\frac{T_{new}/\mu}{T_{old}/\mu}} = \sqrt{\frac{4T_{old}}{T_{old}}} = \sqrt{4} = 2

Therefore:

v_{new} = 2 \times 20 = 40 \text{ m/s}

Let's examine why the other answers are incorrect. Choice A (10 m/s) represents halving the original speed, which would occur if tension were reduced to one-fourth, not quadrupled. Choice B (20 m/s) suggests no change in wave speed, which would only happen if tension remained constant. Choice D (80 m/s) comes from incorrectly assuming wave speed is directly proportional to tension (multiplying by 4 instead of taking the square root), missing the square root relationship in the wave speed equation. Study tip: Remember that wave speed has a square root relationship with tension. When tension changes by a factor, the wave speed changes by the square root of that factor. This square root relationship appears frequently in physics formulas, so always check whether you need to take the square root of your ratio.

Question 2

Two wave pulses of equal amplitude travel toward each other on the same string. Pulse A has a positive displacement and pulse B has a negative displacement of the same magnitude. When they completely overlap, the displacement of the string is:

twice the amplitude of either pulse in the positive direction
twice the amplitude of either pulse in the negative direction
equal to the amplitude of either pulse in the positive direction
equal to the amplitude of either pulse in the negative direction
zero at all points where the pulses overlap (correct answer)

Explanation: When you encounter wave interference problems, you're dealing with the principle of superposition: when two waves meet, their displacements add algebraically at each point. Here's what happens when these pulses overlap completely: Pulse A creates a positive displacement of magnitude A, while Pulse B creates a negative displacement of magnitude -A (same size, opposite direction). When they're perfectly superimposed, the total displacement at every point equals A + (-A) = 0. The string becomes completely flat during the moment of complete overlap. Let's examine why each option fails. Choice A suggests the displacements would add in the positive direction (A + A = 2A), but this ignores that one pulse is negative. Choice B makes the same error in the negative direction (-A + (-A) = -2A), incorrectly assuming both pulses are negative. Choices C and D both suggest one pulse would dominate over the other, giving a net displacement equal to a single amplitude, but there's no physical reason why one would cancel only part of the other when they have equal magnitudes. This complete cancellation is called destructive interference. After the pulses pass through each other, they continue traveling in their original directions, completely unchanged – the cancellation is only temporary during overlap. Remember: interference problems always come down to careful addition of displacements. Pay close attention to the signs (positive vs. negative) and magnitudes. When equal and opposite waves meet, they always produce complete destructive interference.

Question 3

A sinusoidal wave has frequency $f = 60 \text{ Hz}$ and wavelength $\lambda = 0.5 \text{ m}$ . If the frequency is increased to $120 \text{ Hz}$ while the wave speed remains constant, what is the new wavelength?

0.125 m
0.25 m (correct answer)
0.5 m
1.0 m
2.0 m

Explanation: When you encounter wave problems involving frequency and wavelength changes, the fundamental wave equation is your key tool:

v = f\lambda

, where wave speed equals frequency times wavelength. First, find the initial wave speed using the given values:

v = (60 \text{ Hz})(0.5 \text{ m}) = 30 \text{ m/s}

. Since the problem states that wave speed remains constant when frequency changes, this speed stays at 30 m/s. With the new frequency of 120 Hz and the same wave speed, you can solve for the new wavelength:

30 = (120)\lambda

, so

\lambda = 30/120 = 0.25 \text{ m}

. This confirms answer B is correct. Let's examine why the other options are wrong. Choice A (0.125 m) would result from incorrectly dividing the original wavelength by 4 instead of 2, perhaps confusing the frequency ratio. Choice C (0.5 m) incorrectly assumes the wavelength stays the same despite the frequency doubling—this violates the wave equation since speed is constant. Choice D (1.0 m) suggests the wavelength doubled when frequency doubled, which would actually quadruple the wave speed, contradicting the given constraint. Remember this inverse relationship: when wave speed is constant, frequency and wavelength are inversely proportional. If frequency doubles, wavelength must halve to maintain the same wave speed. This pattern appears frequently in wave physics problems, so always check whether wave speed, frequency, or wavelength is held constant, then apply

v = f\lambda

accordingly.

