Harmonics · quiz worked solutions

Tried the quiz first? Good. Here’s how to work through each answer. Follow the link beside a question to jump back to its original spot. Questions marked Extra exploration go beyond this AP Physics 1 course; Book controls questions explain the interface.

Dampened Harmonics · question 1

Back to the question · Extra exploration

Question: For y(t)=10e^[−(1/3 s⁻¹)t]cos[(2 rad/s)t], find the damped angular frequency. The coefficient 10 uses any consistent displacement unit.

Answer: 2 rad/s.

1. Separate the exponential envelope from the cosine phase. Why: The envelope changes amplitude, while the cosine sets oscillation timing.

2. Read the coefficient of t in cos[(2 rad/s)t] as ω_d = 2 rad/s. Why: The −1/3 coefficient is a decay rate, not an angular frequency.

Dampened Harmonics · question 2

Back to the question · Extra exploration

Question: A 2 kg mass follows y(t)=4.5e^[−(3 s⁻¹)t]cos[(3 rad/s)t] under linear viscous damping. Find the spring constant. The coefficient 4.5 uses any consistent displacement unit. Use k=m[ω_d²+β²], with β=3 s⁻¹ and ω_d=3 rad/s.

Answer: 36 N/m.

1. Identify β = 3 s⁻¹ and ω_d = 3 rad/s. Why: For linear viscous underdamping, ω₀² = ω_d² + β².

2. Use k = mω₀² = 2(3² + 3²) = 36 N/m. Why: Ignoring damping would give only 18 N/m and miss the envelope’s effect.

Dampened Harmonics · question 3

Back to the question · Extra exploration

Question: For y(t)=6e^[−(1/6 s⁻¹)t]cos[(5 rad/s)t], determine the decay rate β=b/(2m). No mass is specified. The coefficient 6 uses any consistent displacement unit. The exponent has the form −βt; β is constant and does not depend on an evaluation time.

Answer: 1/6 s⁻¹.

1. Compare the envelope e^[−(1/6 s⁻¹)t] with e^(−βt). Why: The coefficient multiplying time defines the decay rate.

2. Read β = 1/6 s⁻¹. Why: It is positive and constant; no mass is supplied to determine b separately.

Dampened Harmonics · question 4

Back to the question · Extra exploration

Question: A free oscillator with a nonzero viscous damping force F_d=−bv and a decaying amplitude is undergoing ideal simple harmonic motion.

Answer: False.

1. A nonzero −bv term removes mechanical energy and makes the amplitude decay. Why: Ideal SHM has a purely restoring force proportional to displacement.

2. This motion is damped oscillation, not ideal SHM. Why: An approximately sinusoidal appearance does not eliminate its dissipative force.

Dampened Harmonics · question 5

Back to the question · Extra exploration

Question: Given that an oscillating system has a k -value of 100 N/m and a damping constant of 4 × 10 - 3 kg/s , determine the period of the system if the mass is 1 kg .

Answer: 0.63 s.

1. Find β = b/(2m) = 0.004/(2·1) = 0.002 s⁻¹ and ω_d = √(k/m − β²) ≈ 9.9999998 rad/s. Why: Use the underdamped frequency, including the small damping correction.

2. Find T = 2π/ω_d ≈ 0.628319 s. Why: The damping is tiny here, so rounding to 0.63 s also matches the nearly undamped period.

Pendulums · question 1

Back to the question · AP Topic 7.2

Question: For an ideal small-angle simple pendulum with fixed length, what happens to its period if the point-bob mass doubles and all other factors stay the same?

Answer: The period stays the same.

1. Use T = 2π√(L/g) for a point bob and small angles. Why: Mass cancels between inertia and the gravitational restoring force.

2. Doubling bob mass leaves L and g unchanged. Why: It therefore does not change the period under the stated ideal assumptions.

Pendulums · question 2

Back to the question · Extra exploration

Question: The pictured pendulums have the same radius, but their axial lengths, mass distributions and pivot-axis inertias are not all specified. Which conclusion about their small-angle periods is justified from that information?

Answer: Cannot determine the longest period from radius alone.

1. For a physical pendulum, T = 2π√[I/(mgh)]. Why: The pivot-axis inertia and center-of-mass distance both matter.

2. Same radius does not fix I/(mh) for the pictured shapes. Why: Unspecified lengths, distribution and pivots prevent a valid ranking.

Pendulums · question 3

Back to the question · AP Topic 7.2

Question: An ideal small-angle simple pendulum has period T. If its length increases by a factor of 4 with gravity unchanged, what is its new period?

Answer: 2T.

1. Use the simple-pendulum scaling T ∝ √L with g fixed. Why: The model assumes small-angle motion and the same gravity.

2. Multiply the length by 4 to multiply the period by √4 = 2. Why: A linear factor of four in length does not make a factor of four in time.

