MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4a Periodic Motion Mechanical Waves

What this quiz covers

This quiz focuses on 4a Periodic Motion Mechanical Waves, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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.

Question 1

1. Wave 2 travels faster because larger amplitude increases wave speed in the same medium
2. Wave 2 has a longer wavelength because amplitude and wavelength are directly proportional
3. Wave 2 carries more energy, while wave speed and wavelength remain unchanged (correct answer)
4. Wave 2 has a higher frequency because amplitude increases the number of cycles per second

Explanation: This question tests the independence of wave parameters in linear media. In a linear medium, wave speed depends only on medium properties (like density and elasticity), not on wave amplitude. Since both waves travel through the same fluid at the same frequency, they have identical wave speeds and wavelengths. The key difference is energy: wave energy is proportional to amplitude squared, so Wave 2 carries more energy while maintaining the same propagation characteristics. Choice A incorrectly links amplitude to wave speed, confusing energy content with propagation speed. Remember: in linear media, changing amplitude affects energy and intensity but not frequency, wavelength, or wave speed.

Question 2

1. Greater damping causes mechanical energy to be dissipated more rapidly, reducing amplitude over time (correct answer)
2. Greater damping increases the system's total mechanical energy, increasing amplitude over time
3. Greater damping increases the natural frequency, which necessarily increases amplitude
4. Greater damping converts oscillatory motion into constant-velocity linear motion with the same amplitude

Explanation: This question tests the effect of damping on oscillatory motion. Damping represents energy dissipation mechanisms that convert mechanical energy into other forms (typically heat), reducing the system's total mechanical energy over time. Greater damping means energy is lost more rapidly, resulting in smaller oscillation amplitudes for the same initial conditions. The natural frequency may slightly decrease with heavy damping, but this doesn't increase amplitude. Choice B incorrectly claims damping adds energy to the system, contradicting the fundamental nature of dissipative forces. Remember: damping always removes energy from oscillatory systems, reducing amplitude over time—it never adds energy or increases amplitude.

Question 3

1. One antinode at the center and nodes only at the ends
2. Two antinodes with a node at the center and nodes at both ends (correct answer)
3. A uniform displacement along the entire string with no nodes
4. Three antinodes with two interior nodes and nodes at both ends

Explanation: This question tests standing wave patterns for strings fixed at both ends. The second harmonic (n=2) has a wavelength λ = L, where L is the string length, creating two half-wavelengths along the string. This produces two antinodes (maximum displacement points) with a node at the center, plus the required nodes at both fixed ends. The fundamental (n=1) would have only one antinode, while the third harmonic (n=3) would have three antinodes. Choice D describes the third harmonic pattern, not the second. A useful check: for the nth harmonic on a string fixed at both ends, there are n antinodes and (n+1) nodes total.

Question 4

1. The received frequency is lower than $f_0$ because motion toward the probe increases wavelength.
2. The received frequency is higher than $f_0$ because motion toward the probe decreases the effective wavelength between wavefronts. (correct answer)
3. The received frequency equals $f_0$ because sound speed in tissue is constant.
4. The received frequency is higher than $f_0$ only if the emitted amplitude is increased.

Explanation: This question tests the Doppler effect for sound waves when the source (blood) moves toward the observer (probe). When a source moves toward an observer, the wavefronts are compressed, decreasing the effective wavelength between successive crests. Since wave speed in the medium remains constant (v = 1540 m/s in tissue), and v = fλ, a decrease in wavelength must correspond to an increase in frequency. The received frequency is therefore higher than the emitted frequency f₀, confirming choice B. Choice A incorrectly states that motion toward the observer increases wavelength, which would occur if the source moved away. Choice C ignores the Doppler effect entirely. Choice D incorrectly links frequency shift to amplitude, when the Doppler shift depends only on relative motion. For Doppler problems, remember: source moving toward observer → higher frequency; source moving away → lower frequency. The shift magnitude depends on the ratio of source speed to wave speed.

Question 5

1. Constructive interference because a half-wavelength path difference produces in-phase arrival.
2. Destructive interference because a half-wavelength path difference produces a $\pi$ phase shift. (correct answer)
3. No interference because sound waves do not superpose in air.
4. Greater loudness because longer path length increases wave amplitude at the ear.

