Chapter 13 · Oscillations

13.2 Velocity, Acceleration, Force Law and Energy in SHM

In SHM the velocity $v=\pm\omega\sqrt{A^2-x^2}$ is largest at the centre, the acceleration $a=-\omega^2x$ always points back to the centre, the force obeys $F=-kx$, and the energy $\tfrac12kA^2$ is passed steadily between kinetic and potential.

A car hits a bump and its body bounces up and down on the springs of its suspension. When are the passengers thrown about most violently — as the car body passes its normal level at full speed, or at the top and bottom of the bounce, where it stops for an instant? Engineers who design suspensions, earthquake-resistant buildings, loudspeakers and the vibration mounts of spacecraft instruments need exact answers to such questions, because it is the acceleration (the force) that breaks things and the speed that carries energy.

This section finds the velocity and the acceleration of a particle in simple harmonic motion, first from the reference circle and then as rates of change. The acceleration turns out to be simply $-\omega^2$ times the displacement. Through Newton's second law that gives the force law of SHM — a restoring force proportional to the displacement — which is the test you will use to recognise SHM in any system. Finally you will follow the energy of the oscillator as it changes from all potential at the ends to all kinetic in the middle, with the total never changing.

Velocity in SHM

Return to the reference circle of Section 13.1: a particle $P$ moves anticlockwise round a circle of radius $A$ at angular speed $\omega$, and its projection $Q$ on the $x$-axis performs the SHM $x=A\cos(\omega t+\phi)$. The velocity of $P$ has magnitude $v_P=\omega A$ (uniform circular motion, Section 3.4) and is along the tangent, perpendicular to the radius $OP$. When the radius is at angle $\theta=\omega t+\phi$, the tangent velocity makes the angle $\theta$ with the $y$-axis and points to the left in the upper half of the circle, so its $x$-component is $-\omega A\sin\theta$. The shadow $Q$ always has the same $x$-velocity as $P$.

The same result comes from calculus: $v=\dfrac{dx}{dt}$ (Section 2.1), and the rate of change of $\cos(\omega t+\phi)$ is $-\omega\sin(\omega t+\phi)$ (you will prove this in mathematics). Using $\sin^2\theta+\cos^2\theta=1$, the velocity can also be written in terms of the displacement: $A^2\sin^2\theta=A^2-A^2\cos^2\theta=A^2-x^2$.

📐 Result 13.3 — Velocity in SHM

For $x=A\cos(\omega t+\phi)$ the velocity is

$$v=\frac{dx}{dt}=-\omega A\sin(\omega t+\phi),$$

and at displacement $x$ the speed is

$$|v|=\omega\sqrt{A^2-x^2}.$$

The speed is greatest, $v_\text{max}=\omega A$ (the velocity amplitude), as the particle passes through the mean position, and zero at the extremes $x=\pm A$.

The direction of $v$ is not fixed by $x$ alone: at each point inside $(-A,A)$ the particle passes once going each way. On the $v$–$x$ plane the motion traces the ellipse $\dfrac{x^2}{A^2}+\dfrac{v^2}{\omega^2A^2}=1$, clockwise.

✏️ Example 1 — Speeds in a known SHM

A particle moves with $x=0.050\cos\left(4\pi t+\dfrac\pi3\right)$, in metres and seconds (as in Example 3 of Section 13.1). Find (a) its maximum speed; (b) its velocity at $t=0$; (c) its speed when $x=0.030\ \text{m}$.

💡 Show solution▶

(a) $v_\text{max}=\omega A=4\pi\times0.050=0.63\ \text{m/s}$ ($0.628$).

(b) $v(0)=-\omega A\sin\dfrac\pi3=-0.628\times0.866=-0.54\ \text{m/s}$: towards $-x$, as the phase constant in $(0,\pi)$ predicted.

Check At $t=0$, $x=0.025\ \text{m}$: $\omega\sqrt{A^2-x^2}=4\pi\sqrt{0.0025-0.000625}=0.544\ \text{m/s}$, the same as (b) ✓.

✏️ Example 2 — Amplitude and period from two snapshots

A particle in SHM has a speed of $16\ \text{cm/s}$ when it is $3.0\ \text{cm}$ from its mean position and $12\ \text{cm/s}$ when it is $4.0\ \text{cm}$ from it. Find the amplitude, the angular frequency and the period.

