1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
A block hanging from a vertical ideal spring oscillates with period TE on Earth. The same block and spring are taken to the Moon, where the gravitational field is weaker, and the block is set oscillating vertically. How does the period on the Moon compare with TE?
Answer and reasoning
AIt is the same: g shifts the equilibrium position but does not alter m or k.Correct The period of an object–spring oscillator is Ts = 2π√(m/k). On the Moon the block's weight is smaller, so the spring stretches less at equilibrium, but m and k are unchanged and the period is TE.
BIt is longer: weaker gravity adds less to the spring's restoring force there. A student who thinks gravity adds to the restoring force of a vertical oscillator picks this. Gravity is constant, so it shifts the equilibrium position; the net force about that position is still −k times the displacement.
CIt is the same: like a pendulum's period, it does not depend on g at all. A student who thinks a pendulum's period depends only on its length, and so is the same on any planet, picks this. The claim about the spring is right, but the reason is wrong: a pendulum's period, 2π√(l/g), does depend on g. The spring's period is unchanged because Ts = 2π√(m/k) contains neither g nor anything that g alters.
DIt is shorter: the block has less mass on the Moon, so it oscillates more quickly. A student who confuses mass with weight picks this. The block weighs less on the Moon, but its mass, which sets its inertia in Ts = 2π√(m/k), is the same everywhere.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.2.A.1 Period, T Fix
Period, T
The time taken for one complete cycle of an oscillation. Unit: second (s). T = 1/f.
Frequency, f
The number of complete cycles per unit time. Unit: hertz (Hz), equal to one cycle per second (s⁻¹). f = 1/T.
Angular frequency, ω
The quantity 2π/T = 2πf that describes how fast an oscillation proceeds through its cycle, one cycle corresponding to 2π rad. Unit: radian per second (rad/s). It is not the same number as the frequency f.
Period of an object–ideal-spring oscillator
Ts = 2π√(m/k), where m is the mass of the oscillating object and k the spring constant. It does not contain g, so it is the same for horizontal and vertical oscillation and on any planet.
Simple pendulum
An object (bob) small enough to be modeled as a point mass, hanging from a pivot on a light string of length l (also written ℓ) and swinging under gravity.
Period of a simple pendulum displaced by a small angle
Tp = 2π√(l/g), where l is the length from the pivot to the bob and g the gravitational field strength. It does not contain the bob's mass, and it applies only when the angle of swing is small.
Students often think Angular frequency and frequency are the same quantity, so ω (in rad/s) has the same value as f (in Hz). In fact No. ω = 2πf, so ω in rad/s is 2π (about 6.3) times f in Hz. An oscillator with f = 2.0 Hz has ω ≈ 12.6 rad/s.
Students often think The period and the frequency of an oscillation are interchangeable; the value of one can be used where the other is needed. In fact No. They are reciprocals, T = 1/f. A period of 0.50 s means a frequency of 2.0 Hz, not 0.50 Hz.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
The graph shows the position x of an object in simple harmonic motion as a function of time t. What is the angular frequency ω of the motion?
Answer and reasoning
A2.00 rad/s A student who treats angular frequency as the same as frequency picks this: f = 1/(0.50 s) = 2.00 Hz. Each cycle is 2π rad, so ω = 2πf = 12.6 rad/s.
B12.6 rad/sCorrect The curve returns to its maximum, 0.10 m, every 0.50 s, so T = 0.50 s. Then ω = 2π/T = 2π/(0.50 s) = 12.6 rad/s.
C3.14 rad/s A student who uses the period in place of the frequency picks this: ω = 2π(0.50) = 3.14. The frequency is the reciprocal of the period, 2.0 Hz, so ω = 2π(2.0 Hz) = 12.6 rad/s.
D25.1 rad/s A student who reads the period as the time from a maximum to the next minimum, 0.25 s, picks this: 2π/0.25 = 25.1 rad/s. That is half a cycle; the motion repeats completely only after 0.50 s.
Working The motion repeats completely (x = 0.10 m, at rest, about to move in the negative direction) at t = 0, 0.50 s and 1.00 s, so T = 0.50 s. ω = 2π/T = 2π/(0.50 s) = 12.6 rad/s. (Equivalently f = 1/T = 2.0 Hz and ω = 2πf.)
A block attached to a horizontal ideal spring oscillates on a frictionless surface with period T₀. The block is replaced by a block with nine times the mass, attached to the same spring. What is the new period?
