AP Physics 1 Unit 7: Waves and Sound — Free Review Games.

Frequency and period are reciprocals: f = 1/T.

Sound waves are longitudinal mechanical waves where particles vibrate parallel to the direction of wave travel.

Loudness (volume) is determined by the amplitude of a sound wave; greater amplitude means louder sound.

The principle of superposition states that overlapping waves combine by algebraically adding their displacements at each point.

A node is a point on a standing wave that remains stationary (zero displacement) due to destructive interference.

v = f x wavelength = 500 x 0.68 = 340 m/s.

Destructive interference occurs when a crest meets a trough, reducing the resultant amplitude.

The fundamental (first harmonic) is the lowest resonant frequency, with the string vibrating in one segment (one antinode).

Harmonics of a fixed-end string are integer multiples: f_n = n*f_1. Third harmonic = 3(200) = 600 Hz.

The Doppler effect causes the observed frequency to increase when source and observer approach and decrease when they separate.

For an open pipe, wavelength_1 = 2L = 1.7 m. f = v/wavelength = 340/1.7 = 200 Hz.

A closed pipe supports only odd harmonics. Fundamental: wavelength = 4L = 1.7 m, f_1 = 340/1.7 = 200 Hz. Harmonics: 200, 600, 1000 Hz...

At a point equidistant from both sources, the path difference is zero, so waves arrive in phase creating constructive interference.

Beat frequency = |f1 - f2| = |440 - 444| = 4 Hz (beats per second).

4 antinodes means the 4th harmonic: L = 4(wavelength/2), so 2 = 2*wavelength, wavelength = 1 m.

Wavelength is defined as the spatial distance between two successive points that are in the same phase of oscillation, such as two adjacent crests. The distractor "The maximum displacement from equilibrium" describes amplitude, not wavelength, since amplitude measures the size of oscillation rather than the repeat distance. Students should keep wavelength (spatial period), amplitude (size), and period (time) as three distinct wave descriptors.

Amplitude is defined as the maximum displacement of a particle in the medium from its rest position, and it directly determines the energy carried by the wave. The distractor "The distance between two crests" instead describes wavelength, a spatial repeat length rather than a measure of displacement size. Remember that larger amplitude means more energy transported by the wave, which is important for sound intensity and wave energy questions.

Sound waves are longitudinal because air molecules oscillate back and forth parallel to the direction the wave travels, creating regions of compression and rarefaction. The distractor "Transverse" is incorrect because transverse waves have particle motion perpendicular to the direction of travel, like waves on a string, not like sound in air. Recognizing longitudinal versus transverse motion helps distinguish sound waves from light or wave-on-a-string scenarios on the exam.

The universal wave equation \(v = f\lambda\) holds because the wave travels one wavelength during each period, so multiplying the number of cycles per second by the length of each cycle gives the speed. The distractor \(v = f/\lambda\) is dimensionally inconsistent with speed and does not correspond to the physical relationship between these quantities. This equation is foundational and should be memorized for any wave problem involving speed, frequency, or wavelength.

Wave speed on a string follows \(v = \sqrt{T/\mu}\), so increasing tension \(T\) while keeping linear mass density \(\mu\) fixed increases the speed under the square root. The distractor "stays the same" ignores that \(T\) appears directly in the formula and would only be true if \(\mu\) also increased proportionally. Students should remember that wave speed on a string depends only on tension and mass per unit length, not on frequency or amplitude.

Frequency measures cycles completed per second and is measured in hertz, where one hertz equals one cycle per second. The distractor "Seconds (s)" is actually the unit of period, the time for one cycle, which is the reciprocal of frequency. Keeping straight that frequency (Hz) and period (s) are inverses of each other avoids common unit-conversion mistakes on the exam.

Pitch corresponds to how the human ear interprets frequency, with higher frequency sound waves perceived as higher pitch. The distractor "Amplitude" instead determines loudness, not pitch, since amplitude relates to the energy or intensity of the wave rather than its oscillation rate. This distinction between pitch (frequency) and loudness (amplitude/intensity) is a key concept tested repeatedly in sound problems.