💡 Show solution▶

Set-up Square $|v|=\omega\sqrt{A^2-x^2}$ for each snapshot: $16^2=\omega^2(A^2-3.0^2)$ and $12^2=\omega^2(A^2-4.0^2)$ (cm and s).

ω Subtract: $256-144=\omega^2(16-9)$, so $\omega^2=16$ and $\omega=4.0\ \text{rad/s}$.

A, T $A^2=9+\dfrac{256}{16}=25$: $A=5.0\ \text{cm}$. $T=\dfrac{2\pi}{4.0}=1.6\ \text{s}$ ($1.57$).

Check Second snapshot: $4.0\sqrt{25-16}=12\ \text{cm/s}$ ✓. In general $T=2\pi\sqrt{\dfrac{x_2^2-x_1^2}{v_1^2-v_2^2}}$.

Acceleration and the force law

The reference particle $P$ has a centripetal acceleration of magnitude $\omega^2A$ directed towards the centre $O$ (Section 3.4). Its $x$-component is $-\omega^2A\cos\theta=-\omega^2x$. Equivalently, $a=\dfrac{dv}{dt}$ and the rate of change of $-\omega A\sin(\omega t+\phi)$ is $-\omega^2A\cos(\omega t+\phi)$.

📐 Result 13.4 — Acceleration in SHM

For $x=A\cos(\omega t+\phi)$ the acceleration is

$$a=\frac{dv}{dt}=-\omega^2A\cos(\omega t+\phi)=-\omega^2x .$$

The acceleration is proportional to the displacement and always directed towards the mean position. Its magnitude is greatest, $a_\text{max}=\omega^2A$ (the acceleration amplitude), at the extremes $x=\pm A$, and zero at the mean position.

Phase relations. Write all three as cosines: $x=A\cos(\omega t+\phi)$, $v=\omega A\cos\left(\omega t+\phi+\dfrac\pi2\right)$, $a=\omega^2A\cos(\omega t+\phi+\pi)$. The velocity leads the displacement by $\dfrac\pi2$ (a quarter period), and the acceleration leads the velocity by another $\dfrac\pi2$, so the acceleration is in antiphase with the displacement. When $x$ is at its maximum, $v$ is zero and $a$ is at its most negative; a quarter-period later $x=0$, $v$ has its most negative value and $a=0$. Interactive 1 shows the three graphs one above the other.

📘 Definition 13.4 — Linear restoring force and force constant

A force on a particle that is proportional to its displacement $x$ from an equilibrium position and opposite to it,

$$F=-kx\qquad(k\gt0),$$

is a linear restoring force, and the constant $k$ (unit $\text{N/m}$) is its force constant. The force of an ideal spring (Definition 5.6) is the best-known example; then $k$ is the spring constant.

Now apply Newton's second law (Section 4.1) to a particle of mass $m$ in SHM: the net force on it is $F=ma=-m\omega^2x$. This is a linear restoring force with $k=m\omega^2$. The argument also runs backwards: if the net force on a particle is $-kx$, its acceleration is $a=-\dfrac kmx$, and $x=A\cos(\omega t+\phi)$ satisfies this for any $A$ and $\phi$ provided $\omega^2=\dfrac km$. It can be shown that every motion with $a=-\dfrac kmx$ has this form, the constants $A$ and $\phi$ being fixed by the starting position and velocity.

📐 Result 13.5 — The force law for simple harmonic motion

A particle of mass $m$ executes SHM if and only if the net force on it is a linear restoring force,

$$F=-kx .$$

Its angular frequency and period are then

$$\omega=\sqrt{\frac km},\qquad T=2\pi\sqrt{\frac mk},$$

whatever the amplitude. The amplitude and phase constant depend only on how the motion is started.

The force law is the practical test for SHM: find the net force (or the acceleration) when the body is displaced by $x$ from equilibrium. If it is $-(\text{positive constant})\times x$, the motion is simple harmonic, and $\omega^2$ is that constant divided by $m$ (or, for the acceleration, the constant itself). A restoring force that is not proportional to $x$ — such as $-bx^3$ — still gives oscillations, but they are not simple harmonic and their period depends on the amplitude.