Explanation: This question tests wave interference based on path difference. When two coherent sources emit in phase, the phase difference at the observation point depends on the path difference. A path difference of ΔL = λ/2 corresponds to a phase difference of π radians (180°), since phase difference = (2π/λ) × path difference = (2π/λ) × (λ/2) = π. When two waves of equal amplitude arrive with opposite phase (π phase difference), they interfere destructively, producing zero net amplitude at that point. This confirms choice B is correct. Choice A incorrectly claims λ/2 path difference produces constructive interference, which would require integer multiples of λ. Choice C incorrectly denies that sound waves can interfere. Choice D incorrectly links path length to amplitude rather than phase. For interference problems, remember: path difference of nλ → constructive; path difference of (n + 1/2)λ → destructive, where n is an integer.

Question 6

1. The wavelength doubles because frequency and wavelength increase together in a fixed medium.
2. The wavelength is halved because wave speed is constant and $v = f\lambda$. (correct answer)
3. The wave speed doubles because frequency determines wave speed in any medium.
4. The wave amplitude must decrease because higher frequency waves carry less energy.

Explanation: This question tests mechanical wave properties, specifically the relationship between frequency, wavelength, and speed. The wave speed in a medium is constant and determined by the medium's properties, with v = fλ holding for sinusoidal waves. In this fluid-filled tube modeling airway mucus, doubling the frequency while keeping the medium unchanged maintains constant wave speed. The correct answer B follows because halving the wavelength satisfies v = fλ when f doubles and v is fixed. Distractor A fails due to the misconception that wavelength increases with frequency, but they are inversely related at constant speed. To verify in similar questions, calculate wavelength changes using λ = v/f. Always confirm if the medium's properties are altered, as that affects v.

Question 7

1. The pulse travels faster because higher frequency implies higher speed in the same medium.
2. The wavelength decreases, which can improve spatial resolution. (correct answer)
3. The return time decreases because frequency determines time-of-flight.
4. The boundary reflection disappears because higher frequency waves cannot reflect.

Explanation: This question tests mechanical wave propagation and resolution in ultrasound. Higher frequency reduces wavelength via λ = v/f, improving resolution as smaller wavelengths distinguish closer features. In this ultrasound phantom, increasing frequency with constant v shortens λ for better spatial resolution. The correct answer B follows because shorter wavelength directly enhances resolution without altering speed or reflection. Distractor A assumes speed increases with frequency, a misconception ignoring that v depends on the medium. For other wave resolution problems, recall resolution improves with shorter λ. Check if frequency changes affect attenuation, though not relevant here.

Question 8

1. The path length difference is an integer multiple of $\lambda$.
2. The path length difference is a half-integer multiple of $\lambda$. (correct answer)
3. The waves have different speeds, so they cancel regardless of phase.
4. The waves must have different frequencies to cancel at a point.

Explanation: This question tests wave interference principles. Destructive interference for equal-amplitude coherent waves occurs when path difference δ = (m + 1/2)λ, a half-integer multiple. In this audiology setup, the condition for cancellation matches this. The correct answer B follows because destructive interference requires odd multiples of λ/2 path difference. Distractor A describes constructive interference, a common confusion of conditions. For interference problems, use δ = mλ for constructive, (m+1/2)λ for destructive. Assume coherence unless stated otherwise.

Question 9

1. The location of maximal displacement shifts to a region with higher natural frequency. (correct answer)
2. The location of maximal displacement remains fixed because frequency only affects amplitude.
3. The maximal displacement shifts to a region with lower natural frequency because higher frequency waves travel farther.
4. The maximal displacement disappears because resonance cannot occur at higher frequency.

Explanation: This question tests resonance in wave systems like the cochlea. Maximal displacement occurs where local natural frequency matches stimulus frequency. Increasing stimulus frequency shifts peak to higher-frequency regions. The correct answer A follows because resonance mapping places higher f at corresponding sites. Distractor B ignores frequency dependence, assuming fixed location. In tonotopic models, map frequency to position via resonance. Check if system has graded properties affecting local frequencies.

Question 10

1. This is the third harmonic, and the wavelength satisfies $\lambda = 2L/3$. (correct answer)
2. This is the second harmonic, and the wavelength satisfies $\lambda = L$.
3. This is the fundamental, and the wavelength satisfies $\lambda = 2L$.
4. This is not a harmonic because standing waves require only one antinode.

Explanation: This question tests harmonics in standing waves. For fixed ends, nth harmonic has n antinodes, λ=2L/n; three antinodes is n=3, λ=2L/3. With three antinodes, it's the third harmonic. The correct answer A follows because it matches the harmonic and wavelength. Distractor B miscounts as second with wrong λ, confusing node count. In standing wave questions, count antinodes for n. Use f_n = n v /(2L) for frequency scaling.