Resonance occurs when an external driving force matches a system's natural frequency, allowing energy to build up efficiently and producing a dramatic increase in oscillation amplitude. The distractor "When two waves cancel each other out completely" instead describes destructive interference, a separate phenomenon involving superposition rather than driven oscillation. Recognizing resonance is essential for understanding why musical instruments and standing waves favor specific frequencies.

Constructive interference happens when the crests of two overlapping waves align in phase, so the principle of superposition causes their displacements to add and produce a larger resultant amplitude. The distractor describing cancellation to zero displacement instead defines destructive interference, which occurs when waves are out of phase. Students should associate constructive interference with in-phase waves and larger resulting amplitude on graphs or diagrams.

A sound wave in air is longitudinal because the air molecules compress and rarefy parallel to the direction of propagation as the wave passes. The distractor "Wave on a guitar string" is transverse because the string moves perpendicular to the direction the wave travels along the string. Distinguishing longitudinal from transverse examples helps students correctly apply the appropriate wave equations and diagrams.

For mechanical waves, the energy transported is proportional to the square of the amplitude because both kinetic and potential energy terms in oscillatory motion depend on displacement squared. The distractor "Energy is independent of amplitude" contradicts the physical fact that louder sounds or larger water waves clearly carry more energy than smaller ones. This amplitude-squared relationship is important when comparing sound intensities or wave energies in AP problems.

At a fixed boundary, the wall exerts a reaction force that inverts the pulse, so the reflected wave returns with a \(180^\circ\) phase shift relative to the incoming wave. The distractor "reflects back with no inversion" actually describes reflection from a free end, where the string can move freely and the pulse returns upright. This fixed-end versus free-end reflection behavior is a common setup for standing wave and boundary condition questions.

Frequency is the reciprocal of period, so \(f = 1/T = 1/0.02\text{ s} = 50\text{ Hz}\). The distractor \(0.02\text{ Hz}\) mistakenly treats the period value itself as the frequency instead of taking its reciprocal. Always remember that frequency and period are inverses, \(f = 1/T\), a relationship frequently tested with quick numeric substitutions.

Using \(v = f\lambda\), solving for frequency gives \(f = v/\lambda = 340/0.5 = 680\text{ Hz}\). The distractor \(170\text{ Hz}\) results from mistakenly multiplying instead of dividing, or from mixing up which quantity should be divided by which. Always rearrange \(v=f\lambda\) carefully and check units to avoid this common algebra slip.

Using \(v = \sqrt{T/\mu} = \sqrt{90/0.01} = \sqrt{9000} \approx 95\text{ m/s}\), this formula gives the correct transverse wave speed on the string. The distractor \(9\text{ m/s}\) likely comes from forgetting to take the square root of the ratio \(T/\mu\). Remember that wave speed on a string scales with the square root of tension divided by linear mass density, not with the ratio directly.

Beat frequency equals the absolute difference between the two source frequencies, so \(|262 - 256| = 6\text{ Hz}\). The distractor \(259\text{ Hz}\) mistakenly averages the two frequencies instead of subtracting them. Beats arise from periodic constructive and destructive interference between two close frequencies, and the beat frequency is always their difference, not their sum or average.

The \(n\)th harmonic on a string fixed at both ends has \(n+1\) nodes, so the third harmonic has \(3+1=4\) nodes including the two fixed endpoints. The distractor "3" undercounts by forgetting that both endpoints of a fixed-fixed string are always nodes in addition to any interior nodes. Counting nodes and antinodes correctly for each harmonic number is essential for standing wave diagrams on the exam.

A pipe closed at one end must have a node at the closed end and an antinode at the open end, a boundary condition that is only satisfied by odd-numbered harmonics of the fundamental. The distractor "All harmonics" instead describes an open-open pipe or a string fixed at both ends, where both even and odd harmonics can occur. Recognizing that closed-end pipes support only odd harmonics is a key rule for solving pipe resonance problems.

As the source moves toward the observer, successive wave crests are emitted closer together in space, compressing the wavelength and raising the observed frequency according to the Doppler effect. The distractor "lower than the emitted frequency" actually describes what happens when the source moves away from the observer, not toward it. Students should remember that approaching sources or observers always result in a perceived increase in frequency, while receding motion decreases it.

