1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
A block of mass m attached to an ideal spring with spring constant k oscillates on a horizontal surface with negligible friction. Which expression gives the frequency f of the block's oscillation?
Answer and reasoning
A√(k/m)/2πCorrect The period is Ts = 2π√(m/k), and the frequency is its reciprocal: f = 1/(2π√(m/k)). The reciprocal of a product inverts both factors, so the 2π moves to the denominator and m/k becomes k/m: f = √(k/m)/2π.
B2π√(m/k) A student who confuses frequency with period gives the period equation. The frequency is the reciprocal of the period: f = 1/Ts = √(k/m)/2π.
Ck/(2πm) A student who drops the square root picks this. The period equation contains √(m/k), so its reciprocal contains √(k/m): the frequency is proportional to the square root of k/m, not to k/m itself.
D√(k/m) A student who leaves out the 2π picks this. The factor 2π is part of the period equation, so the frequency is √(k/m)/2π, about one-sixth of √(k/m).
Working Ts = 2π√(m/k) and f = 1/T, so f = 1/(2π√(m/k)) = √(k/m)/2π. Taking the reciprocal inverts both factors: the 2π moves to the denominator and the ratio under the root becomes k/m.
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 of SHM Fix
Period of SHM
The time T for an object in SHM to complete one full cycle, for example from one extreme position to the other and back again. Unit: s.
Frequency of SHM
The number of complete cycles an object in SHM makes per unit time, f. Unit: Hz, where 1 Hz = 1 cycle per second = 1 s⁻¹.
Period–frequency relationship
T = 1/f: period and frequency are reciprocals, so when one is multiplied by a factor the other is divided by the same factor. A long period means a low frequency.
Object–ideal-spring oscillator
An object attached to a spring of negligible mass that obeys Hooke's law, oscillating with no friction or air resistance. Its period is Ts = 2π√(m/k), where m is the object's mass and k the spring constant.
Dependence of the spring oscillator's period
Ts is proportional to √m and to 1/√k: four times the mass doubles the period, and a spring four times as stiff halves it. Ts does not depend on g, so the same object and spring have the same period whether they oscillate horizontally or vertically, on Earth or on the Moon.
Linearized graph of T² against m
Squaring Ts = 2π√(m/k) gives T² = (4π²/k)m, so a graph of T² against m for one spring is a straight line through the origin with slope 4π²/k. The spring constant is k = 4π²/slope. Unit of the slope: s²/kg.
Simple pendulum
A small bob on a light string that swings through a small angle about a fixed pivot. Its length ℓ is measured from the pivot to the center of the bob. Its period is Tp = 2π√(ℓ/g).
Dependence of the pendulum's period
Tp is proportional to √ℓ and to 1/√g and does not depend on the mass of the bob: a string four times as long doubles the period, and a location where g is smaller gives a longer period. The equation applies only for small angular displacements.
Gravitational field strength g
The gravitational force per unit mass at a location, which equals the acceleration of a freely falling object there. It is about 9.8 N/kg (9.8 m/s²) at Earth's surface, slightly smaller on a high mountain, and about 1.6 N/kg on the Moon.
Students often think Period and frequency change together: a factor that multiplies the period multiplies the frequency by the same factor, and a longer period means more cycles in a given time. In fact No. T = 1/f, so period and frequency change in opposite directions: doubling the period halves the frequency. A longer period means fewer cycles each second.
Students often think The period is proportional to m (for a spring oscillator) or to ℓ (for a pendulum), so the square root in the period equations can be ignored when predicting a change. In fact No. The period is proportional to the square root: Ts ∝ √m and Tp ∝ √ℓ. Four times the mass, or four times the length, doubles the period rather than quadrupling it.
7 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 7
A block attached to an ideal spring oscillates on a horizontal surface with negligible friction, with frequency f₀. The block is replaced by a block with four times the mass, and the new system is set oscillating on the same spring. By what factor is the frequency of the oscillation multiplied?
Answer and reasoning
A×2 A student who treats frequency like period picks this. Four times the mass does double the period, but f = 1/T, so the frequency is halved, not doubled: a longer period means fewer cycles each second.
B×¼ A student who leaves out the square root treats the period as proportional to the mass, so the frequency falls to one-quarter. Ts = 2π√(m/k) contains √m, so four times the mass doubles the period and halves the frequency.
C×1 A student who carries over the pendulum result that mass does not matter picks this. For an object on a spring, Ts = 2π√(m/k) does depend on the mass: the heavier block oscillates with a longer period and a lower frequency.