Answer and reasoning
A9.00T₀ A student who takes the period to be proportional to the mass picks this. The period depends on the square root of the mass, so it increases by a factor of √9 = 3.
B0.33T₀ A student who recalls the period as 2π√(k/m) picks this: the period would fall by a factor of 3. A heavier block accelerates less for the same spring force, so its period is longer: Ts = 2π√(m/k).
C3.00T₀Correct Ts = 2π√(m/k), so with the same spring the period is proportional to √m. Nine times the mass gives √9 = 3 times the period.
D1.00T₀ A student who thinks the period of a mass–spring oscillator does not depend on the mass, as a pendulum's does not, picks this. The spring force does not grow with the mass, so the heavier block oscillates more slowly.
Working Ts = 2π√(m/k). With k unchanged, T ∝ √m, so Tnew/T₀ = √(9m/m) = 3. New period = 3.00T₀.
When a block hangs at rest from a vertical ideal spring, the spring is stretched by a distance d. The block is then pulled down slightly and released, and it oscillates vertically. Which expression gives the frequency f of the oscillation, in terms of d and the acceleration due to gravity g?
Answer and reasoning
A√(g/d)/(2π)Correct At rest kd = mg, so m/k = d/g, and Ts = 2π√(m/k) = 2π√(d/g). The frequency is the reciprocal of the period: f = 1/(2π√(d/g)) = √(g/d)/(2π).
B2π√(d/g) A student who gives the period where the frequency is asked picks this. 2π√(d/g) is T, in seconds; the frequency is its reciprocal, √(g/d)/(2π).
C1/(2π√(g/d)) A student who recalls the period as 2π√(k/m) gets T = 2π√(g/d) and f = 1/(2π√(g/d)). The period increases with mass, Ts = 2π√(m/k) = 2π√(d/g), so f = √(g/d)/(2π).
D√(g/d) A student who treats angular frequency as frequency picks this: √(k/m) = √(g/d) is ω, in rad/s. The frequency is ω/(2π) = √(g/d)/(2π).
Working At rest: kd = mg, so m/k = d/g. The period of an object–spring oscillator is Ts = 2π√(m/k) = 2π√(d/g) (gravity only shifts the equilibrium position). f = 1/T = 1/(2π√(d/g)) = √(g/d)/(2π).
A 0.50 kg block attached to an ideal spring of spring constant 200 N/m oscillates on a frictionless horizontal surface. What is the frequency of the oscillation?
Answer and reasoning
A0.31 Hz A student who gives the period's value as the frequency picks this. 0.31 s is the period; the frequency is 1/(0.31 s) = 3.2 Hz.
B20 Hz A student who treats angular frequency as frequency picks this: √(k/m) = √(200/0.50) = 20 rad/s is ω. The frequency is ω/(2π) = 3.2 Hz.
C0.0080 Hz A student who recalls the period as 2π√(k/m) gets T = 2π(20) = 126 s and f = 0.0080 Hz. The period is 2π√(m/k) = 0.31 s, so f = 3.2 Hz.
D3.2 HzCorrect Ts = 2π√(m/k) = 2π√(0.50/200) = 2π(0.050 s) = 0.31 s, and f = 1/T = 3.2 Hz.
Working Ts = 2π√(m/k) = 2π√(0.50 kg/200 N/m) = 2π√(0.0025 s²) = 2π(0.050 s) = 0.314 s. f = 1/T = 3.18 Hz ≈ 3.2 Hz.
A simple pendulum swinging through small angles has period T₀ on Earth. The same pendulum is taken to a planet where the gravitational field strength is one-quarter of Earth's and set swinging through small angles. What is its period there?
Answer and reasoning
A2.00T₀Correct Tp = 2π√(l/g), so with the same length T is proportional to 1/√g. With g reduced to g/4, T increases by a factor of √4 = 2.
B0.50T₀ A student who recalls the period as 2π√(g/l) picks this: a quarter of g would halve the period. In a weaker field the restoring force is smaller and the pendulum swings more slowly: Tp = 2π√(l/g) doubles.
C1.00T₀ A student who thinks a pendulum's period depends only on its length picks this. Tp = 2π√(l/g) also depends on g; with g a quarter as large, the period is √4 = 2 times as long.
D4.00T₀ A student who takes the period to be inversely proportional to g, dropping the square root, picks this. The period depends on 1/√g, so a quarter of g gives twice the period, not four times.
