PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Oscillations and simple harmonic motion: the PHYS 101 revision sheet

Q: Does the amplitude change the period of an oscillator?

No, and this is the most commonly tested fact of the chapter. The period of a mass on a spring depends only on the mass and the stiffness of the spring, and the period of a pendulum only on its length and on gravity. Pulling the mass twice as far doubles the maximum speed and multiplies the energy by four, but the time for one complete cycle does not move at all.

Q: Why does a heavier bob not slow a pendulum down?

Because the mass affects the pull and the resistance to being moved in exactly the same proportion. Doubling the bob doubles the force bringing it back towards the centre, and doubles the inertia that force has to overcome, so the two cancel. It is the same cancellation that makes two objects of different masses fall side by side, and it is why the mass does not appear in the pendulum period formula.

Q: Should I use sine or cosine for the position equation?

Look only at where the object is when the stopwatch starts. Released from rest at maximum elongation, the position equals the amplitude at time zero, so you need the cosine. Launched from the equilibrium position with a push, the position is zero at time zero, so you need the sine. The choice describes the launch, never the system, and it changes nothing about the period.

Q: Does hanging a spring vertically change its period?

No. Gravity moves the centre of the oscillation down by an amount equal to the weight divided by the spring constant, and does nothing else. Measured from that new resting position, the resultant force is again proportional to the displacement with the same spring constant, so the period formula gives exactly the same number as on a horizontal table.

Q: What exactly happens at resonance?

A driven oscillator responds most strongly when the driving frequency is close to its own natural frequency. Each push then arrives in step with the motion and adds energy cycle after cycle, so the steady amplitude climbs to a peak. Damping is what keeps that peak finite: the heavier the damping, the lower and the broader the peak, and the slightly lower the frequency at which it sits.

Q: How do I find the phase constant of a simple harmonic motion?

Put time zero into the position equation: the sine of the phase equals the starting position divided by the amplitude. Your calculator gives one angle, but in every cycle there is a second one, pi minus the first. Keep the angle whose cosine has the same sign as the starting velocity: positive if the object starts moving in the positive direction, negative if it starts moving the other way.

This revision sheet is not a summary of the oscillations chapter: you already have the course notes. It answers one question only, what loses marks on this chapter in PHYS 101 at McGill University, and what gesture prevents each loss. Everything here works with algebra and proportions, never with a derivative, exactly as the course requires.

One idea carries the whole sheet. Amplitude and phase describe how the oscillator was LAUNCHED; period, frequency and angular frequency describe the SYSTEM. Pull the mass twice as far, give it a push as you release it, swap the pendulum bob for a heavier one, and $T$ does not move. Sort every quantity of a problem into those two families before you write anything and most of the traps below stop working on you.

The formula sheet you bring to the exam writes the motion $x = A\sin(\omega t + \varphi)$ . It gives no rule for the phase, and that is where marks go: the arcsine has two answers and only the sign of the starting velocity picks one. This sheet adds that gesture, the amplitude of a launch with a push, and the reference circle, a point turning on a circle whose shadow does the oscillation and hands you $v_{max} = A\omega$ and $a_{max} = A\omega^{2}$ without any calculus.

Updated on October 25, 2026

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The thread of the chapter

The period belongs to the SYSTEM, never to the launch: amplitude, initial push and, for a pendulum, the mass of the bob all leave $T$ untouched, and the exam asks you that question in some disguise every single time.

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Simple harmonic motion
Hooke's law
Reference circle
Amplitude and phase constant
Period and frequency
Angular frequency
Elastic potential energy
Mass-spring oscillator
Simple pendulum
Small-angle approximation
Damped oscillations
Resonance

The essentials

On your formula sheet: this chapter's lines, and when each one holds

• $F=-kx$ : The spring pulls or pushes back toward its natural length: the minus sign is a direction, the magnitude is $kx$ .
• $f=1/T$ : $f$ in Hz counts cycles per second; $T$ is the duration of ONE cycle.
• $\omega=2\pi f=2\pi/T$ : $\omega$ in rad/s is not $f$ in Hz: the factor $2\pi$ is the most common slip of the chapter.
• $\omega_{spring}=\sqrt{k/m}$ : Independent of the amplitude and of $g$ : the same spring oscillates at the same rate horizontally, vertically or on the Moon.
• $\omega_{pendulum}=\sqrt{g/L}$ : Small angles only, below about $15^{\circ}$ ; independent of the mass of the bob.
• $x_{max}=A$ : $A$ is measured from equilibrium, not from one end of the motion to the other.
• $v_{max}=A\omega$ : Reached at equilibrium, where the acceleration is zero.
• $a_{max}=A\omega^{2}$ : Reached at the turning points, where the speed is zero.
• $x=A\sin(\omega t+\phi)$ : $\phi$ is fixed by the position AND the direction of motion at $t=0$ : the arcsine gives two candidates, the sign of the velocity chooses.

