Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal

Revision sheet: SHM and mechanical waves (203-SN3-RE)

Simple harmonic motion is the most reused model in all of physics: it describes a spring, a pendulum, a molecule, an LC circuit. Its signature is unique, an acceleration proportional and opposite to the displacement, and it is enough to deduce everything.

Waves are its extension: the same harmonic motion, delayed in time from one point to the next. This sheet gathers the errors that cost the most marks, starting with the confusion between the speed of the wave and that of the medium. It is written for CEGEP students in the Science program, in Waves and Modern Physics, 203-SN3-RE (formerly 203-NYC).

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The thread of the chapter

A wave has TWO speeds that have nothing to do with each other: that of the pattern, set by the medium, and that of a point of the medium, which is only a harmonic motion. And the period of an oscillator never depends on the amplitude.

This chapter is part of 203-SN3-RE, Waves and Modern Physics

The essentials

The signature of SHM, and its three curves

• Definition: $a=-\omega^{2}x$ . Any restoring force proportional to the displacement gives SHM, and nothing else does.
• $x(t)=A\cos(\omega t+\varphi)$ , $v(t)=-A\omega\sin(\omega t+\varphi)$ , $a(t)=-A\omega^{2}\cos(\omega t+\varphi)$ .
• $v_{max}=A\omega$ when passing through EQUILIBRIUM, $a_{max}=A\omega^{2}$ at the EXTREMES. The two never occur at the same instant.
• Constant total energy: $E=\dfrac{1}{2}kA^{2}=\dfrac{1}{2}mv_{max}^{2}$ , and at position $x$ : $v=\omega\sqrt{A^{2}-x^{2}}$ .
• $\omega=2\pi f=\dfrac{2\pi}{T}$ : the angular frequency is in rad/s, the frequency in Hz. Confusing them introduces a factor $2\pi$ everywhere.

The two curves are in quadrature: whenever one passes through an extremum, the other passes through zero. Maximum speed and maximum displacement never coexist.

Where the speed is maximum, the acceleration is zero, and vice versa. It is the quickest way to check an answer: ask where the object is, not which formula to apply.

Mass-spring: the force always points toward equilibrium

• The spring force is $F=-kx$ : it ALWAYS points toward the equilibrium position, and its magnitude grows with the displacement.
• At the extremes, $x=\pm A$ : zero speed, maximum force and acceleration, directed toward the centre.
• Passing through equilibrium: zero force, maximum speed $A\omega$ , once in one direction, once in the other.

Three instants of the same oscillation, without friction: the spring force, in pink, always aims at the equilibrium position, dashed; the speed is maximum where the force vanishes.

“The acceleration is zero, so the block is at rest” is wrong at equilibrium: the block goes through it at its maximum speed.

The periods, and what they do not depend on

• Mass-spring: $\omega=\sqrt{\dfrac{k}{m}}$ and $T=2\pi\sqrt{\dfrac{m}{k}}$ . The period does NOT depend on the amplitude: this is isochronism.
• Simple pendulum, small angles: $T=2\pi\sqrt{\dfrac{L}{g}}$ . Neither the mass nor the amplitude appears in it.
• VERTICAL spring: the weight only shifts the centre of the oscillations by $x_{eq}=\dfrac{mg}{k}$ ; the period stays $2\pi\sqrt{\dfrac{m}{k}}$ .
• Springs in PARALLEL: $k_{eq}=k_{1}+k_{2}$ . In SERIES: $\dfrac{1}{k_{eq}}=\dfrac{1}{k_{1}}+\dfrac{1}{k_{2}}$ . Cutting a spring in half DOUBLES its constant.
• Physical pendulum: $T=2\pi\sqrt{\dfrac{I}{mgd}}$ . Torsion pendulum: $T=2\pi\sqrt{\dfrac{I}{\kappa}}$ , independent of gravity.