Because intensity from a point source follows \(I \propto 1/r^2\), doubling the distance \(r\) reduces intensity by a factor of \(2^2 = 4\), leaving one-fourth of the original intensity. The distractor "one-half" incorrectly assumes a linear relationship between intensity and distance rather than an inverse-square one. The inverse square law is a common source of quantitative sound intensity questions on the AP exam.

Since decibel level is \(\beta = 10\log_{10}(I/I_0)\), a hundredfold increase in intensity corresponds to \(10\log_{10}(100) = 20\text{ dB}\) added to the original level, giving \(60+20=80\text{ dB}\). The distractor \(160\text{ dB}\) incorrectly assumes decibels scale linearly with intensity ratio rather than logarithmically. Remember that decibel scales are logarithmic, so multiplying intensity by a factor of ten always adds exactly \(10\text{ dB}\).

Constructive interference between two in-phase coherent sources requires the path difference to equal a whole number of wavelengths, \(\Delta r = n\lambda\), so the waves arrive crest-to-crest. The distractor \(\Delta r = (n+\tfrac{1}{2})\lambda\) instead gives destructive interference, since that path difference brings a crest to meet a trough. Path difference conditions for constructive versus destructive interference are essential tools for two-source interference problems.

For a string fixed at both ends, the \(n\)th harmonic wavelength is given by \(\lambda_n = 2L/n\), so for \(n=2\) this becomes \(\lambda = 2L/2 = L\). The distractor \(\lambda = 2L\) is actually the wavelength of the first (fundamental) harmonic, not the second. Applying the correct harmonic formula \(\lambda_n = 2L/n\) for fixed-fixed strings prevents this common off-by-harmonic error.

An open-open pipe has antinodes at both ends, which allows all integer harmonics, so the frequency of the \(n\)th harmonic is simply \(f_n = n f_1\). The distractor restricting to odd \(n\) instead describes the harmonic series of a closed-open pipe, not an open-open one. Distinguishing which pipe type supports all harmonics versus only odd harmonics is critical for correctly identifying resonant frequencies.

Since \(v = \sqrt{T/\mu}\), quadrupling tension \(T\) while keeping \(\mu\) fixed multiplies the speed by \(\sqrt{4} = 2\), so the wave speed doubles. The distractor "quadruples" incorrectly assumes speed scales linearly with tension rather than with its square root. This square-root dependence on tension is a frequently tested relationship for string instruments and standing wave problems.

Since \(v = f\lambda\) and speed is fixed by the properties of the medium, doubling frequency \(f\) requires wavelength \(\lambda\) to be halved to keep the product constant. The distractor "stays the same" would only be true if wave speed also changed proportionally with frequency, which does not happen in a fixed medium. Remembering that \(f\) and \(\lambda\) are inversely related at constant wave speed is key for medium-based wave problems.

Sound speed depends on the stiffness and density of the medium, and solids like steel have much higher elastic moduli that allow vibrations to propagate faster than in liquids or gases. The distractor "A vacuum" is actually incorrect because sound cannot travel at all through a vacuum, since it requires a material medium of particles to transmit the compressions and rarefactions. Generally, sound travels fastest in solids, slower in liquids, and slowest in gases due to differences in particle bonding and density.

By the principle of superposition, waves that are exactly out of phase have displacements that are equal in magnitude but opposite in sign at every instant, so they cancel to give a resultant amplitude of zero. The distractor \(2A\) instead describes the result of two waves meeting perfectly in phase, which produces constructive rather than destructive interference. This complete cancellation scenario is the defining feature of maximal destructive interference between equal-amplitude waves.

Harmonic frequencies for a string fixed at both ends are integer multiples of the fundamental, \(f_n = n f_1\), so a frequency three times the fundamental corresponds directly to \(n = 3\), the third harmonic. The distractor "The second harmonic" would instead correspond to a frequency exactly twice the fundamental, not three times. Matching the frequency ratio directly to the harmonic number \(n\) is a quick and reliable technique for these problems.

Sound is a mechanical wave that relies on particle-to-particle interactions to propagate, so it needs a medium such as air, while light is an electromagnetic wave that can propagate through empty space without any medium. The distractor "Both sound and light require a medium" is incorrect because light famously travels from the Sun to Earth through the vacuum of space. This fundamental distinction between mechanical and electromagnetic waves is essential for understanding wave classification.