✏️ Example 3 — A particle under $F=-8.0x$

A particle of mass $0.20\ \text{kg}$ moves along the $x$-axis under a net force $F=-8.0x$ ($F$ in newtons, $x$ in metres). It is released from rest at $x=5.0\ \text{cm}$. Find (a) the angular frequency and the period; (b) the maximum speed; (c) the maximum acceleration and the maximum force; (d) the acceleration at $x=-2.0\ \text{cm}$.

💡 Show solution▶

(a) $k=8.0\ \text{N/m}$: $\omega=\sqrt{\dfrac{8.0}{0.20}}=\sqrt{40}=6.3\ \text{rad/s}$ ($6.32$), $T=\dfrac{2\pi}{6.32}=0.99\ \text{s}$. Released from rest, it starts at an extreme: $A=0.050\ \text{m}$.

(b) $v_\text{max}=\omega A=6.32\times0.050=0.32\ \text{m/s}$.

(d) $a=-\omega^2x=-40\times(-0.020)=+0.80\ \text{m/s}^2$ — towards $+x$, back towards the mean position.

Check Dimensions: $\sqrt{k/m}$ has units $\sqrt{\text{N m}^{-1}\text{kg}^{-1}}=\sqrt{\text{s}^{-2}}=\text{s}^{-1}$ ✓.

✏️ Example 4 — Is it simple harmonic?

For each net force or acceleration, say whether the motion about the equilibrium position is SHM, and if so find $\omega$ (mass $m$; $b$ a positive constant): (a) $F=-bx$ (b) $F=-bx^3$ (c) $F=-b(x-2)$ (d) $F=+bx$ (e) $a=-16\pi^2x$.

💡 Show solution▶

(a) Linear restoring force: SHM, $\omega=\sqrt{\dfrac bm}$.

(b) Restoring (it always points to $x=0$) but not proportional to $x$: the body oscillates, but not in SHM; its period depends on the amplitude.

(d) The force points away from $x=0$: the equilibrium is unstable and the body runs away — no oscillation at all.

(e) $\omega^2=16\pi^2$: SHM with $\omega=4\pi\ \text{rad/s}$, $\nu=2.0\ \text{Hz}$, $T=0.50\ \text{s}$.

A block oscillates in SHM (shown at half speed) along the strip at the top; the green arrow is its velocity, the red arrow its acceleration. The three graphs below share one time axis: $x$ (blue), $v$ (green), $a$ (red), each scaled to its own amplitude $A$, $\omega A$, $\omega^2A$. Follow the cursor and read off the phases: where $x$ peaks, $v$ is zero and $a$ is at its most negative. Change $T$ and watch $v_\text{max}$ and $a_\text{max}$ change.

amplitude $A$:5.0 cmperiod $T$:1.6 s

phase $\omega t$—

displacement $x$—

velocity $v$—

acceleration $a$—

Energy in SHM

A linear restoring force $F=-kx$ is the force of an ideal spring, and it is conservative: as the particle moves from $0$ to $x$ it does work $-\tfrac12kx^2$ whatever the route, so it has a potential energy $U=\tfrac12kx^2$, zero at the mean position (Result 5.6). This holds for any SHM, not only for springs, with $k=m\omega^2$. The kinetic energy follows from Result 13.3.

📐 Result 13.6 — Energy in SHM

For a particle of mass $m$ in SHM with force constant $k=m\omega^2$ and amplitude $A$, at displacement $x$:

$$K=\tfrac12mv^2=\tfrac12k\left(A^2-x^2\right),$$ $$U=\tfrac12kx^2,$$ $$E=K+U=\tfrac12kA^2=\tfrac12m\omega^2A^2 .$$

The total mechanical energy is constant and proportional to the square of the amplitude. At the mean position the energy is all kinetic; at the extremes it is all potential.

As functions of time, $K=E\sin^2(\omega t+\phi)$ and $U=E\cos^2(\omega t+\phi)$. Since $\sin^2\theta=\tfrac12(1-\cos2\theta)$, each energy oscillates between $0$ and $E$ at twice the frequency of the motion (period $\dfrac T2$): the kinetic energy peaks twice per cycle, once going each way through the centre. Over a whole cycle $\sin^2$ and $\cos^2$ each average $\tfrac12$, so the time-averaged kinetic and potential energies are equal, each $\tfrac12E=\tfrac14kA^2$. On a graph of energy against $x$, $U$ is a parabola, $E$ is a horizontal line, and $K$ is the gap between them; the particle is confined between the turning points where the line meets the parabola (compare the potential-energy curves of Section 5.2).