Only one of these periods depends on $g$ : that of the pendulum. A spring oscillates in exactly the same way on the Moon, which makes it the mass-measuring instrument of space stations.

The simple pendulum: only the tangential component restores

• Two forces: the tension $\vec{T}$ , along the string, and the weight $m\vec{g}$ . The tension, perpendicular to the motion, only bends the path.
• The restoring force is the TANGENTIAL component of the weight, $-mg\sin\theta$ . At small angles, $\sin\theta\approx\theta$ in radians, and the motion is an SHM of angular frequency $\omega=\sqrt{\dfrac{g}{L}}$ .
• The mass appears in the force AND in the inertia: it cancels, as in free fall.

At the extreme point, the tension balances the radial component of the weight; what remains is

mg\sin\theta

, tangent to the path, which pulls the bob back toward the vertical.

Beyond about fifteen degrees, $\sin\theta$ departs from $\theta$ and the period lengthens: $2\pi\sqrt{\dfrac{L}{g}}$ is a small-angle formula.

Waves: two speeds, one equation

• PROPAGATION speed on a string: $v=\sqrt{\dfrac{F}{\mu}}$ , where $F$ is the tension and $\mu$ the linear mass density. It is a property of the MEDIUM.
• The FREQUENCY comes from the SOURCE and never changes along the way; it is $\lambda=\dfrac{v}{f}$ that adjusts to the medium.
• Travelling wave toward positive $x$ : $y(x, t)=A\sin(kx-\omega t)$ , with $k=\dfrac{2\pi}{\lambda}$ and $\omega=2\pi f$ ; $v=\dfrac{\omega}{k}=\lambda f$ . The sign $+\omega t$ describes propagation toward negative $x$ .
• TRANSVERSE speed of a point of the medium: it is an SHM, with maximum $A\omega$ , having nothing to do with $v$ .
• Power carried: $P=\dfrac{1}{2}\mu v\omega^{2}A^{2}$ , so proportional to the SQUARE of the amplitude AND to the square of the frequency. Point source: $I=\dfrac{P}{4\pi r^{2}}$ .

On a string stretched at $50$ N with a linear mass density of $5.0$ g/m, the wave runs at $100$ m/s while each point moves at only $6.3$ m/s. These two numbers can be read in the same problem statement and never replace each other.

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. Confusing the speed of the wave with the speed of a point of the medium

the whole question, 5 marks

What not to write

“The wave advances at $100$ m/s, so every point of the string moves at $100$ m/s.”

What to write

“The string does not move horizontally at all. Each point moves up and down at the speed of an SHM, with maximum $v_{trans}=A\omega=0.020\times 314=6.3$ m/s. Only the PATTERN advances at $100$ m/s.”

The pattern moves horizontally at

v=\sqrt{F/\mu}

, but each point of the medium moves only vertically, at the speed of an SHM. Nothing advances along the string except the shape.

Why: This is the central conceptual error of the chapter: a wave carries energy without carrying matter. The two speeds do not even have the same origin, one comes from the medium, the other from the source.

2. Making the period depend on the amplitude

the whole question, 4 marks

What not to write

“The mass is released from twice as far, so it takes twice as long to come back.”

What to write

“ $T=2\pi\sqrt{\dfrac{m}{k}}$ does not contain $A$ : the period is UNCHANGED. The object starts from farther away, but it also goes faster, in exactly the same proportion.”

Why: This is isochronism, the property that makes oscillators clocks. It holds for a spring at any amplitude, and for a pendulum as long as the angles stay small.

3. Changing the frequency by changing the medium

3 marks, and the whole matching question

What not to write

“The wave moves onto a thicker string, it slows down, so its frequency decreases.”

What to write

“The frequency is imposed by the SOURCE and crosses the junction unchanged. It is the wavelength that adjusts: $\lambda=\dfrac{v}{f}$ , so it decreases along with $v$ .”