Question 4

A sound wave travels through air at $20°\text{C}$ with speed $343 \text{ m/s}$ . What happens to the wave speed when the temperature increases to $40°\text{C}$ ?

The wave speed decreases because air density increases with temperature
The wave speed remains constant because it only depends on the medium type
The wave speed increases because sound travels faster in warmer air (correct answer)
The wave speed decreases because molecular motion becomes more random at higher temperatures
The wave speed becomes zero because air expands and loses its ability to transmit waves

Explanation: When you encounter questions about sound wave speed and temperature, focus on how molecular motion affects wave propagation. Sound waves travel by transferring energy through collisions between air molecules, so understanding molecular behavior is key. As temperature increases, air molecules move faster and have more kinetic energy. This increased molecular motion allows sound waves to propagate more efficiently through the medium. The relationship between temperature and sound speed in air follows the equation

v = 331.3\sqrt{\frac{T}{273}}

where

T

is temperature in Kelvin. At

40°C

(313 K) versus

20°C

(293 K), you can calculate that sound speed increases to approximately 355 m/s. Choice A incorrectly suggests that air density increases with temperature, when actually air density decreases as temperature rises (think of hot air balloons). Even if density increased, the net effect would still be faster sound transmission due to increased molecular kinetic energy. Choice B falls into the trap of oversimplification. While the medium type matters, sound speed absolutely depends on the medium's physical properties, including temperature, pressure, and humidity. Choice D correctly identifies that molecular motion increases with temperature but wrongly assumes this makes wave transmission less efficient. Random motion doesn't hinder sound propagation—it enhances it by providing more energetic collisions for wave transmission. Study tip: Remember that faster molecular motion always means faster sound speed in gases. This relationship is fundamental to thermodynamics and appears frequently on physics exams in various contexts.

Question 5

Two coherent sound sources emit waves with the same frequency and amplitude. At a point where the path difference between the two sources is $1.5\lambda$ , the interference is:

constructive, resulting in maximum amplitude
destructive, resulting in zero amplitude (correct answer)
partially constructive, resulting in amplitude between zero and maximum
alternating between constructive and destructive as time progresses
neither constructive nor destructive because the sources are incoherent

Explanation: When two coherent waves meet, the type of interference depends on their phase relationship, which is determined by the path difference between the sources. This is a fundamental concept in wave physics that applies to sound, light, and other wave phenomena. For constructive interference (maximum amplitude), waves must arrive in phase, meaning the path difference must be a whole number of wavelengths:

0, \lambda, 2\lambda, 3\lambda

, etc. For destructive interference (zero amplitude), waves must arrive completely out of phase, requiring a path difference of an odd multiple of half wavelengths:

0.5\lambda, 1.5\lambda, 2.5\lambda

, etc. Since the path difference here is

1.5\lambda = \frac{3\lambda}{2}

, this represents destructive interference. One wave arrives exactly half a wavelength "behind" the other after accounting for the whole wavelength portion, causing the waves to cancel completely and produce zero amplitude. Looking at the wrong answers: (A) incorrectly assumes

1.5\lambda

produces constructive interference—this only occurs at whole wavelength differences. (C) suggests partial interference, but with equal amplitude coherent sources, interference is always either completely constructive or completely destructive, never partial. (D) misunderstands that the interference pattern is determined by the fixed path difference, not changing over time. Study tip: Remember the simple rule: whole wavelengths (

n\lambda

) give constructive interference, while odd half-wavelengths (

\frac{(2n+1)\lambda}{2}

) give destructive interference. When you see 1.5, 2.5, 3.5, etc., think destructive interference immediately.