The sheet prints each equation, never the condition under which it holds: copy the condition next to its line on your own sheet before the exam.

Two families of quantities, and the question that separates them

• The LAUNCH: the amplitude $A$ , that is the largest elongation measured from equilibrium, and the phase, which says where in the cycle your stopwatch was started. You choose both when you release the object.
• The SYSTEM: the period $T$ , the frequency $f = 1/T$ and the angular frequency $\omega = 2\pi f = 2\pi/T$ . Only the stiffness of the restoring force and the inertia can change them.
• Everything else is built from one of each: $v_{max} = \omega A$ , $a_{max} = \omega^{2}A$ , $E = \frac{1}{2}kA^{2}$ . Change the launch and these three change; the clock does not.
• The exam question is always the same one in disguise: what happens to $T$ if I double the amplitude, add a push, use a heavier bob, hang it vertically. The answer is nothing, four times out of four.

Two launches of the same oscillator, 8 cm and 4 cm, one released from the maximum, the other pushed from the centre. Both repeat every

0.8

s, the gap between the grey lines; only height and start differ.

Write the formula for $T$ before answering any qualitative question and look at which letters are actually in it. A letter that is absent cannot change the answer, and saying so explicitly is what earns the mark.

The condition, and the five relations that come out of it

• Condition: the resultant force is proportional to the displacement and opposite to it, $F = -kx$ . Nothing else is needed, and any system that satisfies it oscillates sinusoidally.
• Equations given by the course, never derived here: $x = A\cos(\omega t)$ from a release at maximum elongation, $x = A\sin(\omega t)$ from a launch at the centre, $v = -\omega A\sin(\omega t)$ , and $a = -\omega^{2}x$ .
• Extremes: $v_{max} = \omega A$ at $x = 0$ , and $|a|_{max} = \omega^{2}A$ at $x = \pm A$ . They are a quarter of a cycle apart and never happen together.
• Speed at any elongation, with the time eliminated: $v = \omega\sqrt{A^{2}-x^{2}}$ . Energy: $E = \frac{1}{2}kA^{2} = \frac{1}{2}kx^{2}+\frac{1}{2}mv^{2}$ .
• The argument $\omega t$ is in RADIANS. Everything in this chapter assumes radian mode.

The same block at three points of its motion. At the ends the pink spring force is largest and points back to the centre, while the block is momentarily at rest; at the centre the force is zero and the speed is largest.

The three periods to know cold: $T = 2\pi\sqrt{m/k}$ for a mass on a spring, horizontal or vertical; $T = 2\pi\sqrt{L/g}$ for a simple pendulum at small angles; $T = 2\pi\sqrt{d/g}$ when a vertical spring has been measured with a ruler and $d = mg/k$ is its static stretch.

The simple pendulum: the restoring force is mg sin θ

• Two forces act on the bob: the tension along the string and the weight. Split the weight along the string, $mg\cos\theta$ , and along the arc, $mg\sin\theta$ .
• Only $mg\sin\theta$ pulls the bob back. For small angles $\sin\theta \approx \theta = \frac{s}{L}$ , so the restoring force is $-\frac{mg}{L}s$ : the condition $F = -kx$ with $k = \frac{mg}{L}$ .
• Hence $T = 2\pi\sqrt{\frac{L}{g}}$ : the mass cancels, and the amplitude does not appear as long as it stays small.

The angle is exaggerated for clarity. Along the string,

T

and

mg\cos\theta

; along the arc,

mg\sin\theta

alone, always pointing back toward the lowest point.

Small means about $10^{\circ}$ or less: there $\sin\theta$ and $\theta$ differ by $0.5$ per cent, at $30^{\circ}$ by $4.5$ per cent.

The formula sheet's form: phase constant and reference circle

• The formula sheet writes every simple harmonic motion as $x = A\sin(\omega t + \varphi)$ , and the velocity that goes with it is $v = \omega A\cos(\omega t + \varphi)$ . $A$ and $\varphi$ are the launch; $\omega$ is the system.
• Special launches: through the centre towards $+x$ , $\varphi = 0$ ; released from rest at $x = +A$ , $\varphi = \pi/2$ , which is the cosine of the course, since $A\sin(\omega t + \pi/2) = A\cos(\omega t)$ ; released from rest at $x = -A$ , $\varphi = 3\pi/2$ .
• Any other launch, from $x_{0}$ with a speed $v_{0}$ : the energy gives $A = \sqrt{x_{0}^{2}+\frac{m}{k}v_{0}^{2}}$ ; then $\sin\varphi = x_{0}/A$ has two solutions per cycle, and the sign of $v_{0}$ , which is the sign of $\cos\varphi$ , keeps one.
• Reference circle: a point turning at $\omega$ on a circle of radius $A$ , seen edge-on, moves exactly like the oscillator. The formula sheet's $v = \omega r$ and $a = v^{2}/r$ give its speed $\omega A$ and its acceleration $\omega^{2}A$ : they are $v_{max}$ and $a_{max}$ , and the angle of the point at $t = 0$ is $\varphi$ .