Why: The two pieces of string are tied together: they move up and down together at the junction, so at the same rate. A different frequency on either side would tear the string.

4. Halving the stiffness of a spring that is cut in two

3 marks, and a period that is wrong by a factor

\sqrt{2}

What not to write

“The $100$ N/m spring cut in half gives two $50$ N/m springs.”

What to write

“Each half is TWICE as stiff: $k=200$ N/m. For the same total extension, each half stretches by only half as much, so it takes twice the force for the same deformation per coil.”

Why: Stiffness is INVERSELY proportional to length, just as the resistance of a wire is directly proportional to its length. This is also why two springs end to end, hence in series, are softer than either one.

5. Swapping the rules for springs in series and in parallel

3 marks

What not to write

“Two springs side by side, so in parallel: $\dfrac{1}{k_{eq}}=\dfrac{1}{k_{1}}+\dfrac{1}{k_{2}}$ , like resistors.”

What to write

“In PARALLEL, the stiffnesses ADD, $k_{eq}=k_{1}+k_{2}$ : two springs pull together, the assembly is stiffer. In SERIES, it is the sum of the inverses.”

Same block, same springs: in parallel the extension is half as large,

k

doubled; in series it is twice as large,

k

halved.

Why: Stiffness behaves like a CONDUCTANCE, not like a resistance: two paths in parallel offer more restoring force. The physical check is immediate, two springs side by side are necessarily harder to stretch than one.

6. Letting the weight change the period of a vertical spring

3 marks, often on a justification question

What not to write

“The spring is vertical, so the weight adds to the restoring force and the period changes.”

What to write

“The weight only SHIFTS the centre of the oscillations by $x_{eq}=\dfrac{mg}{k}$ . Around this new equilibrium position, the restoring force is still $-kx$ and the period is still $2\pi\sqrt{\dfrac{m}{k}}$ .”

The weight stretches the spring by

x_{eq}=\dfrac{mg}{k}

, then the block oscillates around this NEW equilibrium position, with the same period

2\pi\sqrt{\dfrac{m}{k}}

Why: The weight is a CONSTANT force: it shifts the equilibrium but changes neither the stiffness nor the mass. That is why $T=2\pi\sqrt{\dfrac{x_{eq}}{g}}$ makes it possible to measure $g$ with a simple spring and a ruler.

7. Believing that the power of a wave follows the amplitude

3 marks, and a factor 4 on the whole end of the problem

What not to write

“The amplitude is doubled, so the wave carries twice as much energy.”

What to write

“ $P=\dfrac{1}{2}\mu v\omega^{2}A^{2}$ : doubling the amplitude MULTIPLIES the power by four. Doubling the frequency quadruples it too.”

Why: The energy of an oscillator goes as the square of its amplitude, $\dfrac{1}{2}kA^{2}$ , and a wave is only a line of oscillators. The same square explains why sound intensity drops so fast with distance.

8. Believing that damping mainly changes the period

2 marks on the discussion question

What not to write

“With friction, the oscillator is slowed down, so its period increases a lot.”

What to write

“ $\omega'=\sqrt{\omega_{0}^{2}-\gamma^{2}}$ : for weak damping, $\omega'$ is almost equal to $\omega_{0}$ . It is the AMPLITUDE that collapses as $e^{-\gamma t}$ , long before the rhythm moves.”

The envelope collapses exponentially, but the zero crossings stay equally spaced: damping eats the amplitude long before it touches the rhythm.

Why: The damped angular frequency changes noticeably only near critical damping $b_{c}=2\sqrt{km}$ . That is why a pendulum clock keeps time even as its amplitude shrinks.