Question 6

A wave pulse traveling on a rope encounters a boundary where the rope connects to a heavier rope. What happens to the wave speed, wavelength, and frequency in the heavier rope?

Speed decreases, wavelength decreases, frequency remains constant (correct answer)
Speed decreases, wavelength increases, frequency decreases
Speed increases, wavelength increases, frequency remains constant
Speed remains constant, wavelength decreases, frequency increases
Speed increases, wavelength decreases, frequency increases

Explanation: When a wave encounters a boundary between two different media, you need to understand which properties change and which remain constant. This is a fundamental principle of wave transmission across boundaries. The key insight is that frequency always remains constant when a wave crosses a boundary - this is because the source continues oscillating at the same rate. However, wave speed changes based on the properties of the new medium. For mechanical waves on ropes, speed depends on tension and linear density:

v = \sqrt{T/\mu}

. Since the heavier rope has greater linear density (

\mu

), the wave speed decreases when entering it. Since

v = f\lambda

and frequency stays constant while speed decreases, the wavelength must also decrease proportionally to maintain this relationship. Looking at the wrong answers: Option B incorrectly suggests wavelength increases - this would violate the wave equation since both speed and frequency can't both decrease while wavelength increases. Option C wrongly claims speed increases in the denser medium, which contradicts the physics of wave propagation in heavier materials. Option D incorrectly states that speed remains constant, but wave speed definitely changes when the medium's properties change. Option A correctly identifies that speed decreases (due to higher density), wavelength decreases (to compensate for constant frequency), and frequency remains constant (universal principle at boundaries). Study tip: Remember the boundary rule - frequency never changes across boundaries, but speed changes based on medium properties. Then use

v = f\lambda

to determine what happens to wavelength.

Question 7

A sinusoidal wave is described by $y(x,t) = A \sin(kx - \omega t + \phi)$ . If $k = 2.0 \text{ rad/m}$ and $\omega = 10 \text{ rad/s}$ , what is the wavelength?

$\pi \text{ m}$ (correct answer)
$2\pi \text{ m}$
$0.5\pi \text{ m}$
$5 \text{ m}$
$0.2 \text{ m}$

Explanation: When you encounter a sinusoidal wave equation, you're working with the fundamental relationship between wave parameters. The general form

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

contains the wave number

k

, which directly determines the wavelength. The wave number

k

and wavelength

\lambda

are related by

k = \frac{2\pi}{\lambda}

. This means that

\lambda = \frac{2\pi}{k}

. With

k = 2.0 \text{ rad/m}

, you get:

\lambda = \frac{2\pi}{2.0} = \pi \text{ m}

This confirms that answer A is correct. Let's examine why the other options are wrong. Answer B (

2\pi \text{ m}

) would result if you mistakenly used

\lambda = \frac{4\pi}{k}

or confused the relationship entirely. Answer C (

0.5\pi \text{ m}

) comes from incorrectly using

\lambda = \frac{\pi}{k}

, missing the factor of 2 in the correct formula. Answer D (

5 \text{ m}

) might tempt you if you tried to use the frequency relationship

\lambda = \frac{v}{f}

without knowing the wave speed, or if you somehow divided

\omega

k

directly. Remember this key strategy: wave number problems always use

k = \frac{2\pi}{\lambda}

. The

2\pi

factor is crucial and appears because one complete wavelength corresponds to

2\pi

radians of phase change. Don't confuse this with frequency relationships or try to use

\omega

directly for wavelength calculations.