The point turns at

\omega

; its shadow sits at

A\sin\theta

\theta = \omega t + \varphi

. It keeps the vertical parts of

v = \omega A

and

a = \omega^{2}A

: all of

v

at the centre, all of

a

at the ends.

Quadrant check: $\varphi$ between $0$ and $\pi/2$ , the object starts on the positive side moving outwards; between $\pi/2$ and $\pi$ , positive side moving back in; between $\pi$ and $3\pi/2$ , negative side moving out; between $3\pi/2$ and $2\pi$ , negative side moving back in.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Doubling the amplitude to change the period

the whole question, 3 or 4 marks

What not to write

“The mass is pulled twice as far, so it has twice as far to travel, so the period doubles.”

What to write

“ $T = 2\pi\sqrt{m/k}$ contains neither $A$ nor the way the mass was released, so the period is unchanged at $0.99$ s. What doubles is $v_{max} = \omega A$ , and the energy $\frac{1}{2}kA^{2}$ is multiplied by four.”

Why: Twice as far also means twice the restoring force at every point of the path, so the object covers the longer route proportionally faster. The cancellation is exact, and it is what makes an oscillator a clock.

2. Half the amplitude, half the speed

2 marks, and every later number of the problem

What not to write

“At $x = A/2$ the object is halfway, so it is going at half of its maximum speed.”

What to write

“ $v = \omega\sqrt{A^{2}-x^{2}}$ , so at $x = A/2$ the speed is $\omega A\sqrt{1-\frac{1}{4}} = 0.87\,v_{max}$ . At half the amplitude the object still has $87\%$ of its top speed.”

Speed against elongation, both as fractions of their maximum. The true relation is the quarter circle: at half the amplitude it reads

0.87

, while the straight-line guess drawn dashed reads

0.50

Why: The energy goes with the SQUARE of the elongation, so at half the amplitude only a quarter of the energy has been handed back to the spring and three quarters is still kinetic. Speed against elongation is a quarter circle, not a straight line.

3. No speed at the turning point, so no acceleration

2 marks, and it makes the following part impossible

What not to write

“At $x = A$ the object has stopped, so $v = 0$ and $a = 0$ .”

What to write

“At $x = A$ the velocity is zero for one instant, but $a = -\omega^{2}x$ gives the LARGEST acceleration of the whole motion, $\omega^{2}A$ , directed back towards the centre.”

Why: If both were zero the object would stay at the end of its path forever. Stopping and being unaccelerated are two different statements, and the turning point of an oscillation is the standard example that separates them.

4. The calculator left in degree mode

every numerical value of the question

What not to write

“ $x = 0.12\cos(12.6 \times 0.10) = 0.12\cos(1.26) = 0.1200$ m.”

What to write

“In radian mode, $\cos(1.26) = 0.306$ , so $x = 0.037$ m. The argument $\omega t$ is an angle in radians because $\omega$ is in rad/s.”

Why: In degree mode a small argument gives a cosine very close to 1, so the answer comes out suspiciously near the amplitude and looks plausible. Only the mode itself, and a check at $t = T/2$ , reveal it.

5. Sine written where the launch calls for a cosine

1 to 2 marks, more when a value at a given time is then asked

What not to write

“The motion is simple harmonic, so $x = A\sin(\omega t)$ .”

What to write

“The object is released from rest at maximum elongation, so $x = A$ at $t = 0$ , and the function that starts at its own maximum is the cosine: $x = A\cos(\omega t)$ .”

Why: Sine and cosine differ only by where the stopwatch was started, which is the phase, a LAUNCH quantity. Test $t = 0$ against the situation described in the question and the choice is made in five seconds. On the formula sheet the cosine is simply $A\sin(\omega t + \pi/2)$ : choosing between sine and cosine IS choosing the phase.

6. A heavier bob for a slower pendulum

the whole question, and it is asked almost every term

What not to write

“The bob is twice as heavy, so it is harder to move and the period increases.”

What to write

“ $T = 2\pi\sqrt{L/g}$ contains no mass at all, so the period is unchanged. Only the length and the local $g$ can change it.”

Why: A heavier bob is pulled back by a proportionally larger force and needs a proportionally larger force to be accelerated, exactly as in free fall. Do not import this from the spring, where the mass genuinely does appear.

7. The weight carried through a vertical spring problem

2 to 3 marks, and an answer that no check can rescue

What not to write

“The spring is vertical, so the energy is $\frac{1}{2}kx^{2}+mgx+\frac{1}{2}mv^{2}$ and the resultant force is $kx - mg$ .”

What to write

“Measuring $x$ from the HANGING position, the resultant is $k(d+x)-mg = kx$ because $kd = mg$ , and the energy balance is $\frac{1}{2}kA^{2} = \frac{1}{2}kx^{2}+\frac{1}{2}mv^{2}$ with no gravity term.”