D×½Correct Ts = 2π√(m/k), so four times the mass multiplies the period by √4 = 2. Frequency is the reciprocal of period, T = 1/f, so the frequency is multiplied by ½: the heavier block completes half as many cycles each second.
Working Ts = 2π√(m/k), so Ts ∝ √m: four times the mass multiplies the period by √4 = 2. T = 1/f, so the frequency is multiplied by 1/2: the new frequency is f₀/2.
A block attached to an ideal spring oscillates on a horizontal surface with negligible friction, with period TH. The same block and spring are then hung vertically, and the block oscillates up and down with period TV. Air resistance is negligible. Which statement is correct?
Answer and reasoning
ATV is greater than TH, as the spring is stretched further when vertical. A student who thinks the period depends on how far the spring is stretched picks this. The extra stretch only sets the new equilibrium position; the period depends on m and k, which are the same in both set-ups.
BTV equals TH, as gravity shifts only the block's equilibrium position.Correct Gravity is a constant force: it moves the equilibrium position down to where the spring force balances the block's weight, but measured from there the net force is still −kΔx. The mass and spring constant are unchanged, so Ts = 2π√(m/k) gives the same period.
CTV is less than TH, as gravity adds to the spring's restoring force. A student who thinks gravity strengthens the restoring force picks this. Gravity pulls down with the same force whichever way the block is displaced, so it does not change the restoring force per unit displacement, k, and the period is unchanged.
DTV cannot be found from m and k, as gravity stops the motion from being SHM. A student who thinks an extra force breaks the SHM model picks this. Measured from the new equilibrium position, the net force on the block is −kΔx, so the vertical motion is SHM and TV = 2π√(m/k).
Students hang objects of different mass m from the same vertical ideal spring, set each one oscillating, and measure the period T. The graph shows T² as a function of m for their data, with a best-fit straight line through the origin. What is the spring constant of the spring?
Answer and reasoning
A0.80 N/m A student who takes the slope of the graph to be the spring constant picks this. The slope, 0.80 s²/kg, equals 4π²/k, as T² = (4π²/k)m shows; its unit, s²/kg, is not even the unit of k.
B1.3 N/m A student who leaves out the 2π writes T² = m/k and gets k = 1/0.80 = 1.3 N/m. Squaring Ts = 2π√(m/k) gives a factor of 4π², so k = 4π²/0.80 = 49 N/m.
C49 N/mCorrect The best-fit line passes through (0.50 kg, 0.40 s²), so its slope is 0.80 s²/kg. Squaring Ts = 2π√(m/k) gives T² = (4π²/k)m, so the slope equals 4π²/k and k = 4π²/0.80 = 49 N/m.
D0.020 N/m A student who rearranges slope = 4π²/k the wrong way divides the slope by 4π² and gets 0.80/39.5 = 0.020 N/m, which is 1/k. Since k is in the denominator, k = 4π²/slope.
Working Slope of the best-fit line = 0.40 s² / 0.50 kg = 0.80 s²/kg. Squaring Ts = 2π√(m/k) gives T² = (4π²/k)m, so the slope is 4π²/k and k = 4π²/slope = 39.5/0.80 = 49 N/m.
A block hangs at rest from a vertical ideal spring; in this position the spring is stretched a distance d beyond its relaxed length. The block is pulled down slightly and released, and it oscillates vertically. Air resistance is negligible. Which expression gives the period of the oscillation, in terms of d and g?
Answer and reasoning
A√(g/d)/2π A student who uses the frequency expression, √(k/m)/2π, as the period picks this. The period is the reciprocal of the frequency, Ts = 2π√(m/k) = 2π√(d/g); a check is that a softer spring, which stretches more (larger d), should give a longer period, and √(g/d)/2π gets smaller as d grows.
B2π√(d/g)Correct At equilibrium the spring's force balances the weight, kd = mg, so m/k = d/g. Substituting into Ts = 2π√(m/k) gives Ts = 2π√(d/g): the period can be found from the static stretch alone.
C√(d/g) A student who leaves the factor 2π out of the period equation picks this. The ratio m/k = d/g is right, but Ts = 2π√(m/k), so the period is 2π√(d/g), about six times √(d/g).
D2π(d/g) A student who treats the period as proportional to m/k, dropping the square root, gets 2π(d/g). The period depends on the square root of m/k, so Ts = 2π√(d/g); 2π(d/g) does not even have the unit of time.