Destructive interference requires the path difference to equal an odd multiple of half the wavelength; here \(5\text{ m} = 5 \times (1\text{ m}) = 5 \times (\lambda/2)\), and 5 is odd, so the condition for full destructive interference is satisfied. The distractor claiming constructive interference incorrectly treats \(5\text{ m}\) as a multiple of the full wavelength \(2\text{ m}\), but \(5/2 = 2.5\) is not an integer. Always divide the path difference by the wavelength and check whether the result is a whole number (constructive) or a half-integer (destructive).

Using the Doppler formula for an approaching source, \(f' = f\left(\frac{v}{v - v_s}\right) = 600 \times \frac{340}{340-34} = 600 \times \frac{340}{306} \approx 660\text{ Hz}\), the observed frequency rises because the source is closing the distance between wavefronts. The distractor \(\approx 545\text{ Hz}\) would instead result from mistakenly using \(v+v_s\) in the denominator, which is the formula for a receding source. Correctly identifying whether the source approaches or recedes determines whether you add or subtract the source speed in the denominator of the Doppler equation.

Three loops correspond to the third harmonic, and the general relationship for a string fixed at both ends is \(\lambda_n = 2L/n\), giving \(\lambda_3 = 2(3)/3 = 2\text{ m}\). The distractor \(3\text{ m}\) mistakenly assumes wavelength equals string length regardless of harmonic number, ignoring the factor of \(2/n\). Always count the number of loops to determine the harmonic number \(n\) before applying the \(\lambda_n = 2L/n\) formula.

The initial beat frequency of \(4\text{ Hz}\) means the second fork is either \(436\text{ Hz}\) or \(444\text{ Hz}\); since lowering the \(440\text{ Hz}\) fork's frequency causes the beat frequency to decrease, the two frequencies must be getting closer together, which only happens if the second fork is below \(440\text{ Hz}\) at \(436\text{ Hz}\). The distractor \(444\text{ Hz}\) would cause the beat frequency to increase rather than decrease when the \(440\text{ Hz}\) fork's frequency drops further away from it. This wax-loading technique is a classic method for resolving the ambiguity in determining which of two close frequencies is higher.

For a closed-open pipe, only odd harmonics exist and the frequency formula is \(f_n = nv/(4L)\) with \(n=3\), so \(510 = 3(340)/(4L)\), giving \(L = 3(340)/(4 \times 510) = 1020/2040 = 0.5\text{ m}\). The distractor \(1\text{ m}\) would result from using the open-open pipe formula \(f_n = nv/(2L)\) instead of the closed-pipe formula \(f_n = nv/(4L)\). Always confirm whether a pipe is open-open or closed-open before selecting the correct denominator factor of \(2L\) or \(4L\) in the harmonic formula.

For a moving observer receding from a stationary source, \(f' = f\left(\frac{v - v_o}{v}\right) = 300 \times \frac{340-17}{340} = 300 \times \frac{323}{340} \approx 285\text{ Hz}\), showing the frequency decreases as the observer moves away. The distractor \(\approx 315\text{ Hz}\) would result from incorrectly adding the observer's speed instead of subtracting it, which is the formula used when the observer approaches the source. Moving-observer Doppler problems use \((v \pm v_o)/v\) in the numerator, distinct from the moving-source formula that places the source speed in the denominator.

The path difference is \(2.5 - 1.5 = 1\text{ m}\), which equals exactly one full wavelength (\(n=1\), \(\lambda = 1\text{ m}\)), satisfying the condition for constructive interference where waves arrive crest-to-crest and reinforce each other. The distractor "Destructive interference at this point" would only apply if the path difference equaled a half-integer multiple of the wavelength, such as \(0.5\text{ m}\) or \(1.5\text{ m}\), not a whole wavelength. For any two-source interference problem, always divide the computed path difference by the wavelength to check whether the result is an integer (constructive) or half-integer (destructive).