Why: Gravity is a constant force: it moves the centre of the oscillation down by $d = mg/k$ and does nothing else. Once you measure from that new centre it has already been counted, and adding it again counts it twice.

8. The angular frequency written as one over T

1 mark here, then every speed and acceleration of the problem

What not to write

“ $T = 0.80$ s, so $\omega = 1/0.80 = 1.25$ rad/s.”

What to write

“ $f = 1/T = 1.25$ Hz and $\omega = 2\pi f = 2\pi/T = 7.85$ rad/s. The frequency counts cycles, the angular frequency counts radians, and one cycle is $2\pi$ radians.”

Why: Both are called a frequency and both are written with an f sound in speech, so the two get mixed. The damage is a factor of $6.28$ on $v_{max}$ and of $39.5$ on $a_{max}$ , which is large enough to spot at the end.

9. Damping described as a slowing down

2 marks, and it wrecks the resonance part that follows

What not to write

“The oscillation is damped, so each swing takes longer than the one before until the object stops.”

What to write

“Damping shrinks the AMPLITUDE by a constant factor each cycle, here $0.80$ , while the period stays $0.50$ s from the first cycle to the last. The energy, going as the square, falls by $36\%$ per cycle.”

Why: Light damping does lengthen the period, but by far too little to read on a trace. Keeping the period fixed is also what makes resonance intelligible: the system still has one natural frequency to be driven at.

10. A push at release, and the amplitude read off the release point

2 marks, then the phase, the maximum speed and the energy

What not to write

“The block is let go at $6.0$ cm, so $A = 6.0$ cm.”

What to write

“It is let go with a speed, so it goes further: $\frac{1}{2}kA^{2} = \frac{1}{2}kx_{0}^{2}+\frac{1}{2}mv_{0}^{2}$ gives $A = \sqrt{x_{0}^{2}+\frac{m}{k}v_{0}^{2}} = \sqrt{0.0036+0.0064} = 0.100$ m, that is $10.0$ cm.”

Why: The amplitude is a turning point, where the speed is zero. A block that still moves at the release has not reached its turning point, so the amplitude is always LARGER than $x_{0}$ , and only the energy says by how much. Here $k = 200$ N/m, $m = 0.50$ kg, $x_{0} = 6.0$ cm and $v_{0} = -1.60$ m/s.

11. The calculator's arcsine taken as the phase

1 to 2 marks, and every position computed afterwards

What not to write

“ $\sin\varphi = 0.600$ , so $\varphi = \arcsin(0.600) = 0.644$ rad.”

What to write

“ $\sin\varphi = 0.600$ gives $\varphi = 0.644$ rad or $\pi - 0.644 = 2.50$ rad. The block starts towards the centre, so $v_{0} = \omega A\cos\varphi < 0$ , $\cos\varphi < 0$ , and $\varphi = 2.50$ rad.”

Both curves start at

x = 6.0

cm with

A = 10.0

cm. The solid one,

\varphi = 2.50

rad, heads for the centre as the push requires; the calculator's

0.644

rad, dashed, moves away first.

Why: The calculator returns an angle between $-\pi/2$ and $\pi/2$ , where the cosine is positive: a launch towards $+x$ , every time. Half of all launches go the other way, and only the sign of the velocity tells the two angles apart.

Which method to choose

Which period formula, read off the set-up

Look at what is attached to what, and at which numbers the question gives you. The period is decided before any launch detail is read.

Read the set-up before the numbers: a block on a spring, lying flat or hanging, keeps

T = 2\pi\sqrt{m/k}

with the same

k

; only a bob on a string swaps the spring for gravity,

T = 2\pi\sqrt{L/g}

If a block attached to a spring, horizontal on a table → $T = 2\pi\sqrt{m/k}$

Example: $m = 0.60$ kg on $k = 24$ N/m gives $T = 0.99$ s

the amplitude in the question is there for the speed and the energy, not for the period
If the same block hanging from the same spring, vertically → $T = 2\pi\sqrt{m/k}$ again, with the same $k$ , the centre of the motion having moved down by $d = mg/k$

Example: $m = 0.25$ kg on $k = 40$ N/m gives $0.497$ s hanging as well as lying flat

gravity changes where the oscillation happens, never how fast
If a vertical spring whose static stretch $d$ is given but not $k$ → $T = 2\pi\sqrt{d/g}$

Example: $d = 8.0$ cm gives $T = 2\pi\sqrt{0.080/9.8} = 0.568$ s, with neither $m$ nor $k$ needed
If a bob on a string or a light rod, swinging through a small angle → $T = 2\pi\sqrt{L/g}$ , with no mass anywhere

Example: $L = 0.60$ m gives $1.55$ s whatever the bob
If two springs on the same block → side by side, $k = k_{1}+k_{2}$ ; end to end, $\frac{1}{k} = \frac{1}{k_{1}}+\frac{1}{k_{2}}$ , then the usual formula

Example: $300$ and $600$ N/m give $900$ N/m side by side and $200$ N/m end to end

end to end the answer must be SMALLER than either spring; if it is not, the reciprocals were not inverted back

No branch mentions the amplitude, the initial speed or the release angle, and that is the point of the tree. If the only thing a question changes is one of those, the period has already been answered: unchanged.