Which method to choose

Which period formula

The device described by the problem, not the quantity asked for

If mass at the end of a spring, horizontal or vertical → $T=2\pi\sqrt{\dfrac{m}{k}}$ , whatever the amplitude and whatever the orientation

Example: $0.250$ kg on $100$ N/m gives $0.314$ s
If point mass on a string, small angles → $T=2\pi\sqrt{\dfrac{L}{g}}$ , without the mass

Example: a $1.00$ m string gives $2.01$ s
If EXTENDED body swinging about an axis → physical pendulum: $T=2\pi\sqrt{\dfrac{I}{mgd}}$ , with $d$ from the axis to the centre of mass

Example: a rod hung by one end
If disc or bar hung from a TORSION WIRE → $T=2\pi\sqrt{\dfrac{I}{\kappa}}$ , independent of $g$

Example: the Cavendish balance
If several springs → compute $k_{eq}$ first: sum in parallel, sum of the inverses in series

Example: a spring cut in half is worth $2k$
If the problem mentions friction or a periodic external force → damped, $\omega'=\sqrt{\omega_{0}^{2}-\gamma^{2}}$ , or driven with resonance at $\omega_{0}$

Example: critical damping for $b=2\sqrt{km}$

None of these periods contains the amplitude, and only one contains $g$ . These two observations let you eliminate half the answers of a multiple-choice question without doing a single calculation.

Reading a wave equation

The expression $y(x, t)=A\sin(kx-\omega t)$ as written

If the amplitude is asked → it is the factor in front of the sine, with its unit

Example: $y=0.020\sin(3.14x-314t)$ gives $A=2.0$ cm
If the wavelength is asked → $\lambda=\dfrac{2\pi}{k}$ , where $k$ is the coefficient of $x$

Example: $k=3.14$ rad/m gives $\lambda=2.0$ m
If the frequency or the period is asked → $\omega$ is the coefficient of $t$ , then $f=\dfrac{\omega}{2\pi}$ and $T=\dfrac{1}{f}$

Example: $\omega=314$ rad/s gives $f=50$ Hz
If the PROPAGATION speed is asked → $v=\dfrac{\omega}{k}=\lambda f$ , that is, the ratio of the two coefficients

Example: $\dfrac{314}{3.14}=100$ m/s
If the DIRECTION of propagation is asked → minus sign between the two terms: toward positive $x$ ; plus sign: toward negative $x$

Example: $\sin(kx+\omega t)$ travels backward
If the speed of a POINT of the medium is asked → differentiate with respect to TIME: $v_{y}=-A\omega\cos(kx-\omega t)$ , with maximum $A\omega$

Example: $0.020\times 314=6.3$ m/s, unrelated to the $100$ m/s

The last two branches are the ones that pay off: the direction of propagation is read off a sign, and the transverse speed is obtained by differentiating with respect to $t$ , never with respect to $x$ . A derivative with respect to $x$ would give the slope of the string.

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.

Extracting every quantity from a wave equation

When to use it: The problem gives $y(x, t)$ and asks for amplitude, wavelength, frequency, speed and direction.

1 Put the expression in the canonical form $A\sin(kx\pm\omega t)$ , factoring if necessary to isolate the coefficients of $x$ and $t$ .
2 Read the three constants one by one, with their units: $A$ in metres, $k$ in rad/m, $\omega$ in rad/s.
3 Convert: $\lambda=\dfrac{2\pi}{k}$ , $f=\dfrac{\omega}{2\pi}$ , $T=\dfrac{1}{f}$ .
4 Compute the propagation speed as the RATIO $\dfrac{\omega}{k}$ , and check that it gives back $\lambda f$ .
5 State the direction from the sign, then, if asked, differentiate with respect to time for the transverse speed.

Concluding sentence

“In the form $y=0.020\sin(3.14x-314t)$ , we read $A=2.0$ cm, $k=3.14$ rad/m and $\omega=314$ rad/s, hence $\lambda=2.0$ m, $f=50$ Hz and $v=\dfrac{\omega}{k}=100$ m/s toward positive $x$ . The maximum speed of a point of the medium is $A\omega=6.3$ m/s.”