Question 11

1. The period decreases by a factor of 2 because a longer pendulum swings faster.
2. The frequency increases by a factor of 4 because the length is quadrupled.
3. The period increases by a factor of 2 because $T \propto \sqrt{L}$ for small angles. (correct answer)
4. The period is unchanged because only the mass affects pendulum motion.

Explanation: This question tests periodic motion of simple pendulums. The period T of a simple pendulum for small angles is given by T = 2π√(L/g), showing T proportional to √L. In this motor-control experiment, increasing length from L to 4L quadruples the argument under the square root. The correct answer C follows because T increases by √4 = 2, matching the formula. Distractor A reflects the misconception that longer pendulums swing faster, but actually longer lengths increase the period. For other pendulum problems, check if angles are small to apply the approximation. Compute ratios like T_new/T_old = √(L_new/L_old) to predict changes.

Question 12

1. Signal 2 leads Signal 1 by $\pi/2$ radians.
2. Signal 2 lags Signal 1 by $\pi/2$ radians. (correct answer)
3. Signal 2 is in phase with Signal 1 because they share the same period.
4. Signal 2 is $\pi$ radians out of phase with Signal 1 because $T/4$ is half a cycle.

Explanation: This question tests phase relationships in periodic motion. Phase difference φ = (Δt / T) * 2π, so Δt = T/4 gives φ = π/2 lag if the second signal peaks later. In this vibrating membrane comparison, Signal 2 peaking after T/4 indicates a lag. The correct answer B follows because the delay corresponds to a π/2 phase lag. Distractor D miscalculates T/4 as half-cycle (π), a common error in phase conversion. For phase problems, use φ = 2π Δt / T. Check if signals have the same frequency, essential for phase comparison.

Question 13

1. The transmitted wave speed increases because larger $\mu$ increases $v$.
2. The transmitted wave speed decreases because $v=\sqrt{T/\mu}$. (correct answer)
3. The wave cannot transmit at all because speed must be the same on both sides.
4. The frequency changes at the boundary because the medium sets the oscillation rate.

Explanation: This question tests wave transmission across media. Wave speed v=√(T/μ) decreases with larger μ at fixed T. Transitioning to thicker string (higher μ) slows the transmitted wave. The correct answer B follows because v decreases per the formula. Distractor D wrongly claims frequency changes, but frequency is continuous across boundaries. In boundary problems, note frequency constant, speed and wavelength adjust. Use v1/v2 = √(μ2/μ1) for predictions.

Question 14

1. The steady-state displacement amplitude increases, while the oscillation frequency remains the driving frequency. (correct answer)
2. The steady-state displacement amplitude remains constant, but the oscillation frequency increases.
3. The oscillator switches to its natural frequency because amplitude determines frequency.
4. The period decreases because larger driving amplitude shortens each cycle.

Explanation: This question tests driven oscillators away from resonance. In linear regime, steady-state amplitude increases with driving force amplitude, frequency remains driving frequency. For this medical device model, larger drive increases response amplitude. The correct answer A follows because amplitude is proportional to driving force in linear systems. Distractor C assumes switch to natural frequency, but driven systems lock to driver. In driven problems, distinguish transient (natural) from steady-state (driven) frequency. Check linearity by ensuring small oscillations.

Question 15

1. The period decreases by a factor of 2 because heavier masses oscillate faster.
2. The period increases by a factor of 2 because $T \propto \sqrt{m}$. (correct answer)
3. The period increases by a factor of 4 because it is directly proportional to mass.
4. The period is unchanged because the amplitude is unchanged.

Explanation: This question tests periodic motion in mass-spring systems. The period T = 2π√(m/k) shows T proportional to √m for fixed k. In this organ oscillation model, increasing mass from m to 4m quadruples the term under the square root. The correct answer B follows because T increases by √4 = 2, directly from the formula. Distractor A reflects the misconception that heavier masses oscillate faster, but increased inertia lengthens the period. For other SHM problems, use T ratios like T_new/T_old = √(m_new/m_old). Ensure oscillations are small to apply the harmonic approximation.