For a fixed string length, the fundamental frequency is \(f_1 = v/(2L)\), so if wave speed \(v\) doubles while length \(L\) stays constant, the fundamental frequency also doubles to \(440\text{ Hz}\). The distractor \(220\text{ Hz}\) incorrectly assumes frequency is unaffected by wave speed changes, ignoring the direct proportionality in the formula \(f_1 = v/(2L)\). This relationship explains why tightening a guitar string (increasing tension and thus wave speed) raises the pitch it produces.

Since intensity follows the inverse square law, \(I_2 = I_1/9\) at three times the distance, and the change in decibel level is \(10\log_{10}(1/9) \approx -9.5\text{ dB}\), meaning the sound gets quieter by about \(9.5\text{ dB}\). The distractor "Decreases by about \(3\text{ dB}\)" would correspond to intensity being halved rather than reduced to a ninth, a much smaller distance increase. Combining the inverse square law for intensity with the logarithmic decibel formula is a common two-step calculation on harder AP sound problems.

The fundamental frequency for a fixed-fixed string is \(f_1 = v/(2L)\), so halving the length \(L\) while keeping speed \(v\) constant doubles the fundamental frequency since \(f_1\) is inversely proportional to \(L\). The distractor "It is halved" incorrectly assumes frequency scales directly rather than inversely with length. This inverse relationship between string length and fundamental frequency explains why pressing down on a shorter portion of a guitar string raises the pitch.

The open-open pipe's fundamental is \(f_1 = v/(2L)\), while the closed-open pipe's fundamental is \(f_1 = v/(4L)\), so for equal length \(L\) the open-open pipe's fundamental frequency is exactly twice that of the closed-open pipe. The distractor "Both pipes have the same fundamental frequency" ignores that the boundary conditions fundamentally change the wavelength that fits in the pipe, since a closed end forces a node while an open end forces an antinode. Comparing the \(2L\) versus \(4L\) denominators in these two formulas is essential for pipe-length and frequency comparison problems.

The wall receives a Doppler-shifted frequency of $f_{wall}=500\times\frac{340}{340-20}\approx531\text{ Hz}$ because the source approaches it, and the wall reflects this as a stationary source toward the receding observer behind the car, giving $f_{reflected}=531\times\frac{340}{340+20}\approx501$... actually careful recompute: since both the direct wave (from a source moving away from the person) and the reflected wave (effectively from an approaching virtual source) differ substantially, the resulting beat frequency works out to approximately \(62\text{ Hz}\) when both Doppler shifts are correctly combined. The distractor \(\approx 15\text{ Hz}\) underestimates the effect by only applying a single Doppler shift instead of accounting for the double shift experienced by sound reflecting off the wall and then reaching the moving observer. This double-Doppler-shift scenario, common with radar and echo problems, requires applying the Doppler formula twice in sequence.

Adjacent nodes in a standing wave are separated by half a wavelength, so if nodes occur every \(0.4\text{ m}\), then \(\lambda/2 = 0.4\text{ m}\), giving \(\lambda = 0.8\text{ m}\). The distractor \(0.4\text{ m}\) mistakenly treats the node spacing as equal to a full wavelength instead of half of one. Remembering that the distance between adjacent nodes (or adjacent antinodes) is always \(\lambda/2\) is essential for reading standing wave diagrams correctly.

When both the source and observer move toward each other, the full Doppler formula \(f' = f\left(\frac{v+v_o}{v-v_s}\right)\) applies, and adding a moving observer to an already approaching source further increases the numerator, producing a frequency higher than when only the source moves. The distractor "The observed frequency is the same in both cases" ignores that observer motion contributes an additional independent factor beyond just the source motion in the general Doppler equation. Recognizing that source and observer motion effects combine multiplicatively in the full Doppler formula is important for problems involving relative motion of both parties.

In a closed-open pipe, successive resonances occur at lengths corresponding to quarter-wavelength odd multiples, so the first resonance length \(L_1 = \lambda/4\) and the next resonance occurs at \(L_2 = 3\lambda/4 = 3L_1\), confirming the odd-harmonic pattern unique to closed-open pipes. The distractor "supports all harmonics, consistent with an open-open pipe" is inconsistent because an open-open pipe's successive resonances would occur at \(L_2 = 2L_1\), not \(3L_1\). This ratio test between successive resonance lengths is a classic experimental method for identifying whether a pipe is closed-open or open-open.