Oscillations practice problems AP Physics 1, Unit 7

Written by STEM-Tutor's tutors · Updated October 2026 · 10 questions and 1 free-response question

Simple harmonic motion of masses on springs and pendulums: period, frequency, energy and graphs. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page

Key equations
Warm-up questions
Exam-style questions
Free-response practice
Common mistakes
Keep going

Key equations for oscillations

T = 2π√(m/k)	period of a mass on a spring
T = 2π√(L/g)	period of a simple pendulum, for small angles
f = 1/T	frequency and period
E = ½kA²	total energy of a mass-spring oscillator

All questions use g = 9.8 m/s², as on the AP Physics 1 equation sheet. A calculator is allowed, as in the exam. The full sheet is in our AP Physics 1 equation sheet.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-up

A 0.50 kg mass oscillates on a spring with k = 200 N/m. What is the period?

Hint

Use T = 2π√(m/k).

Worked solution

T = 2π√(0.50 / 200) = 2π√0.0025 = 2π × 0.050.
T = 0.314 s.

Q2Warm-up

What is the period of a simple pendulum 1.0 m long?

Hint

Use T = 2π√(L/g).

Worked solution

T = 2π√(1.0 / 9.8) = 2π × 0.319.
T = 2.01 s.

Q3Warm-up

A mass on a spring is in simple harmonic motion. At the equilibrium position, its:

Hint

What is the net force on the mass at equilibrium?

Worked solution

At equilibrium the net force is zero, so the acceleration is zero.
All the energy is kinetic there, so the speed is at its maximum.
The answer is B. At the turning points it is the other way round.

Q4Warm-up

An oscillator has a period of 0.25 s. What is its frequency?

Hint

Frequency is the number of cycles each second.

Worked solution

f = 1/T = 1 / 0.25.
f = 4.0 Hz.

Exam-style questions

Questions at the level of the AP Physics 1 exam. The ones marked Challenge combine more than one idea.

Q5Exam-style

The mass on a spring oscillator is doubled. What happens to the period?

Hint

T depends on the square root of m.

Worked solution

T = 2π√(m/k), so T is proportional to √m.
Doubling m multiplies T by √2, about 1.41.
The answer is B.

Q6Exam-style

A 0.50 kg mass on a spring (k = 50 N/m) oscillates with amplitude 0.20 m. What is its maximum speed?

Hint

At equilibrium, all of the spring energy ½kA² has become kinetic energy.

Worked solution

½kA² = ½mv_max², so v_max = A√(k/m).
v_max = 0.20 × √(50 / 0.50) = 0.20 × 10.
v_max = 2.0 m/s.

Q7Exam-style

For the oscillator in the previous question (k = 50 N/m, amplitude 0.20 m), what is its total mechanical energy?

Hint

At the amplitude the mass is momentarily at rest, so all the energy is stored in the spring.

Worked solution

E = ½kA² = ½ × 50 × 0.20².
E = 1.0 J. Check: ½ × 0.50 × 2.0² = 1.0 J too.

Q8Exam-style

A pendulum clock is taken to the Moon, where g is about one sixth of its value on Earth. How does its period change?

Hint

T depends on 1/√g.

Worked solution

T = 2π√(L/g), so T is proportional to 1/√g.
Dividing g by 6 multiplies T by √6, about 2.45.
The answer is B: the clock runs slow.

Q9Exam-style

What length of simple pendulum has a period of 1.5 s?

Hint

Rearrange T = 2π√(L/g) for L.

Worked solution

Square both sides: T² = 4π²L/g, so L = gT² / 4π².
L = 9.8 × 1.5² / 39.48 = 22.05 / 39.48.
L = 0.559 m, about 56 cm.

Q10Challenge

Hanging a mass on a vertical spring stretches it by 4.9 cm. The mass is then pulled down and released. What is the period of its oscillation?

Hint

You do not need the mass or k separately: the stretch tells you m/k.

Worked solution

At equilibrium kx = mg, so m/k = x/g = 0.049 / 9.8 = 0.0050 s².
T = 2π√(m/k) = 2π√0.0050.
T = 0.444 s.

Free-response practice

One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.

Experimental design and analysis style · 6 points · about 20 minutes

A student wants to find the value of g using a simple pendulum, a stopwatch, a metre rule and string of adjustable length.

(a) Describe a procedure that would let the student find g, including how to reduce uncertainty in the timing.
(b) Which quantities should the student graph to get a straight line, and how is g found from it?
(c) The student's slope is 4.10 s²/m. Calculate g.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Set the length L (to the centre of the bob) with the metre rule. Release the bob from a small angle (under about 10°). Time 10 full oscillations and divide by 10 to find T. Repeat for at least five different lengths, and repeat each timing to average out reaction time.

1 point for: Measures L and T for several different lengths
1 point for: Times many oscillations and divides, or repeats and averages, to reduce timing uncertainty
1 point for: Uses small angles so the period formula applies

Part (b)

From T = 2π√(L/g), T² = (4π²/g)L. Plot T² against L: a straight line through the origin with slope 4π²/g, so g = 4π² / slope.

1 point for: Graphs T² against L (or an equivalent linearisation)
1 point for: g = 4π² / slope

Part (c)

g = 4π² / 4.10 = 39.48 / 4.10 = 9.63 m/s².

1 point for: About 9.6 m/s², with units

Your mark: 0 of 6

Common mistakes in oscillations

Thinking the period of a pendulum depends on its mass. It does not.
Thinking the period depends on the amplitude. For small angles it does not.
Measuring a pendulum's length to the top of the bob instead of its centre.
Confusing the period (seconds per cycle) with the frequency (cycles per second).

Keep going

Revise the ideas behind these questions:

Then try real exam questions: AP Physics 1 past papers, with official scoring guidelines. · All AP Physics 1 units