Which relation gives a speed, and which gives a position

Read what the question hands you: an elongation, an instant, or an energy. Each one has its own relation, and mixing them is what produces impossible numbers.

If the speed at a NAMED elongation $x$ → $v = \omega\sqrt{A^{2}-x^{2}}$ , or the energy balance, which gives the same thing

Example: $A = 0.10$ m, $x = 0.05$ m, $\omega = 20$ rad/s gives $1.73$ m/s
If the maximum speed, or the speed at the centre → $v_{max} = \omega A$

Example: $\omega = 20$ rad/s and $A = 0.10$ m give $2.00$ m/s
If the position or the speed at a given TIME → the time equations, $x = A\cos(\omega t)$ or $A\sin(\omega t)$ according to the launch, and $v = -\omega A\sin(\omega t)$ , calculator in radians

Example: $A = 0.12$ m, $\omega = 12.6$ rad/s, $t = 0.10$ s give $x = 0.037$ m
If an energy, or a speed asked from an energy → $E = \frac{1}{2}kA^{2}$ , then split it as $\frac{1}{2}kx^{2}+\frac{1}{2}mv^{2}$

Example: $k = 200$ N/m and $A = 0.10$ m give $E = 1.00$ J and $v_{max} = 2.00$ m/s

the two forms are equal at $x = A/\sqrt{2} \approx 0.71\,A$ , not at $A/2$
If the acceleration anywhere → $a = -\omega^{2}x$ , so it is proportional to $x$ and largest at the ends

Example: at half the amplitude the acceleration is exactly half of $a_{max}$ , while the speed is $0.87\,v_{max}$
If the phase constant, after a launch from $x_{0}$ with a speed $v_{0}$ → $A$ from the energy, then $\sin\varphi = x_{0}/A$ , and the sign of $v_{0} = \omega A\cos\varphi$ to choose between $\varphi$ and $\pi - \varphi$

Example: $x_{0} = 6.0$ cm, $v_{0} = -1.60$ m/s, $\omega = 20$ rad/s give $A = 10.0$ cm and $\varphi = 2.50$ rad

the calculator only ever returns the angle of a launch towards $+x$

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Answering a qualitative period question

When to use it: The question changes one thing and asks what happens to the period, the frequency or the time per swing. It is worth three or four marks and it is answered in four lines.

1 Name the system and write the period formula that belongs to it, $T = 2\pi\sqrt{m/k}$ or $T = 2\pi\sqrt{L/g}$ , before touching the numbers.
2 List out loud the letters the formula contains, and say that the quantity the question changed is not one of them, or is one of them.
3 If it is one of them, give the proportionality rather than recomputing from scratch: $T \propto \sqrt{m}$ , so four times the mass doubles the period.
4 State what DOES change as a consequence of the change made, since the marker is checking that you have not confused unchanged with unaffected: $v_{max}$ , $a_{max}$ and the energy usually move.
5 Conclude in one sentence naming the two quantities that set the period.

Concluding sentence

“The period is $T = 2\pi\sqrt{m/k}$ , in which the amplitude does not appear: it is therefore unchanged at $0.99$ s. The maximum speed $v_{max} = \omega A$ doubles to $0.94$ m/s and the energy $\frac{1}{2}kA^{2}$ is four times larger.”

The trap: Answering only with the word unchanged. Half the marks sit in the reason, that is in naming the letters present in the formula, and the other half in what does change.

Marking: 1 mark for the formula, 1 for the absent letter named explicitly, 1 or 2 for the quantities that do change.

Getting a speed out of an energy balance

When to use it: The question gives an elongation and asks for a speed, or the reverse, with no time anywhere in the wording. Do not reach for the time equations.

1 Write the total energy at the most convenient instant, usually the release: $E = \frac{1}{2}kA^{2}$ , since the object is at rest at $x = A$ .
2 State that there is no friction, therefore $E$ is the same at the second instant, and write the balance $\frac{1}{2}kA^{2} = \frac{1}{2}kx^{2}+\frac{1}{2}mv^{2}$ .
3 Solve for $v$ , keeping the substitution symbolic one line longer than feels necessary so the marker can follow the isolation.
4 Check the answer against a limiting case you already know: at $x = 0$ it must return $v_{max} = \omega A$ , and at $x = A$ it must return zero.

Concluding sentence

“With no friction the mechanical energy is conserved, so $\frac{1}{2}kA^{2} = \frac{1}{2}kx^{2}+\frac{1}{2}mv^{2}$ , which gives $v = \sqrt{\frac{k}{m}(A^{2}-x^{2})} = \omega\sqrt{A^{2}-x^{2}} = 1.73$ m/s.”