Question 8

A rope has a discontinuity where a light section connects to a heavy section. A wave pulse traveling from the light section toward the heavy section will:

be completely transmitted with no reflection and maintain its original amplitude
be completely reflected with no transmission and maintain its original amplitude
be partially transmitted and partially reflected, with the reflected pulse inverted (correct answer)
be partially transmitted and partially reflected, with the reflected pulse not inverted
be completely absorbed at the boundary with no transmission or reflection

Explanation: When waves encounter a boundary between different media, you're dealing with wave transmission and reflection at an interface. This is a fundamental concept in wave physics that applies to all types of waves - sound, light, and mechanical waves like those on ropes. When a wave pulse travels from a lighter rope section to a heavier section, the boundary creates an impedance mismatch. The heavier rope has greater linear mass density, which affects how waves propagate through it. At this interface, some of the wave's energy continues forward (transmission) while some bounces back (reflection). This partial transmission and reflection occurs because the wave speeds differ in the two media - waves travel slower in the denser, heavier section. The key insight is the phase relationship of the reflected wave. When a wave reflects off a boundary where it encounters a denser medium (light to heavy rope), the reflected pulse undergoes a phase inversion - it flips upside down. This happens because the boundary acts like a "fixed end" relative to the incoming wave. Answer A is wrong because complete transmission with no reflection only occurs when media have identical properties. Answer B is incorrect because complete reflection only happens at perfect boundaries like a truly fixed end. Answer D makes the common error of forgetting about phase inversion - while it correctly identifies partial transmission and reflection, it misses that the reflected pulse inverts when going from less dense to more dense media. Remember this pattern: light-to-heavy boundaries always produce inverted reflections, while heavy-to-light boundaries produce non-inverted reflections.

Question 9

A wave pulse on a string has width $w$ and travels with speed $v$ . The time required for the pulse to completely pass a fixed point is:

$w/v$ (correct answer)
$v/w$
$2w/v$
$w \cdot v$
$\sqrt{w/v}$

Explanation: When analyzing wave motion, you need to think about the relationship between distance, speed, and time. This question tests your understanding of how long it takes for an entire wave pulse to move past a stationary observer. To find the time for the pulse to completely pass a fixed point, consider what "completely pass" means. The pulse has a width

w

, so for it to entirely move past the observation point, every part of the pulse must travel a distance equal to its width. Since the pulse moves at speed

v

, you can use the fundamental relationship: time = distance/speed. Here, the relevant distance is the pulse width

w

, giving you

t = w/v

. This makes answer A correct. Let's examine why the other options are wrong. Answer B (

v/w

) inverts the relationship, giving you speed divided by distance, which yields units of inverse time (1/s) rather than time. This is a common algebraic error when students confuse which variable goes in the numerator. Answer C (

2w/v

) suggests the pulse needs to travel twice its width, which would only be true if you incorrectly thought about the pulse needing to travel past and then return, or made an error in visualizing the geometry. Answer D (

w \cdot v

) multiplies width and speed, giving units of distance²/time, which is dimensionally incorrect for a time measurement. Remember: for wave problems involving "time to pass," always identify the relevant distance (often the wave's physical extent) and divide by the wave's speed. Check your units to catch common mistakes.

Question 10

A standing wave on a string has nodes at $x = 0, 0.5, 1.0, 1.5 \text{ m}$ . What is the wavelength of the traveling waves that form this standing wave pattern?

0.5 m
1.0 m (correct answer)
1.5 m
2.0 m
3.0 m

Explanation: Standing waves form when two identical waves traveling in opposite directions interfere with each other. The key insight is understanding the relationship between node spacing and wavelength in the resulting pattern. When you examine the given node positions at

x = 0, 0.5, 1.0, 1.5

m, notice that consecutive nodes are separated by exactly 0.5 m. This consistent spacing is crucial because in any standing wave pattern, the distance between adjacent nodes equals half a wavelength (

\lambda/2

). This happens because nodes occur where the two traveling waves destructively interfere completely, and these points of total cancellation are spaced

\lambda/2

apart. Since the node separation is 0.5 m, we can write:

\lambda/2 = 0.5

m, which gives us

\lambda = 1.0

m. This confirms answer choice B. Let's examine why the other options are incorrect. Choice A (0.5 m) represents a common misconception where students confuse the node spacing with the actual wavelength—remember, node spacing is only half the wavelength. Choice C (1.5 m) might tempt students who incorrectly think about the distance spanning three node intervals, but this doesn't correspond to any meaningful wave relationship. Choice D (2.0 m) could arise from doubling the wavelength instead of halving the node spacing. For standing wave problems, always remember this key relationship: node spacing = λ/2. When you see evenly spaced nodes, measure the distance between adjacent ones and double it to find the wavelength of the constituent traveling waves.