✏️ Example 5 — Energy book-keeping for a block on a spring

A $0.50\ \text{kg}$ block on a smooth horizontal surface is attached to a spring with $k=50\ \text{N/m}$ and oscillates with amplitude $4.0\ \text{cm}$. Find (a) the total energy; (b) the potential and kinetic energies and the speed at $x=2.0\ \text{cm}$; (c) the displacement at which $K=U$; (d) the shortest time from the mean position to that point.

💡 Show solution▶

(a) $E=\tfrac12kA^2=\tfrac12\times50\times(0.040)^2=0.040\ \text{J}$.

(b) $U=\tfrac12\times50\times(0.020)^2=0.010\ \text{J}$; $K=E-U=0.030\ \text{J}$; $v=\sqrt{\dfrac{2K}{m}}=\sqrt{\dfrac{0.060}{0.50}}=0.35\ \text{m/s}$ ($0.346$).

(d) $\omega=\sqrt{\dfrac{50}{0.50}}=10\ \text{rad/s}$, $T=0.628\ \text{s}$. From $0$ to $\dfrac{A}{\sqrt2}$ the reference radius turns $45^\circ$: $\dfrac T8=0.079\ \text{s}$.

Check $\omega\sqrt{A^2-x^2}=10\sqrt{0.0016-0.0004}=0.346\ \text{m/s}$ ✓. Interactive 2 starts with these values.

✏️ Example 6 — What changes when the amplitude doubles?

An oscillator is set swinging with twice its previous amplitude. By what factors do (a) the period, (b) the maximum speed, (c) the maximum acceleration and (d) the total energy change? (e) In the original motion, at what displacement is the speed half its maximum value?

💡 Show solution▶

(a) $T=2\pi\sqrt{\dfrac mk}$ does not contain $A$: unchanged (factor $1$).

(b), (c) $v_\text{max}=\omega A$ and $a_\text{max}=\omega^2A$ are proportional to $A$: both double.

(d) $E=\tfrac12kA^2$: four times as large.

(e) $\omega\sqrt{A^2-x^2}=\tfrac12\omega A$ gives $A^2-x^2=\tfrac14A^2$, $x=\pm\dfrac{\sqrt3}{2}A\approx\pm0.87A$.

Check In (e) $K=\tfrac14E$, so $U=\tfrac34E=\tfrac12kx^2$, $x^2=\tfrac34A^2$ ✓. With twice the amplitude, the block covers twice the distance in the same time — so it must go twice as fast.

Left: the potential energy $U=\tfrac12kx^2$ (parabola) and the total energy $E$ (horizontal line); the dot is the oscillator, and the green gap above it is its kinetic energy. Right: $K$ and $U$ against time over two periods of the motion (played in slow motion). Notice that each energy goes through two cycles while the block goes through one, and that $K+U$ never changes.

mass $m$:0.50 kg$k$:50 N/m$A$:4.0 cm

displacement $x$—

$U=\tfrac12kx^2$—

$K=E-U$—

$E$, period $T$—

⚠️ Common mistakes

1. "The acceleration is greatest where the speed is greatest." The opposite: at the mean position $v$ is maximum and $a=0$; at the extremes $v=0$ and $|a|$ is maximum.

2. Dropping the minus sign in $a=-\omega^2x$. The sign is the physics: the acceleration always points back towards the mean position.

3. $E\propto A$. Energy goes as the square of the amplitude: double $A$, four times $E$.

4. Period of the energy. $K$ and $U$ repeat every $\dfrac T2$ (frequency $2\nu$), not every $T$.

5. $K=U$ at $x=\dfrac A2$. At $x=\dfrac A2$, $U=\tfrac14E$ and $K=\tfrac34E$. Equal shares occur at $x=\dfrac{A}{\sqrt2}$.

6. Any restoring force gives SHM. Only a restoring force proportional to the displacement does; $F=-bx^3$ gives oscillations whose period depends on the amplitude.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — more systems that obey the force law

In JEE Main, NEET and JEE Advanced (as applications of simple harmonic motion); not required by the CBSE syllabus.

The force-law test of Result 13.5 works for any system: displace it by a small $x$, find the restoring force, and if it is $-kx$ read off $\omega=\sqrt{k/m}$, where $m$ is the whole moving mass.