Working Tp = 2π√(l/g). With l unchanged, T ∝ 1/√g, so Tnew/T₀ = √(g/(g/4)) = √4 = 2. New period = 2.00T₀.
The diagram shows three simple pendulums, P, Q and R, with their string lengths and bob masses. Each is set swinging through a small angle. Which ranks the periods of the pendulums from longest to shortest?
Answer and reasoning
AQ > R > P A student who thinks a heavier bob gives a longer period, as on a spring, picks this: with T ∝ √(ml), Q's four-times-heavier bob outweighs R's doubled length. The mass cancels for a pendulum, so P and Q have equal periods.
BR > P > Q A student who thinks a heavier bob swings faster picks this, putting Q below P. The period of a simple pendulum does not depend on the bob's mass, so P and Q have equal periods.
CP = Q > R A student who recalls the period as 2π√(g/l) picks this: the longer pendulum would have the shorter period. A longer pendulum swings more slowly: Tp = 2π√(l/g) is largest for R.
DR > P = QCorrect Tp = 2π√(l/g) depends on the length but not on the bob's mass. P and Q have the same length, so their periods are equal; R is twice as long, so its period is √2 times theirs and is the longest.
Working Tp = 2π√(l/g) does not depend on the bob's mass. P and Q (l = 0.50 m) have equal periods, 2π√(0.050 s²) ≈ 1.4 s with g = 10 m/s²; R (l = 1.0 m) has a period √2 times longer, ≈ 2.0 s. Ranking: R > P = Q.
A simple pendulum swinging through small angles has frequency f₀ at a place where the acceleration due to gravity is g. Which expression gives the length ℓ of the pendulum?
Answer and reasoning
Agf₀²/(4π²) A student who substitutes the frequency where the period belongs, f₀ = 2π√(ℓ/g), picks this. The period is the reciprocal of the frequency, 1/f₀, which puts f₀² in the denominator.
Bg/(2πf₀)²Correct The period is 1/f₀, and Tp = 2π√(ℓ/g). Squaring, 1/f₀² = 4π²ℓ/g, so ℓ = g/(4π²f₀²) = g/(2πf₀)².
C4π²gf₀² A student who recalls the period as 2π√(g/ℓ) picks this: 1/f₀ = 2π√(g/ℓ) gives ℓ = 4π²gf₀². A longer pendulum has a longer period, Tp = 2π√(ℓ/g), so ℓ = g/(2πf₀)².
Dg/f₀² A student who treats the angular frequency √(g/ℓ) as the frequency picks this: f₀ = √(g/ℓ) gives ℓ = g/f₀². The angular frequency is 2πf₀, so ℓ = g/(2πf₀)².
Working T = 1/f₀ and Tp = 2π√(ℓ/g). Squaring: 1/f₀² = 4π²ℓ/g, so ℓ = g/(4π²f₀²) = g/(2πf₀)².
The table shown gives the period T of a simple pendulum of length 1.00 m for four different amplitudes (the largest angle the string makes with the vertical). For this pendulum, Tp = 2π√(ℓ/g) gives 1.99 s, using g = 10 m/s². Which claim do the data in the table support?
Answer and reasoning
AT is the same at all four amplitudes, within the precision of the data given. A student who thinks 2π√(ℓ/g) holds at any amplitude picks this. The data are given to 0.01 s, and 2.13 s at 60° differs from 1.99 s by 0.14 s, far more than that precision.
BT is proportional to amplitude, since the bob has farther to travel each swing. A student who thinks the period is proportional to the amplitude picks this. From 5° to 60° the amplitude increases twelvefold while T increases by only 7%, so T is not proportional to amplitude.
CAt 5° and 10°, T matches 2π√(ℓ/g); at 60° it is about 7% longer than that.Correct At 5° and 10° the table gives 1.99 s, the value of 2π√(ℓ/g). At 60° it gives 2.13 s, and 2.13/1.99 ≈ 1.07, so the period is about 7% longer: the equation applies to small angles of swing.
DAt 60° the bob moves faster, so 2π√(ℓ/g) overestimates the period there. A student who thinks a larger swing has a shorter period because the bob moves faster picks this. The table shows the period at 60° is 2.13 s, longer than the 1.99 s the equation gives, so the equation underestimates it.
Working 2π√(1.00/10) = 1.99 s. At 5° and 10° the table gives 1.99 s, equal to the prediction; at 20°, 2.00 s (0.5% longer); at 60°, 2.13 s, which is 2.13/1.99 = 1.07 times, about 7% longer than predicted.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