Pendulums · question 4

Back to the question · AP Topic 3.4

Question: A point bob on a light inextensible string of length L is released from rest at angular displacement θ from the downward vertical. Assume the string remains taut throughout the swing (for example, a release no more than 90° from the downward vertical). With no dissipation and gravitational magnitude g, what is its maximum speed?

Answer: √[2gL(1 − cosθ)].

1. The bob drops by Δh = L − L cosθ = L(1 − cosθ) to the bottom. Why: The assumed taut string keeps the bob a distance L from the pivot, so this circular-path geometry gives the gravitational potential-energy decrease.

2. Set mgΔh = ½mv_max², giving v_max = √[2gL(1 − cosθ)]. Why: With a taut string, tension is perpendicular to the motion and does no work. With no dissipation, mechanical energy is conserved; potential energy is minimum and speed maximum at the bottom, without a small-angle approximation.

Pendulums · question 5

Back to the question · Extra exploration

Question: A uniform solid cylinder has length L=1.00 m, diameter 0.500 cm and mass 0.500 g. It is pivoted at the centre of one end about an axis perpendicular to the cylinder’s length, as a physical pendulum. Thus h=L/2. Find the inertia about that pivot in kg·m² and the small-angle period in seconds, using g=9.8 m/s². For this axis, I=mL²/3+mR²/4.

Answer: I ≈ 1.67×10⁻⁴ kg·m²; T ≈ 1.64 s.

1. Convert mass to 0.000500 kg and radius to 0.00250 m. Why: The diameter is 0.500 cm, so radius is half that value.

2. Use I = mL²/3 + mR²/4 ≈ 0.00016666745 kg·m². Why: The specified transverse end pivot differs from a central axis.

3. With h = 0.500 m, T = 2π√[I/(mgh)] ≈ 1.639 s. Why: The physical-pendulum period uses the pivot inertia and center-of-mass distance together.

Simple Harmonics · question 1

Back to the question · Extra exploration

Question: How long is a string if it has 4 nodes of the following wave and is affixed at both ends?​ y ( x , t ) = 6 sin ( 4 π x ) cos ( 3 t )

Answer: 0.75 m.

1. Nodes satisfy sin(4πx) = 0, so adjacent nodes are π/(4π) = 0.25 m apart. Why: Both fixed ends are nodes.

2. Four nodes include three gaps, giving L = 3(0.25) = 0.75 m. Why: Counting four gaps would incorrectly treat the number of intervals as the number of points.

Simple Harmonics · question 2

Back to the question · Extra exploration

Question: An equal-amplitude counterpropagating wave y₂=4sin(4x+8t) combines with y₁=4sin(4x−8t), with x in m and t in s. Which expression gives their standing-wave sum?

Answer: y = 8 sin(4x)cos(8t).

1. Add y₁ + y₂ = 4[sin(4x − 8t) + sin(4x + 8t)]. Why: Linear superposition adds the displacements.

2. Apply sin(A − B) + sin(A + B) = 2sinA cosB. Why: The result has stationary nodes and an 8-unit maximum antinode amplitude.

Simple Harmonics · question 3

Back to the question · Extra exploration

Question: What is the wavelength of the following standing wave?​ y ( x , t ) = 6 sin ( 6 x ) cos ( 6 t )

Answer: 1.047 m.

1. Read the spatial wave number k = 6 rad/m from sin(6x). Why: The time factor does not determine wavelength.

2. Use λ = 2π/k = 2π/6 ≈ 1.0472 m. Why: A complete spatial cycle changes phase by 2π.

Simple Harmonics · question 4

Back to the question · Extra exploration

Question: What frequency does the following standing wave posses? y ( x , t ) = 4 sin ( 4 x ) cos ( 4 t )

Answer: 0.637 Hz.

1. Read angular frequency ω = 4 rad/s from cos(4t). Why: The coefficient is angular frequency, not cycles per second.

2. Find f = ω/(2π) = 4/(2π) ≈ 0.6366 Hz. Why: The conversion divides by one full cycle’s 2π radians.

Simple Harmonics · question 5

Back to the question · Extra exploration

Question: How many nodes are on the doubly-fixed 1m string with the following standing wave?​ y ( x , t ) = 6 sin ( 2 π x ) cos ( 8 t )

Answer: 3 nodes.

1. Nodes occur where sin(2πx) = 0, so x = n/2 metres. Why: The spatial factor sets permanent zero-displacement locations.

2. Within 0 ≤ x ≤ 1, the nodes are x = 0, 0.5 and 1 m. Why: Include both fixed ends as well as the one interior node.

Model references

These original worked steps use the stated question models. Physics references: Pendulums, Damped oscillations, Standing waves. AP topic labels follow the supplied-edition College Board CED, Effective Fall 2024, ©2026, pinned in the course audit.