Liquid in a U-tube. A liquid column of total length $L$ and density $\rho$ fills a U-tube of uniform cross-section $a$. Push the liquid down by $y$ in one arm: it rises by $y$ in the other, so the levels differ by $2y$, and the weight of that unbalanced column, $\rho a(2y)g$, pushes the whole mass $\rho aL$ back. So $F=-2\rho agy$ and

$$\omega=\sqrt{\frac{2g}{L}},\qquad T=2\pi\sqrt{\frac{L}{2g}} .$$

For $L=40\ \text{cm}$ of water (or mercury — the density cancels), $T=0.90\ \text{s}$.

A tunnel through the Earth. Inside a uniform Earth, $g$ falls in proportion to the distance $r$ from the centre (Section 7.2): the force on a body of mass $m$ is $-\dfrac{mg}{R_E}r$. A body dropped into a straight frictionless tunnel through the centre therefore oscillates in SHM with $T=2\pi\sqrt{\dfrac{R_E}{g}}=84$ minutes — the same as the period of a satellite skimming the surface.

Exercises

Take $g=9.8\ \text{m/s}^2$. Phases are in radians. Give answers to 2–3 significant figures.

Velocity in SHM.

1 An SHM has amplitude $5.0\ \text{cm}$ and period $0.20\ \text{s}$. Find the maximum speed.
2 For $x=0.10\cos8t$ (m, s), find the velocity at $t=0.10\ \text{s}$.
3 An SHM has amplitude $6.0\ \text{cm}$ and angular frequency $5.0\ \text{rad/s}$. Find the speed at $x=3.6\ \text{cm}$.
4 A particle in SHM has a maximum speed of $2.0\ \text{m/s}$ and an amplitude of $0.10\ \text{m}$. Find $\omega$ and $T$.
5 At what displacement (as a fraction of $A$) is the speed of a particle in SHM half its maximum value?
6 A particle in SHM has speeds of $8.0\ \text{cm/s}$ at $x=3.0\ \text{cm}$ and $6.0\ \text{cm/s}$ at $x=4.0\ \text{cm}$. Find the amplitude and the period.
7 For $x=4.0\sin\pi t$ (cm, s), find the velocity at $t=\tfrac13\ \text{s}$.
8 Use the reference circle to explain why the speed in SHM is greatest at the mean position and zero at the extremes.

💡 Show solutions▶

1 $\omega=\dfrac{2\pi}{0.20}=10\pi\ \text{rad/s}$; $v_\text{max}=10\pi\times0.050=1.6\ \text{m/s}$ ($1.57$).

2 $v=-0.80\sin(0.80)=-0.57\ \text{m/s}$ (the angle $0.80$ is in radians).

3 $|v|=5.0\sqrt{36-12.96}=5.0\times4.8=24\ \text{cm/s}$.

4 $\omega=\dfrac{v_\text{max}}{A}=20\ \text{rad/s}$; $T=\dfrac{2\pi}{20}=0.31\ \text{s}$.

5 $A^2-x^2=\tfrac14A^2$: $x=\pm\dfrac{\sqrt3}2A=\pm0.87A$.

6 $64=\omega^2(A^2-9)$, $36=\omega^2(A^2-16)$: $28=7\omega^2$, $\omega=2.0\ \text{rad/s}$; $A^2=9+16=25$, $A=5.0\ \text{cm}$; $T=\pi\ \text{s}=3.1\ \text{s}$.

7 $x=4.0\cos\left(\pi t-\dfrac\pi2\right)$, so $v=4.0\pi\cos\pi t$; at $t=\tfrac13$: $v=4.0\pi\times\tfrac12=6.3\ \text{cm/s}$, towards $+x$.

8 The reference particle always moves at $\omega A$ along the tangent. At the mean position the tangent is parallel to the line of motion, so the shadow moves at the full $\omega A$; at an extreme the tangent is perpendicular to it, so the shadow is momentarily at rest.

Acceleration and the force law.