Question 11

When two wave pulses of different shapes but equal amplitudes meet and overlap completely, the principle that determines the resulting displacement is:

The pulses maintain their individual identities and do not affect each other
The larger pulse dominates and the smaller pulse is absorbed completely
The displacement at each point equals the sum of the individual displacements (correct answer)
The displacement at each point equals the product of the individual displacements
The displacement oscillates between the two pulse shapes as time progresses

Explanation: When you encounter wave interference problems, you're dealing with one of the most fundamental principles in wave physics: superposition. This principle governs what happens when two or more waves occupy the same space simultaneously. The principle of superposition states that when waves overlap, the displacement at any point equals the algebraic sum of the individual displacements from each wave. This means you simply add the amplitudes at each point, taking into account their directions (positive or negative). So if one pulse causes a +3 unit displacement and another causes a -2 unit displacement at the same point, the resulting displacement is +1 unit. Looking at the incorrect options: Choice A is wrong because waves do interact when they overlap—they don't just pass through each other without effect during the overlap period. Choice B incorrectly suggests that wave amplitude determines dominance, but superposition applies regardless of relative sizes when amplitudes are equal as stated in the problem. Choice D represents a common misconception—waves add their displacements, they don't multiply them. Multiplication would create completely different physics and wouldn't conserve energy properly. The correct answer is C because superposition requires linear addition of displacements at each point in space. Study tip: Remember "superposition = super-position = adding positions." When you see any wave interference problem, immediately think addition of displacements. This principle applies whether you're dealing with sound waves, light waves, or waves on strings—it's universal for linear wave systems.

Question 12

A longitudinal wave travels through a spring. The wavelength is $\lambda = 0.8 \text{ m}$ and the frequency is $f = 5 \text{ Hz}$ . At a certain instant, a coil has maximum compression. The nearest coil that has maximum extension (rarefaction) is located at a distance of:

0.2 m
0.4 m (correct answer)
0.8 m
1.2 m
1.6 m

Explanation: When analyzing longitudinal waves, you need to understand the relationship between compressions, rarefactions, and wavelength. In a longitudinal wave, compressions (maximum density) and rarefactions (minimum density/maximum extension) are the wave's extreme points, analogous to peaks and troughs in transverse waves. The key insight is that compressions and rarefactions are separated by half a wavelength (

\lambda/2

). This is because they represent opposite phases of the wave motion - when one point has maximum compression, the point half a wavelength away has maximum rarefaction. Given

\lambda = 0.8 \text{ m}

, the distance between maximum compression and the nearest maximum rarefaction is:

\frac{\lambda}{2} = \frac{0.8 \text{ m}}{2} = 0.4 \text{ m}

Answer B (0.4 m) is correct because it represents this half-wavelength separation. Answer A (0.2 m) represents

\lambda/4

, which would be the distance to a point with zero displacement, not maximum rarefaction. Answer C (0.8 m) represents one full wavelength, which would bring you back to another compression point, not a rarefaction. Answer D (1.2 m) represents

1.5\lambda

, which would be the distance to the second nearest rarefaction (after going through one complete wavelength plus another half wavelength). Remember this pattern: compressions and rarefactions in longitudinal waves are always separated by

\lambda/2

, just like peaks and troughs in transverse waves. The frequency information given here is extra data - wavelength alone determines the spatial relationships.

Question 13

Two identical waves traveling in opposite directions on a string create a standing wave pattern. If the original traveling waves each have amplitude $A$ , what is the maximum amplitude of the standing wave?