The trap: Forgetting that the elastic energy at $x$ is not zero. Subtracting nothing there is what produces the famous $v = v_{max}/2$ at half the amplitude.

Marking: 1 mark for the conserved total, 1 for the complete balance, 1 for the isolation, 1 for the value with its unit.

Finding the phase constant from a launch

When to use it: The question says where the object is and how fast it moves at $t = 0$ , and asks for $x(t)$ in the form of the formula sheet, $x = A\sin(\omega t + \varphi)$ .

1 Compute $\omega$ from the system, $\omega = \sqrt{k/m}$ or $\sqrt{g/L}$ , before reading the launch.
2 Get $A$ from the energy, $\frac{1}{2}kA^{2} = \frac{1}{2}kx_{0}^{2}+\frac{1}{2}mv_{0}^{2}$ , never from the release point alone.
3 Write $\sin\varphi = x_{0}/A$ and give BOTH solutions in a cycle, $\varphi_{1}$ from the calculator and $\pi - \varphi_{1}$ .
4 Write the sign of $v_{0} = \omega A\cos\varphi$ that the wording demands, and keep the solution whose cosine has that sign.
5 Put $t = 0$ back into $x$ and $v$ : both must return the launch, with its sign.

Concluding sentence

“ $\sin\varphi = \frac{0.060}{0.100} = 0.600$ gives $\varphi = 0.644$ rad or $\pi - 0.644 = 2.50$ rad. The block starts towards the centre, so $v_{0} = \omega A\cos\varphi < 0$ requires $\cos\varphi < 0$ : $\varphi = 2.50$ rad, and $x = 0.100\sin(20t + 2.50)$ in metres.”

The trap: Writing only the calculator's angle. The marker reads that line as a launch towards $+x$ , whatever the question said.

Marking: 1 mark for $A$ from the energy, 1 for the two candidates, 1 for the choice justified by the sign of $v_{0}$ , 1 for the equation with its units.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The unit inside the square root

Check $m/k$ before trusting a period: kilograms divided by newtons per metre is $\text{kg}\cdot\text{m}/\text{N}$ , and one newton is $\text{kg}\cdot\text{m}/\text{s}^{2}$ , so the fraction is in s $^{2}$ and its root is a time. Do the same with $L/g$ and with $d/g$ .

A period that comes out at $40$ instead of $0.99$ is the signature of $k/m$ written upside down, since $\sqrt{24/0.60} = 6.3$ and the $2\pi$ then brings it to $39.7$ .
The two limiting cases of the speed formula

Put $x = 0$ into whatever expression you obtained for the speed: it must return $\omega A$ . Then put $x = A$ : it must return zero. Two substitutions, five seconds, and they catch a sign error or a swapped $A$ and $x$ immediately.

$v = \omega\sqrt{x^{2}-A^{2}}$ fails on the spot, since $x = 0$ would ask for the root of a negative number.
Radian mode, tested at half a period

Before any calculation with $\cos(\omega t)$ , evaluate it at $t = T/2$ . The answer must be exactly $-1$ , since half a period is $\pi$ radians. Anything else means the calculator is in degrees.

With $\omega = 11.07$ rad/s and $T/2 = 0.284$ s, radian mode gives $\cos(3.14) = -1.000$ and degree mode gives $0.9985$ , a value that would then pass unnoticed into the rest of the page.
The order of magnitude of a period

A laboratory spring with a few hundred grams oscillates in a few tenths of a second, and a pendulum around a metre long takes about two seconds. Compare your answer with those two anchors before writing it down.

A pendulum period of $12$ s means a length of $36$ m, which no laboratory has: the mistake is almost always $L$ left in centimetres.
Energy counted twice

In a vertical spring problem, look for a $mgh$ term in your energy balance. If $x$ is measured from the hanging position, that term must not be there. If it is, delete it and start the balance again.

$\frac{1}{2}kA^{2} = 0.0613$ J against $\frac{1}{2}mv_{max}^{2} = 0.0613$ J: the two agree only when gravity has been left out of both sides.
A result in letters: its unit, then one letter sent to zero

When the question asks for an answer in letters, check that every term carries the unit of the quantity, then send one letter to zero and see whether you recover a case you already know. Two lines, and they catch a missing square or a swapped $m$ and $k$ .

$A = \sqrt{x_{0}^{2}+\frac{m}{k}v_{0}^{2}}$ : $\frac{m}{k}v_{0}^{2}$ is in $\text{s}^{2}\cdot\text{m}^{2}/\text{s}^{2} = \text{m}^{2}$ ; $v_{0} \to 0$ returns $A = x_{0}$ , the release from rest, and $x_{0} \to 0$ returns $v_{0} = \omega A$ .
The phase put back at t = 0

Once $\varphi$ is chosen, evaluate $A\sin\varphi$ and $\omega A\cos\varphi$ . The first must return the starting position, the second the starting velocity WITH its sign. A sign that comes out wrong means you kept the calculator's angle instead of $\pi$ minus it.