Question 16

1. The wavelength in air decreases by a factor of 2 while wave speed remains constant (correct answer)
2. The wavelength in air increases by a factor of 2 because frequency increased
3. The wave speed increases by a factor of 2 because frequency increased
4. The amplitude must decrease by a factor of 2 to conserve energy in the wave

Explanation: This question tests understanding of the wave equation v = fλ and how frequency changes affect wavelength when wave speed is constant. The fundamental principle is that in a given medium at constant temperature, the speed of sound remains fixed at v = 340 m/s. Since v = fλ must hold true, when frequency doubles from 500 Hz to 1000 Hz, the wavelength must halve to maintain the constant wave speed. This gives λ₁ = 340/500 = 0.68 m initially, and λ₂ = 340/1000 = 0.34 m after the frequency increase, confirming wavelength decreases by a factor of 2. Choice B incorrectly assumes wavelength increases with frequency, while choice C wrongly suggests wave speed depends on frequency in air. A key check for wave problems: when wave speed is constant (determined by medium properties), frequency and wavelength are inversely proportional.

Question 17

1. The wave frequency changes to match the resonance of the tissue
2. The wave speed and wavelength may change, but the frequency remains $f$ (correct answer)
3. The wavelength remains constant because the source frequency is fixed
4. The amplitude must remain constant to conserve energy across the boundary

Explanation: This question tests understanding of wave behavior at boundaries between different media. When a wave crosses from one medium to another, frequency must remain constant because it's determined by the source, not the medium. However, wave speed typically changes because v = √(elastic property/inertial property) depends on medium properties. Since v = fλ, if frequency f stays constant and speed v changes, then wavelength λ must change proportionally with speed. In this case, sound typically travels faster in soft tissue than in air, so both speed and wavelength increase while frequency remains at f. Choice A incorrectly suggests frequency changes at boundaries, while choice C wrongly assumes wavelength stays constant. Remember: frequency is conserved across boundaries, but speed and wavelength change together.

Question 18

1. The wave speed increased, implying tension must have increased
2. The wave speed increased, implying the medium change did not reduce speed as much as the frequency increase raised $f\lambda$ (correct answer)
3. The wave speed decreased, implying wavelength must have increased
4. The wave speed is unchanged, implying frequency cannot change if wavelength is constant

Explanation: This question tests understanding of the wave equation v = fλ under changing conditions. If frequency increases and wavelength stays approximately constant, then wave speed v = fλ must increase. Normally, decreasing string tension would decrease wave speed (v = √(T/μ)), but the frequency increase more than compensates for this effect. The net result is higher wave speed despite lower tension. This scenario demonstrates that multiple factors can affect wave properties simultaneously. Choice A incorrectly attributes speed increase to tension increase, choice C wrongly suggests speed decreased, and choice D misunderstands the relationship between parameters. When analyzing wave changes, consider all factors: both medium properties and wave parameters.

Question 19

1. Incorrect; in a transverse wave, string elements oscillate perpendicular to the direction of propagation (correct answer)
2. Correct; all mechanical waves require particle motion parallel to propagation
3. Correct; wave speed equals particle speed for a transverse wave
4. Incorrect; string elements remain stationary while only energy propagates

Explanation: This question tests understanding of particle motion in transverse waves. In a transverse wave, particles of the medium oscillate perpendicular to the direction of wave propagation. If the wave travels in the +x direction along a string, string elements move up and down (in the ±y direction), not forward and backward. This is the defining characteristic that distinguishes transverse waves from longitudinal waves. The wave pattern moves in the +x direction, but individual string particles only move transversely. Choice B incorrectly claims all waves require parallel motion (only longitudinal waves do), choice C wrongly equates wave speed with particle speed, and choice D incorrectly suggests particles don't move at all. Understanding this distinction is crucial for interpreting wave behavior in biological systems.

Question 20

1. Correct, because larger $k$ means the wave advances more quickly in space
2. Incorrect, because increasing $k$ at fixed $\omega$ decreases wave speed (correct answer)
3. Correct, because $k$ and $v$ are directly proportional when $\omega$ is fixed
4. Incorrect, because wave speed depends only on amplitude and not on $k$ or $\omega$

Explanation: This question tests understanding of wave parameters and the wave speed equation v = ω/k. For a sinusoidal wave y(x,t) = A sin(kx - ωt), the wave number k represents 2π/λ (spatial frequency) and ω represents 2πf (temporal frequency). Wave speed is v = ω/k = fλ. If k increases while ω stays constant, then v = ω/k decreases because k appears in the denominator. Physically, increasing k means decreasing wavelength, and at fixed frequency, shorter wavelengths mean slower wave speed. The student's conclusion is incorrect. Choice A wrongly suggests larger k means faster propagation, while choice D incorrectly claims wave speed depends on amplitude. Always check: v = ω/k shows inverse relationship between wave speed and wave number.