$A/2$
$A$
$A\sqrt{2}$
$2A$ (correct answer)
$4A$

Explanation: When two waves traveling in opposite directions interfere, they create a standing wave pattern through superposition. The key insight is understanding how wave amplitudes add when waves are perfectly in phase. For two identical waves with amplitude

A

traveling in opposite directions, we can write them as

y_1 = A\sin(kx - \omega t)

and

y_2 = A\sin(kx + \omega t)

. Using the trigonometric identity for the sum of sines, their superposition becomes:

y = y_1 + y_2 = 2A\sin(kx)\cos(\omega t)

This equation reveals that the standing wave has an amplitude of

2A\sin(kx)

, which varies with position. At antinodes (where

\sin(kx) = \pm 1

), the amplitude reaches its maximum value of

2A

. This occurs because at these points, both traveling waves are always perfectly in phase, so their amplitudes add constructively to give the maximum possible reinforcement. Choice A (

A/2

) incorrectly suggests the waves partially cancel rather than reinforce. Choice B (

A

) represents no change from the original amplitude, missing the constructive interference effect entirely. Choice C (

A\sqrt{2}

) might result from incorrectly applying root-mean-square calculations, which don't apply to direct wave superposition. The correct answer is D (

2A

). Remember: in standing wave problems, maximum amplitude always occurs at antinodes where constructive interference is complete. For identical waves, this means the amplitudes simply add:

A + A = 2A

. This doubling effect is a hallmark of perfect constructive interference.

Question 14

A wave equation is given by $y(x,t) = 0.05 \sin(4x - 20t)$ where $x$ is in meters, $t$ is in seconds, and $y$ is in meters. What is the speed of this wave?

2.5 m/s
5.0 m/s (correct answer)
15 m/s
20 m/s
80 m/s

Explanation: When you encounter a wave equation in the standard form

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

, you're looking at a sinusoidal wave where the wave speed can be found using the relationship

v = \frac{\omega}{k}

. In your given equation

y(x,t) = 0.05 \sin(4x - 20t)

, you can identify the wave parameters by comparing to the standard form. The coefficient of

x

gives you the wave number

k = 4 \text{ rad/m}

, and the coefficient of

t

gives you the angular frequency

\omega = 20 \text{ rad/s}

. Therefore, the wave speed is

v = \frac{\omega}{k} = \frac{20}{4} = 5.0 \text{ m/s}

, which is answer B. Let's examine why the other answers are incorrect. Answer A (2.5 m/s) would result if you mistakenly calculated

\frac{k}{\omega} = \frac{4}{20}

- you've inverted the correct formula. Answer C (15 m/s) might come from incorrectly subtracting the coefficients:

20 - 4 - 1 = 15

, which has no physical basis. Answer D (20 m/s) represents just the angular frequency

\omega

, not the wave speed - this confuses the temporal oscillation rate with spatial propagation speed. Remember this key relationship: wave speed always equals

\frac{\omega}{k}

for sinusoidal waves. The angular frequency tells you how fast the wave oscillates in time, while the wave number tells you how many wavelengths fit in

2\pi

meters. Their ratio gives the physical speed of wave propagation.

Question 15

A standing wave is formed on a string fixed at both ends. The string length is $L = 2.0 \text{ m}$ and the wave speed is $v = 400 \text{ m/s}$ . What is the frequency of the third harmonic?

100 Hz
200 Hz
300 Hz (correct answer)
400 Hz
600 Hz

Explanation: When you encounter standing wave problems with strings fixed at both ends, you're dealing with harmonic frequencies where the string length must accommodate complete wave patterns with nodes at both endpoints. For a string fixed at both ends, the wavelength of the nth harmonic follows the relationship

\lambda_n = \frac{2L}{n}

, where L is the string length and n is the harmonic number. For the third harmonic (n = 3), the wavelength becomes

