Syllabus Edition

First teaching 2023

First exams 2025

Simple Harmonic Motion (HL IB Physics)

Topic Questions

Easy

Medium

Hard

1 mark

Which equation is used for calculating the displacement as a function of time for an oscillator that begins its oscillation from the equilibrium position?

$x = x_{0} \sin ω t$
$v = ω x_{0} \cos ω t$
$a = - ω^{2} x_{0} \sin ω t$
$a = - ω x$

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1 mark

The following graph shows the acceleration as a function of time for a simple pendulum oscillating isochronously.

gz4QtOVG_9-1-ib-hl-hmcq-2-q-stem

Which of the following gives the times at which kinetic energy is at a maximum?

$\frac{T}{2}, T$
$\frac{T}{4}, \frac{3 T}{2}$
$\frac{T}{4}, \frac{T}{2}$
$\frac{3 T}{2}, T$

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1 mark

A point on a guitar string S oscillates about its equilibrium position Z in simple harmonic motion.

9-1-hl-mcq-medium-1-diagram

The amplitude of the oscillation is YZ.

Which positions show when the acceleration of point S is at a maximum and the velocity of point S is at zero?

	Acceleration	Velocity
A.	Z	Y
B.	Y	Y
C.	Z	Z
D.	Y	Z

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	X	Y	Z
A.	Max	Zero	Max
B.	Zero	Max	Zero
C.	Max	Max	Zero
D.	Zero	Zero	Max

	Force F	Spring constant k	Displacement x
A.	N	N m⁻¹	m
B.	Nm	N	m²
C.	N m⁻¹	Nm	ml
D.	N	N kg⁻¹	m

	Total energy	Period
A.	$\frac{E}{16}$	$\frac{T}{4}$
B.	$\frac{E}{8}$	$\frac{T}{2}$
C.	$\frac{E}{16}$	$T$
D.	$\frac{E}{4}$	$4 T$

A.	$x = r \cos ω t$	$v = - r ω \sin ω t$	$a = r ω^{2} \cos ω t$
B.	$x = r \cos ω t$	$v = - r ω \sin ω t$	$a = - r ω^{2} \cos ω t$
C.	$x = r \sin ω t$	$v = - r ω \cos ω t$	$a = r ω^{2} \sin ω t$
D.	$x = r \sin ω t$	$v = - r ω \cos ω t$	$a = - r ω^{2} \sin ω t$

a (mm s⁻²)	16	8	0	− 8	− 16
x (mm)	− 4	− 2	0	2	4

	Force	Acceleration	Velocity
A.	zero	zero	max
B.	max	max	zero
C.	max	zero	max
D.	zero	max	zero

	Simple Pendulum	Mass-spring System
A.	T	Greater than T
B.	T	T
C.	Greater than T	Greater than T
D.	Greater than T	T

	Acceleration	Displacement	Velocity
A.	Z	Y	X
B.	Y	X	Y
C.	X	Z	Z
D.	Y	X	Z