The trap: Announcing $v=\omega k$ or $v=\lambda k$ . The speed is the RATIO of the two coefficients, which the unit check confirms immediately: rad/s divided by rad/m gives m/s.

Marking: Usually 1 mark per quantity read, 2 marks for the justified speed, 1 mark for the direction.

Solving an SHM by energy

When to use it: The problem asks for a speed at a given position, or a position for a given energy.

1 Write the total energy from the amplitude: $E=\dfrac{1}{2}kA^{2}$ , stating that it is CONSTANT throughout the motion.
2 Write the sharing at the requested position: $E=\dfrac{1}{2}kx^{2}+\dfrac{1}{2}mv^{2}$ .
3 Isolate the unknown, which gives $v=\omega\sqrt{A^{2}-x^{2}}$ for a speed, or $x=A\sqrt{1-\dfrac{E_{k}}{E}}$ for a position.
4 Check the two limiting cases: $x=0$ must give back $v_{max}=A\omega$ , and $x=A$ must give $v=0$ .

Concluding sentence

“The total energy is $\dfrac{1}{2}kA^{2}=0.720$ J. At $x=6.00$ cm, the potential energy is $0.180$ J, so the kinetic energy is $0.540$ J and the speed is $v=\omega\sqrt{A^{2}-x^{2}}=2.08$ m/s.”

The trap: Looking for the speed through kinematics, by first computing the instant at which the object passes through $x$ . Energy gives the answer without any trigonometric function, and without having to know the phase.

Marking: 1 mark for the total energy, 2 for the sharing written out, 2 for isolating the unknown, 1 for the check of the limiting cases.

Check before you hand in

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

The two speeds told apart on the paper

Write “pattern” or “point” next to each speed. The first is computed with the medium, the second with $A\omega$ , and they generally differ by a factor of ten or more.

A transverse speed of $100$ m/s on a guitar string: the speed of the wave has been copied.
No amplitude in the periods

Reread the period formula used and check that $A$ does not appear in it. Any answer in which the period varies with the amplitude is wrong at the level of the course.

A period that doubles because the amplitude doubled: isochronism forbids it.
The limiting-case check in energy

Check that the formula obtained gives $v_{max}$ at $x=0$ and $v=0$ at $x=A$ . These two substitutions take five seconds and catch a sign error under the root.

A $v=\omega\sqrt{x^{2}-A^{2}}$ : the root becomes imaginary, the two terms have been swapped.
The units of $k$ and $\omega$

$k$ is in rad per METRE, $\omega$ in rad per SECOND. Their ratio must come out in m/s, and their product has no physical meaning.

A speed of $986$ m/s obtained from $\omega k$ : the units alone disqualify the result.
The order of magnitude of a period

A laboratory spring oscillates in a few tenths of a second, a one-metre pendulum in two seconds. A period of several minutes or a few milliseconds signals a unit error on $k$ or on $L$ .

A $1$ m pendulum quoted at $0.20$ s: it would need a string of $1$ cm.

The typical problem, taken apart

A complete oscillator, from kinematics to energy

A mass $m=0.250$ kg is attached to a spring of constant $k=100$ N/m and oscillates horizontally without friction with an amplitude $A=12.0$ cm.

Find the angular frequency, the period, the maximum speed and the maximum acceleration, then the total energy and the speed at the point $x=6.00$ cm. Finally, say what changes if the same spring is hung vertically.

The two parabolas complement each other: their sum is the horizontal line of the total energy. They cross at

x=\dfrac{A}{\sqrt{2}}=8.49

cm, not at the middle of the amplitude.

Step 1

Angular frequency and period: $\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{100}{0.250}}=20.0$ rad/s, so $T=\dfrac{2\pi}{\omega}=0.314$ s and $f=3.18$ Hz.

Why

Neither the amplitude nor the initial position comes into it: the angular frequency depends only on the mass-spring pair. That is what makes it possible to compute it before even knowing how the motion was started.