\lambda_3 = \frac{2(2.0)}{3} = \frac{4}{3} \text{ m}

. Using the wave equation

f = \frac{v}{\lambda}

, the frequency of the third harmonic is

f_3 = \frac{400}{\frac{4}{3}} = 400 \times \frac{3}{4} = 300 \text{ Hz}

. This confirms answer C is correct. Looking at the wrong answers: A) 100 Hz would be the first harmonic (fundamental frequency), calculated as

f_1 = \frac{400}{4} = 100 \text{ Hz}

. B) 200 Hz represents the second harmonic, where

f_2 = \frac{400}{2} = 200 \text{ Hz}

. D) 400 Hz might tempt you if you mistakenly think the wave speed equals the frequency, but this ignores the wavelength relationship entirely. Remember that for strings fixed at both ends, harmonic frequencies follow the pattern

f_n = n \times f_1

, where the fundamental frequency

f_1 = \frac{v}{2L}

. Always identify which harmonic the question asks for, then either use this multiplication rule or calculate the wavelength directly. The harmonic number tells you how many half-wavelengths fit in the string length.

Question 16

The intensity of a sound wave is proportional to the square of which wave property?

The wavelength of the sound wave
The frequency of the sound wave
The amplitude of the sound wave (correct answer)
The period of the sound wave
The wave speed of the sound wave

Explanation: When you encounter questions about sound wave intensity, remember that intensity measures how much energy passes through a given area per unit time. The key insight is understanding how wave energy relates to the wave's physical properties. Sound wave intensity is proportional to the square of the amplitude. This relationship comes from the fact that wave energy depends on how far particles are displaced from their equilibrium position (the amplitude), and energy scales with the square of displacement. Think of it this way: if you double the amplitude of a sound wave, you quadruple its intensity, making it much louder. Looking at why the other options are incorrect: Option A (wavelength) is wrong because intensity doesn't depend on wavelength in this direct mathematical relationship. You can have high-intensity sound at any wavelength. Option B (frequency) is also incorrect - while frequency affects pitch, intensity is independent of frequency for a given amplitude. A low-frequency bass note and high-frequency treble note can have identical intensities if their amplitudes are the same. Option D (period) is simply the inverse of frequency, so it suffers from the same issue as option B. The correct answer is C because the mathematical relationship is

I \propto A^2

, where I is intensity and A is amplitude. Remember this pattern: when you see questions about wave intensity or energy, look for amplitude-related answers. The square relationship between intensity and amplitude is fundamental across all types of waves, not just sound waves. This same principle applies to electromagnetic waves, water waves, and other wave phenomena you'll encounter in physics.

Question 17

When a sound wave passes from air into water, which property changes the most significantly?

The frequency increases by a factor of approximately 4
The wavelength increases by a factor of approximately 4 (correct answer)
The amplitude increases by a factor of approximately 4
The period decreases by a factor of approximately 4
The phase velocity decreases by a factor of approximately 4

Explanation: When sound waves cross boundaries between different media, you need to understand which wave properties are conserved and which change based on the medium's characteristics. The key principle is that frequency remains constant when a wave crosses into a new medium - this is determined by the source, not the medium. However, since wave speed changes dramatically between air and water, and we know that

v = f\lambda

, the wavelength must adjust to maintain this relationship. Sound travels much faster in water (about 1500 m/s) than in air (about 340 m/s) - roughly 4.4 times faster. Since frequency stays the same but velocity increases by approximately a factor of 4, the wavelength must also increase by that same factor to satisfy

v = f\lambda

. This makes B correct. Looking at the wrong answers: A incorrectly suggests frequency changes, but frequency is always preserved across boundaries - it's an intrinsic property of the source. C about amplitude is wrong because amplitude depends on energy transfer and reflection coefficients, not the 4:1 speed ratio, and typically decreases due to impedance mismatch. D claims period decreases, but since period equals

1/f

and frequency doesn't change, the period also remains constant. Study tip: Remember the wave boundary rule: frequency (and therefore period) never changes when crossing media boundaries, but wavelength and speed change together to maintain

v = f\lambda

. When you see "factor of 4" in wave problems, think about the speed difference between the two media.