1 An SHM has period $0.50\ \text{s}$. Find the acceleration at $x=2.0\ \text{cm}$.
2 A particle vibrates in SHM at $10\ \text{Hz}$ with amplitude $3.0\ \text{cm}$. Find its maximum acceleration.
3 A particle in SHM has $v_\text{max}=2.0\ \text{m/s}$ and $a_\text{max}=8.0\ \text{m/s}^2$. Find $\omega$, $T$ and $A$.
4 A $0.25\ \text{kg}$ body moves under a net force $F=-25x$ (N, m). Find $\omega$, $T$ and $\nu$.
5 Which of these describe SHM? Give $\omega$ where they do (mass $1.0\ \text{kg}$, SI units). (a) $a=-9x$ (b) $a=-9x^2$ (c) $F=4-2x$ (d) $a=9x$
6 A particle in SHM is at $x=+\dfrac A2$, moving towards $+x$. Give the directions of its velocity and acceleration. Is it speeding up or slowing down?
7 A $2.0\ \text{kg}$ body executes SHM with amplitude $0.20\ \text{m}$ and period $2.0\ \text{s}$. Find the maximum restoring force on it.
8 Show that the graph of $a$ against $x$ for an SHM is a straight line through the origin. If its slope is $-25\ \text{s}^{-2}$, find the period.

💡 Show solutions▶

1 $\omega=4\pi$: $a=-16\pi^2\times0.020=-3.2\ \text{m/s}^2$ (towards the mean position).

2 $\omega=20\pi$: $a_\text{max}=(20\pi)^2\times0.030=1.2\times10^2\ \text{m/s}^2$ ($118$).

3 $\omega=\dfrac{a_\text{max}}{v_\text{max}}=4.0\ \text{rad/s}$, $T=1.6\ \text{s}$ ($1.57$), $A=\dfrac{v_\text{max}}{\omega}=0.50\ \text{m}$.

4 $\omega=\sqrt{\dfrac{25}{0.25}}=10\ \text{rad/s}$; $T=0.63\ \text{s}$; $\nu=1.6\ \text{Hz}$.

5 (a) SHM, $\omega=3.0\ \text{rad/s}$. (b) Not SHM (and $-9x^2$ is not even restoring for $x\lt0$). (c) $F=-2(x-2)$: SHM about $x=2\ \text{m}$, $\omega=\sqrt2=1.4\ \text{rad/s}$. (d) Not SHM: the acceleration points away from equilibrium.

6 $v$ is towards $+x$ (given); $a=-\omega^2x$ is towards $-x$. They are opposite, so it is slowing down, on its way out to $+A$.

7 $\omega=\pi\ \text{rad/s}$: $F_\text{max}=m\omega^2A=2.0\times\pi^2\times0.20=3.9\ \text{N}$.

8 $a=-\omega^2x$ is of the form $y=mx$ with slope $-\omega^2$. $\omega^2=25$, $\omega=5.0\ \text{rad/s}$, $T=1.3\ \text{s}$ ($1.26$).

III

Energy in SHM.

1 A $0.20\ \text{kg}$ block on a spring of $k=80\ \text{N/m}$ oscillates with amplitude $5.0\ \text{cm}$. Find the total energy and the maximum speed.
2 For the oscillator in part 1, find $U$ and $K$ at $x=3.0\ \text{cm}$.
3 At what displacement are the kinetic and potential energies of an SHM equal?
4 What fraction of the total energy is kinetic when $x=\dfrac A2$? When $x=\dfrac{A}{3}$?
5 The amplitude of an oscillator is tripled. By what factors do its total energy, maximum speed and period change?
6 An SHM has period $0.80\ \text{s}$. Find the period and frequency of the oscillation of its kinetic energy.
7 An oscillator has total energy $0.10\ \text{J}$. What is its kinetic energy averaged over one complete period?
8 A $0.10\ \text{kg}$ particle in SHM has total energy $0.020\ \text{J}$, and its kinetic energy is $0.012\ \text{J}$ when $x=4.0\ \text{cm}$. Find the force constant, the amplitude and the period.

💡 Show solutions▶

1 $E=\tfrac12(80)(0.050)^2=0.10\ \text{J}$; $v_\text{max}=\sqrt{\dfrac{2E}{m}}=\sqrt{1.0}=1.0\ \text{m/s}$ (check: $\omega=20\ \text{rad/s}$, $\omega A=1.0$ ✓).

2 $U=\tfrac12(80)(0.030)^2=0.036\ \text{J}$; $K=0.10-0.036=0.064\ \text{J}$.

3 $\tfrac12kx^2=\tfrac14kA^2$: $x=\pm\dfrac A{\sqrt2}\approx\pm0.71A$.

4 $\dfrac KE=1-\dfrac{x^2}{A^2}$: $\dfrac34$ at $x=\dfrac A2$; $\dfrac89$ at $x=\dfrac A3$.