$\varphi = 2.50$ rad gives $0.100\sin(2.50) = 0.060$ m and $20 \times 0.100 \times \cos(2.50) = -1.60$ m/s, as launched; $0.644$ rad gives $+1.60$ m/s, the wrong way.

The typical problem, taken apart

A full vertical spring laboratory, from a ruler to a maximum speed

A spring hangs from a stand. A mass $m = 0.40$ kg is attached to its lower end and, once at rest, the spring is found to be $8.0$ cm longer than when it was unloaded. The mass is then pulled $5.0$ cm below that resting position and released from rest. Take $g = 9.8$ m/s $^{2}$ .

Find the spring constant, the period, the maximum speed and the maximum acceleration, and check that the motion really is simple harmonic throughout.

The three states of the experiment: the unloaded spring, the mass hanging at rest

d

lower, and the mass pulled a further

A

below. The oscillation happens about the middle position, not about the unloaded one.

Step 1

At rest the spring force balances the weight: $kd = mg$ , so $k = \frac{mg}{d} = \frac{0.40 \times 9.8}{0.080} = 49.0$ N/m.

Why

The static stretch measures the SPRING, not the motion. One ruler reading and one known mass give $k$ before any timing, and this is the step markers most often see missing.

Step 2

$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.40}{49.0}} = 2\pi \times 0.0904 = 0.568$ s, so $f = 1.76$ Hz and $\omega = \frac{2\pi}{T} = 11.07$ rad/s.

Why

Notice that nothing about the launch has been read yet. The period is fixed by the system alone, so it can be written before the question even says how far the mass is pulled.

Step 3

Check with the ruler alone: $T = 2\pi\sqrt{\frac{d}{g}} = 2\pi\sqrt{\frac{0.080}{9.8}} = 0.568$ s.

Why

The two routes must agree, because $d/g = m/k$ is exactly what $kd = mg$ says. An independent check costing one line, and it catches a mass entered in grams.

Step 4

The launch: released from rest $5.0$ cm below equilibrium, so $A = 0.050$ m. Then $v_{max} = \omega A = 11.07 \times 0.050 = 0.553$ m/s, reached as the mass passes through the hanging position, and $|a|_{max} = \omega^{2}A = 122.5 \times 0.050 = 6.13$ m/s $^{2}$ , reached at the two ends of the travel.

Why

This is where the launch finally enters, and it enters only through $A$ . Releasing it from rest is also what makes the release point the amplitude; a push would have made the amplitude larger than $5.0$ cm.

Step 5

Energy cross-check: $E = \frac{1}{2}kA^{2} = 0.5 \times 49.0 \times 0.0025 = 0.0613$ J, and $\frac{1}{2}mv_{max}^{2} = 0.5 \times 0.40 \times 0.553^{2} = 0.0613$ J.

Why

The two agree with no $mgh$ term anywhere, which is the practical proof that measuring from the hanging position has already absorbed gravity.

Step 6

Validity: at the highest point of the motion the spring is still stretched by $8.0 - 5.0 = 3.0$ cm, so it never goes slack, and $F = -kx$ holds all the way round.

Why

The condition for simple harmonic motion is about the FORCE, and a slack spring supplies none. With $A$ larger than $d$ the mass would go into free fall for part of each cycle and every formula above would fail.

The conclusion, written out

“The static stretch gives $k = 49.0$ N/m, so the mass oscillates about its hanging position with a period of $0.568$ s, a maximum speed of $0.553$ m/s at that position and a maximum acceleration of $6.13$ m/s $^{2}$ at the two ends. Since $A = 5.0$ cm is smaller than $d = 8.0$ cm, the spring stays stretched throughout and the motion is simple harmonic.”

The classic mistake on this problem: Writing $\frac{1}{2}kA^{2}+mgA$ for the energy at the lowest point, which counts gravity twice and gives a maximum speed near $1.1$ m/s instead of $0.553$ m/s.