Question 18

A wave pulse reflects from a fixed end of a string. Compared to the incident pulse, the reflected pulse has:

the same amplitude and the same phase orientation
the same amplitude but opposite phase orientation (correct answer)
twice the amplitude and the same phase orientation
half the amplitude and opposite phase orientation
twice the amplitude but opposite phase orientation

Explanation: When a wave pulse encounters a boundary, the type of boundary determines how it reflects. This question tests your understanding of reflection from fixed versus free boundaries, which is fundamental to wave behavior on strings. At a fixed end, the string cannot move vertically at that point. This creates a constraint that fundamentally affects the reflected wave. When the incident pulse arrives, it tries to displace the end of the string, but the fixed boundary prevents this motion. The only way to satisfy this boundary condition is for the reflected pulse to have the opposite phase orientation (inverted) compared to the incident pulse. However, energy conservation requires that the amplitude remains the same - no energy is lost in this idealized reflection. Looking at the wrong answers: Choice A incorrectly suggests the phase stays the same, which would occur at a free end, not a fixed end. At a free end, the string can move freely, so the reflected pulse maintains its original orientation. Choice C doubles the amplitude while keeping the same phase - this doesn't occur in wave reflection from any boundary. Choice D halves the amplitude, which would only happen if energy were being absorbed or transmitted, but at a fixed end, all the wave energy is reflected back. The key insight is that fixed ends always invert the phase (create a 180° phase shift) while preserving amplitude, whereas free ends preserve phase while still preserving amplitude. Study tip: Remember "fixed flips" - a fixed boundary always flips the phase orientation of a reflected wave pulse, while a free boundary preserves it.

Question 19

A sinusoidal wave on a string has frequency $f = 60 \text{ Hz}$ and wavelength $\lambda = 0.8 \text{ m}$ . The string tension is then increased by a factor of 4 while keeping the frequency constant. What is the new wavelength?

$0.4 \text{ m}$ because wavelength decreases when tension increases
$0.8 \text{ m}$ because wavelength depends only on frequency
$1.6 \text{ m}$ because wave speed doubles when tension quadruples (correct answer)
$3.2 \text{ m}$ because wavelength increases proportionally with tension

Explanation: The wave speed on a string is

v = \sqrt{T/\mu}

, where T is tension and μ is linear mass density. When tension increases by a factor of 4, the wave speed increases by a factor of

\sqrt{4} = 2

. The original wave speed was

v_1 = f\lambda = 60 \times 0.8 = 48 \text{ m/s}

. The new wave speed is

v_2 = 2 \times 48 = 96 \text{ m/s}

. Since frequency remains constant at 60 Hz, the new wavelength is

\lambda_2 = v_2/f = 96/60 = 1.6 \text{ m}

. Choice A incorrectly assumes wavelength decreases. Choice B ignores the change in wave speed. Choice D uses incorrect proportionality.

Question 20

A standing wave is established on a string of length $L = 1.2 \text{ m}$ fixed at both ends. The wave has frequency $f = 440 \text{ Hz}$ and corresponds to the third harmonic. If the string tension is reduced by 25% while maintaining the same harmonic mode, what is the new frequency?

$330 \text{ Hz}$ because frequency decreases proportionally with tension
$381 \text{ Hz}$ because frequency is proportional to the square root of tension (correct answer)
$495 \text{ Hz}$ because reduced tension increases the effective wavelength
$440 \text{ Hz}$ because harmonic frequencies depend only on string length

Explanation: For a string fixed at both ends, the frequency of the nth harmonic is

f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}

. Since the harmonic number n and string length L remain constant, frequency is proportional to

\sqrt{T}

. When tension is reduced by 25%, the new tension is

T' = 0.75T

. Therefore, the new frequency is

f' = f\sqrt{\frac{T'}{T}} = 440\sqrt{0.75} = 440 \times 0.866 = 381 \text{ Hz}

. Choice A incorrectly assumes direct proportionality. Choice C incorrectly suggests frequency increases. Choice D ignores the tension dependence entirely.