5 $E\times9$, $v_\text{max}\times3$, period unchanged.

6 $\dfrac T2=0.40\ \text{s}$; frequency $2.5\ \text{Hz}$ (twice $1.25\ \text{Hz}$).

7 $\tfrac12E=0.050\ \text{J}$.

8 $U=0.020-0.012=0.008\ \text{J}=\tfrac12k(0.040)^2$, so $k=10\ \text{N/m}$. $\tfrac12kA^2=0.020$: $A=\sqrt{0.0040}=0.063\ \text{m}$. $T=2\pi\sqrt{\dfrac{0.10}{10}}=0.63\ \text{s}$.

Applications and harder problems.

1 (Modelling) After hitting a bump, a car body bounces on its suspension at $1.2\ \text{Hz}$ with an amplitude of $3.0\ \text{cm}$. Find the maximum vertical acceleration of the passengers as a multiple of $g$.
2 (Modelling) A loudspeaker cone vibrates in SHM at $1.0\ \text{kHz}$ with an amplitude of $0.10\ \text{mm}$. Find its maximum speed and maximum acceleration (as a multiple of $g$).
3 A block in SHM passes through the mean position at $0.50\ \text{m/s}$ and has an acceleration of $-2.0\ \text{m/s}^2$ at $x=+8.0\ \text{cm}$. Find $\omega$, the amplitude and the period.
4 A coin rests on a horizontal platform that moves up and down in SHM with amplitude $5.0\ \text{cm}$. Find the greatest frequency at which the coin stays in contact with the platform throughout the motion. Where does it first lose contact if the frequency is exceeded?
5 A box rests on a horizontal platform that oscillates horizontally in SHM with period $1.0\ \text{s}$. The coefficient of static friction is $0.40$. Find the largest amplitude for which the box does not slip.
6 At what displacement is the kinetic energy of an SHM three times its potential energy? Sketch $K$, $U$ and $E$ against $x$.

💡 Show solutions▶

1 $a_\text{max}=(2\pi\times1.2)^2\times0.030=1.7\ \text{m/s}^2\approx0.17g$.

2 $\omega=2000\pi$: $v_\text{max}=2000\pi\times1.0\times10^{-4}=0.63\ \text{m/s}$; $a_\text{max}=(2000\pi)^2\times10^{-4}=3.9\times10^3\ \text{m/s}^2\approx400g$.

3 $\omega^2=\dfrac{2.0}{0.080}=25$, $\omega=5.0\ \text{rad/s}$; $A=\dfrac{0.50}{5.0}=0.10\ \text{m}$; $T=1.3\ \text{s}$ ($1.26$).

4 The platform can push but not pull, so the coin can accelerate downward by at most $g$. The largest downward acceleration, $\omega^2A$, occurs at the top. Contact is kept if $\omega^2A\le g$: $\omega\le\sqrt{\dfrac{9.8}{0.050}}=14\ \text{rad/s}$, $\nu\le2.2\ \text{Hz}$ ($2.23$). Above this, the coin leaves the platform on the way up, before the top, at the point where the required downward acceleration $\omega^2x$ first exceeds $g$ — a height $x=\dfrac{g}{\omega^2}$ above the mean position.

5 Friction must supply $m\omega^2A\le\mu_smg$: $A\le\dfrac{\mu_sg}{\omega^2}=\dfrac{0.40\times9.8}{(2\pi)^2}=0.099\ \text{m}\approx9.9\ \text{cm}$.

6 $K=3U$ means $U=\tfrac14E$: $x^2=\tfrac14A^2$, $x=\pm\dfrac A2$. Sketch: $U$ an upward parabola from $0$ at $x=0$ to $E$ at $\pm A$; $K$ an inverted parabola from $E$ at $x=0$ to $0$ at $\pm A$; $E$ a horizontal line.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks A particle of mass $m$ executes SHM $x=A\cos(\omega t+\phi)$. Obtain expressions for its kinetic and potential energies at displacement $x$, and show that its total energy is constant. Sketch both energies against $x$.