Learn by heart

• $T = 2\pi\sqrt{m/k}$ and $T = 2\pi\sqrt{L/g}$ : no amplitude in either, and NO MASS in the pendulum.
• $\omega = 2\pi f = 2\pi/T$ , in rad/s. Never $\omega = 1/T$ , which is the frequency.
• $v_{max} = \omega A$ at the CENTRE; $|a|_{max} = \omega^{2}A$ at the two ENDS. Never at the same place.
• $v = \omega\sqrt{A^{2}-x^{2}}$ : at half the amplitude the speed is $0.87\,v_{max}$ , and the acceleration is exactly half of $a_{max}$ .
• $E = \frac{1}{2}kA^{2} = \frac{1}{2}mv_{max}^{2}$ , and the two forms are equal at $x = A/\sqrt{2} \approx 0.71\,A$ .
• Vertical spring: the centre drops by $d = mg/k$ , the period does NOT change, and $T = 2\pi\sqrt{d/g}$ .
• Springs side by side: $k = k_{1}+k_{2}$ . End to end: the RECIPROCALS add, and the result is softer than either spring.
• Small angles means radians: at $10^{\circ}$ , $\sin\theta$ and $\theta$ differ by $0.5\%$ ; at $30^{\circ}$ , by $4.5\%$ , and the true period is about $1.7\%$ long.
• Damping cuts the amplitude by a constant factor per cycle and leaves the period alone; resonance is driving at $f \approx f_{0}$ , and heavier damping gives a lower, broader peak.
• $x = A\sin(\omega t + \varphi)$ on the formula sheet: $\sin\varphi = x_{0}/A$ has TWO answers, $\varphi$ and $\pi - \varphi$ , and the sign of $v_{0} = \omega A\cos\varphi$ chooses.
• A push at release: $A = \sqrt{x_{0}^{2}+\frac{m}{k}v_{0}^{2}}$ , always more than $x_{0}$ .
• Reference circle: radius $A$ , turning at $\omega$ ; its speed $\omega A$ and its centripetal acceleration $\omega^{2}A$ are $v_{max}$ and $a_{max}$ , and its starting angle is $\varphi$ .

Frequently asked questions

Does the amplitude change the period of an oscillator?

No, and this is the most commonly tested fact of the chapter. The period of a mass on a spring depends only on the mass and the stiffness of the spring, and the period of a pendulum only on its length and on gravity. Pulling the mass twice as far doubles the maximum speed and multiplies the energy by four, but the time for one complete cycle does not move at all.

Why does a heavier bob not slow a pendulum down?

Because the mass affects the pull and the resistance to being moved in exactly the same proportion. Doubling the bob doubles the force bringing it back towards the centre, and doubles the inertia that force has to overcome, so the two cancel. It is the same cancellation that makes two objects of different masses fall side by side, and it is why the mass does not appear in the pendulum period formula.

Should I use sine or cosine for the position equation?

Look only at where the object is when the stopwatch starts. Released from rest at maximum elongation, the position equals the amplitude at time zero, so you need the cosine. Launched from the equilibrium position with a push, the position is zero at time zero, so you need the sine. The choice describes the launch, never the system, and it changes nothing about the period.

Does hanging a spring vertically change its period?

No. Gravity moves the centre of the oscillation down by an amount equal to the weight divided by the spring constant, and does nothing else. Measured from that new resting position, the resultant force is again proportional to the displacement with the same spring constant, so the period formula gives exactly the same number as on a horizontal table.

What exactly happens at resonance?

A driven oscillator responds most strongly when the driving frequency is close to its own natural frequency. Each push then arrives in step with the motion and adds energy cycle after cycle, so the steady amplitude climbs to a peak. Damping is what keeps that peak finite: the heavier the damping, the lower and the broader the peak, and the slightly lower the frequency at which it sits.

How do I find the phase constant of a simple harmonic motion?

Put time zero into the position equation: the sine of the phase equals the starting position divided by the amplitude. Your calculator gives one angle, but in every cycle there is a second one, pi minus the first. Keep the angle whose cosine has the same sign as the starting velocity: positive if the object starts moving in the positive direction, negative if it starts moving the other way.

Practise it

Corrected exercises: Oscillations and simple harmonic motion, PHYS 101 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/phys101-oscillations. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Oscillations and simple harmonic motion: the PHYS 101 revision sheet

The essentials

On your formula sheet: this chapter's lines, and when each one holds

Two families of quantities, and the question that separates them

The condition, and the five relations that come out of it

The simple pendulum: the restoring force is mg sin θ

The formula sheet's form: phase constant and reference circle

The mistakes that cost marks

1. Doubling the amplitude to change the period

2. Half the amplitude, half the speed

3. No speed at the turning point, so no acceleration

4. The calculator left in degree mode

5. Sine written where the launch calls for a cosine

6. A heavier bob for a slower pendulum

7. The weight carried through a vertical spring problem

8. The angular frequency written as one over T

9. Damping described as a slowing down

10. A push at release, and the amplitude read off the release point

11. The calculator's arcsine taken as the phase

Which method to choose

Which period formula, read off the set-up

Which relation gives a speed, and which gives a position

How the answer is expected to be written

Answering a qualitative period question

Getting a speed out of an energy balance

Finding the phase constant from a launch

Check before you hand in

The typical problem, taken apart

A full vertical spring laboratory, from a ruler to a maximum speed

Learn by heart

Frequently asked questions

Does the amplitude change the period of an oscillator?

Why does a heavier bob not slow a pendulum down?

Should I use sine or cosine for the position equation?

Does hanging a spring vertically change its period?

What exactly happens at resonance?

How do I find the phase constant of a simple harmonic motion?

Corrected exercises: Oscillations and simple harmonic motion, PHYS 101 at McGill

See also

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