Step 2

Maximum speed and acceleration: $v_{max}=A\omega=0.120\times 20.0=2.40$ m/s, reached at $x=0$ ; $a_{max}=A\omega^{2}=0.120\times 400=48.0$ m/s $^{2}$ , reached at $x=\pm A$ .

Why

The two maxima occur at the two OPPOSITE places of the motion. Saying where they are reached is part of the expected answer, and avoids adding them by mistake in an energy question.

Step 3

Total energy: $E=\dfrac{1}{2}kA^{2}=\dfrac{1}{2}(100)(0.120)^{2}=0.720$ J. Check: $\dfrac{1}{2}mv_{max}^{2}=\dfrac{1}{2}(0.250)(2.40)^{2}=0.720$ J.

Why

The two expressions must give the same number since the energy is entirely potential at the extremes and entirely kinetic at the centre. It is a free check, using two results already obtained.

Step 4

At $x=6.00$ cm: $E_{p}=\dfrac{1}{2}kx^{2}=0.180$ J, so $E_{k}=0.720-0.180=0.540$ J and $v=\sqrt{\dfrac{2E_{k}}{m}}=2.08$ m/s.

Why

At half the amplitude, the potential energy is only a QUARTER of the total, since it varies as the square. The speed there is therefore still $87\,\%$ of the maximum speed, which often comes as a surprise.

Step 5

Check by the direct formula: $v=\omega\sqrt{A^{2}-x^{2}}=20.0\sqrt{0.120^{2}-0.060^{2}}=20.0\times 0.1039=2.08$ m/s.

Why

The same number is found by a path that uses neither the mass nor the spring constant: two independent methods that agree validate the result.

Step 6

Where the energy is shared equally: $\dfrac{1}{2}kx^{2}=\dfrac{E}{2}$ gives $x=\dfrac{A}{\sqrt{2}}=8.49$ cm, and not $6.00$ cm.

Why

The equal sharing does not occur at the middle of the travel, because the potential energy is quadratic. It is the classic trick question of this chapter.

Step 7

Vertical spring: the centre of the oscillations drops by $x_{eq}=\dfrac{mg}{k}=\dfrac{(0.250)(9.8)}{100}=2.45$ cm, but $\omega$ , $T$ , $v_{max}$ and $E$ stay EXACTLY the same.

Why

The weight is a constant force: it shifts the equilibrium position without changing the stiffness or the mass. The period can then also be written $2\pi\sqrt{\dfrac{x_{eq}}{g}}$ , which gives a way to measure $g$ with a simple ruler.

The conclusion, written out

“The oscillator has an angular frequency $\omega=20.0$ rad/s, hence a period $T=0.314$ s. Its maximum speed is $2.40$ m/s at the centre and its maximum acceleration $48.0$ m/s $^{2}$ at the extremes. Its total energy is $0.720$ J, and at $x=6.00$ cm its speed is still $2.08$ m/s. Hung vertically, the same spring oscillates in the same way, around a centre lowered by $2.45$ cm.”

The classic mistake on this problem: Believing that at half the amplitude the energy is shared half and half. The potential energy there is $\dfrac{1}{4}$ of the total and not $\dfrac{1}{2}$ , because it depends on the SQUARE of the position.