Working At the equilibrium position the net force on the block is zero: kd = mg, so m/k = d/g. The period of a block–spring oscillator is Ts = 2π√(m/k), and gravity only shifts the equilibrium position, so Ts = 2π√(d/g).
The diagram shows two simple pendulums, A and B, labeled with the mass of each bob and the length of each light string. Both pendulums swing through small angles at the same location. What is the ratio TB/TA of the period of B to the period of A?
Answer and reasoning
A4.0 A student who treats the period as proportional to the length picks this. Tp contains √ℓ, so four times the length multiplies the period by √4 = 2.
B2.8 A student who expects the heavier bob to lengthen the period, as a heavier object on a spring would, multiplies by √2 for the mass as well: √(4 × 2) = 2.8. The pendulum's period does not depend on the bob's mass.
C1.4 A student who thinks the heavier bob swings faster divides by √2 for its mass: √(4/2) = 1.4. The larger weight of B's bob is matched by its larger mass, so the mass has no effect; only the length ratio matters.
D2.0Correct Tp = 2π√(ℓ/g) does not depend on the bob's mass, and g is the same for both pendulums. B's string is 4 times as long, so TB/TA = √4 = 2.0.
Working Tp = 2π√(ℓ/g), independent of the bob's mass; g is the same for both. TB/TA = √(4ℓ/ℓ) = √4 = 2.0. The bob of B having twice the mass does not affect the ratio.
A simple pendulum of length 2.5 m has a period of 3.1 s on Earth. Use g = 10 m/s². The pendulum is taken to the Moon, where g = 1.6 m/s², and set swinging through a small angle. What is the period of the pendulum on the Moon?
Answer and reasoning
A3.1 s A student who thinks a pendulum's period depends only on its length keeps the Earth value. Tp = 2π√(ℓ/g) also depends on g, which is about six times smaller on the Moon, so the period is longer.
B7.9 sCorrect Tp = 2π√(ℓ/g) = 2π√(2.5/1.6) = 2π × 1.25 s = 7.9 s. The Moon's weaker gravitational field gives a smaller restoring torque, so the same pendulum swings more slowly than on Earth.
C9.8 s A student who leaves out the square root calculates 2π(2.5/1.6) = 9.8 s. The period is proportional to √(ℓ/g): 2π√(2.5/1.6) = 7.9 s.
D1.3 s A student who leaves out the 2π calculates √(2.5/1.6) = 1.3 s. The factor 2π is part of the period equation, Tp = 2π√(ℓ/g), so the period is 2π times larger.
Working Tp = 2π√(ℓ/g) = 2π√(2.5 m / 1.6 m/s²) = 2π√(1.5625 s²) = 2π(1.25 s) = 7.9 s. (Check: with g = 10 m/s², the Moon's g is 10/1.6 = 6.25 times smaller, so the period is √6.25 = 2.5 times the Earth value; 2.5 × 3.14 s = 7.85 s ≈ 7.9 s, using the unrounded Earth period 2π√(2.5/10) = 3.14 s.)
A pendulum clock keeps correct time at sea level. The clock counts one tick for each period of its pendulum, which can be modeled as a simple pendulum swinging through a small angle. The clock is taken to the top of a high mountain, where g is slightly smaller than at sea level. Which statement about the clock on the mountain is correct?
Answer and reasoning
AIt runs slow, since the pendulum's period is longer where g is smaller.Correct Tp = 2π√(ℓ/g), so a smaller g gives a slightly longer period. Each tick then takes longer than it should, the clock counts fewer ticks each hour, and it falls behind: it runs slow.
BIt keeps correct time, since a pendulum's period depends only on its length. A student who thinks a pendulum's period depends only on its length picks this. Tp = 2π√(ℓ/g) depends on g as well, so the smaller g on the mountain lengthens the period and the clock loses time.
CIt runs fast, since a longer period means more swings each hour. A student who treats a longer period as a higher frequency picks this. The period is longer, but f = 1/T, so the pendulum completes fewer swings each hour and the clock runs slow.
DIt runs fast, since its bob weighs less and so swings more quickly. A student who takes the bob's smaller weight to mean it is easier to move picks this. The bob's mass is unchanged; only the pull of gravity on it is smaller, so the restoring torque is smaller while its inertia is the same. Tp = 2π√(ℓ/g) is longer where g is smaller, and the clock runs slow.
Compiled from the AP Physics 1 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