Learn by heart

• Signature of SHM: $a=-\omega^{2}x$ , and nothing else gives SHM.
• $v_{max}=A\omega$ at the centre, $a_{max}=A\omega^{2}$ at the extremes: never in the same place.
• $E=\dfrac{1}{2}kA^{2}$ and $v=\omega\sqrt{A^{2}-x^{2}}$ .
• $T=2\pi\sqrt{\dfrac{m}{k}}$ and $T=2\pi\sqrt{\dfrac{L}{g}}$ : neither contains the amplitude.
• Springs: $k_{eq}=k_{1}+k_{2}$ in parallel, sum of the inverses in series; cutting in half DOUBLES $k$ .
• Vertical spring: the weight shifts the equilibrium by $\dfrac{mg}{k}$ , the period does not change.
• $v=\sqrt{\dfrac{F}{\mu}}$ comes from the MEDIUM, $f$ comes from the SOURCE, and $\lambda=\dfrac{v}{f}$ adjusts.
• In $A\sin(kx-\omega t)$ : $\lambda=\dfrac{2\pi}{k}$ , $f=\dfrac{\omega}{2\pi}$ , $v=\dfrac{\omega}{k}$ , and the minus sign means toward the right.
• Speed of a point: $A\omega$ , unrelated to the speed of the wave.
• $P=\dfrac{1}{2}\mu v\omega^{2}A^{2}$ : the square of the amplitude AND the square of the frequency.

Frequently asked questions

What is the difference between the speed of a wave and the speed of a point of the medium?

The speed of the wave is that of the moving pattern, and it depends only on the medium, for example on the tension and the mass per metre of a string. The speed of a point is that of its back-and-forth motion, equal to the product of the amplitude and the angular frequency. The two are often in a ratio of ten or more.

Does the period of a spring depend on the amplitude?

No, and it is the most useful property of harmonic motion. The period depends only on the mass and the stiffness. Releasing the mass from twice as far simply gives it twice the speed, so the path, twice as long, is covered in the same time.

What happens to a wave that moves onto a thicker string?

It slows down, since its speed depends on the mass per metre of the medium, and its wavelength decreases in the same proportion. Its frequency, however, does not change at all: it is imposed by the source, and the two pieces of string oscillate together at the junction.

What happens to the stiffness of a spring cut in half?

It doubles. Stiffness is inversely proportional to length: half a spring has half as many coils to absorb the same extension, so each coil works twice as hard. The same logic makes two springs end to end softer than either one.

Does a vertical spring oscillate more slowly than a horizontal one?

No, the period is exactly the same. The weight is a constant force: it lowers the equilibrium position by the ratio of the weight to the stiffness, but around this new centre, the restoring force stays proportional to the displacement, with the same constant.

How do you find the speed of an oscillator at a given position?

By energy, without going through time. The total energy, equal to half the stiffness times the square of the amplitude, is shared between potential and kinetic energy. From it you deduce the speed, which equals the angular frequency times the square root of the difference of the squares of the amplitude and the position.

Practise it

Corrected exercises: Simple harmonic motion and mechanical waves, 203-SN3-RE

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.

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Revision sheet: SHM and mechanical waves (203-SN3-RE)

The essentials

The signature of SHM, and its three curves

Mass-spring: the force always points toward equilibrium

The periods, and what they do not depend on

The simple pendulum: only the tangential component restores

Waves: two speeds, one equation

The mistakes that cost marks

1. Confusing the speed of the wave with the speed of a point of the medium

2. Making the period depend on the amplitude

3. Changing the frequency by changing the medium

4. Halving the stiffness of a spring that is cut in two

5. Swapping the rules for springs in series and in parallel

6. Letting the weight change the period of a vertical spring

7. Believing that the power of a wave follows the amplitude

8. Believing that damping mainly changes the period

Which method to choose

Which period formula

Reading a wave equation

How the answer is expected to be written

Extracting every quantity from a wave equation

Solving an SHM by energy

Check before you hand in

The typical problem, taken apart

A complete oscillator, from kinematics to energy

Learn by heart

Frequently asked questions

What is the difference between the speed of a wave and the speed of a point of the medium?

Does the period of a spring depend on the amplitude?

What happens to a wave that moves onto a thicker string?

What happens to the stiffness of a spring cut in half?

Does a vertical spring oscillate more slowly than a horizontal one?

How do you find the speed of an oscillator at a given position?

Corrected exercises: Simple harmonic motion and mechanical waves, 203-SN3-RE

See also

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