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K $v=-\omega A\sin(\omega t+\phi)$, so $K=\tfrac12m\omega^2A^2\sin^2(\omega t+\phi)=\tfrac12m\omega^2(A^2-x^2)$. [1]

U The force is $F=-m\omega^2x=-kx$; the work done against it from $0$ to $x$ is $\tfrac12kx^2$, so $U=\tfrac12m\omega^2x^2$. [1]

E $E=K+U=\tfrac12m\omega^2A^2$, independent of $x$ and $t$. Sketch: $U$ a parabola rising from zero at $x=0$; $K$ an inverted parabola, zero at $x=\pm A$; their sum a horizontal line. [1]

Exam tip Mark the points $x=\pm\dfrac A{\sqrt2}$ where the curves cross ($K=U=\tfrac12E$) — examiners look for them.

E2. NEET For a particle in SHM of amplitude $A$, the ratio of kinetic energy to potential energy at $x=\dfrac A2$ is (a) $1:3$ (b) $3:1$ (c) $1:1$ (d) $4:1$

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Idea $\dfrac KU=\dfrac{A^2-x^2}{x^2}=\dfrac{A^2-A^2/4}{A^2/4}=3$.

Answer $3:1$: (b).

Exam tip (a) inverts the ratio; (d) is $E:U$. At half the amplitude three-quarters of the energy is still kinetic.

E3. JEE Main A particle in SHM has speed $v_1$ at displacement $x_1$ and speed $v_2$ at displacement $x_2$. Its period is (a) $2\pi\sqrt{\dfrac{x_2^2-x_1^2}{v_1^2-v_2^2}}$ (b) $2\pi\sqrt{\dfrac{v_1^2-v_2^2}{x_2^2-x_1^2}}$ (c) $2\pi\sqrt{\dfrac{x_1^2+x_2^2}{v_1^2+v_2^2}}$ (d) $2\pi\sqrt{\dfrac{x_2^2-x_1^2}{v_1^2+v_2^2}}$

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Set-up $v_1^2=\omega^2(A^2-x_1^2)$, $v_2^2=\omega^2(A^2-x_2^2)$. Subtract: $v_1^2-v_2^2=\omega^2(x_2^2-x_1^2)$.

Period $\omega=\sqrt{\dfrac{v_1^2-v_2^2}{x_2^2-x_1^2}}$, $T=\dfrac{2\pi}\omega$.

Answer (a).

Exam tip (b) is $2\pi\omega$, not $\dfrac{2\pi}{\omega}$ — check dimensions: (a) has $\sqrt{\text{m}^2/(\text{m/s})^2}=\text{s}$ ✓, (b) has $\text{s}^{-1}$. Test with Example 2: $2\pi\sqrt{\dfrac{16-9}{256-144}}=\dfrac{2\pi}{4}$ ✓.

E4. JEE Main (Numerical value) A small coin is placed on a horizontal platform that oscillates vertically in SHM at $5.0\ \text{Hz}$. Find the largest amplitude, in millimetres (to the nearest integer), for which the coin never leaves the platform. Take $g=10\ \text{m/s}^2$ and $\pi^2=9.87$.

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Idea At the top the platform accelerates downward at $\omega^2A$; the coin can follow only if this does not exceed $g$ (the normal force cannot be negative).

Numbers $\omega=10\pi\ \text{rad/s}$, $\omega^2=987\ \text{s}^{-2}$; $A_\text{max}=\dfrac{g}{\omega^2}=\dfrac{10}{987}=1.01\times10^{-2}\ \text{m}$.

Answer $10$ (mm).

Exam tip The critical point is the top; at the bottom the platform pushes harder than $mg$, never less.

E5. JEE Advanced (One or more correct) A particle of mass $m$ moves in SHM, $x=A\cos\omega t$, with frequency $\nu$. Which statements are correct? (A) Its speed is greatest at the instants its acceleration is zero. (B) Its kinetic energy and its potential energy each oscillate with frequency $2\nu$. (C) Its acceleration leads its displacement by $\dfrac\pi2$. (D) The kinetic energy averaged over one period equals the potential energy averaged over one period.

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(A) $a=-\omega^2x=0$ at $x=0$, where $|v|=\omega A$ is greatest ✓.

(B) $K=E\sin^2\omega t=\tfrac E2(1-\cos2\omega t)$ and $U=\tfrac E2(1+\cos2\omega t)$: angular frequency $2\omega$, frequency $2\nu$ ✓.

(C) $a=\omega^2A\cos(\omega t+\pi)$: it leads by $\pi$ (antiphase), not $\dfrac\pi2$ ✗. It is the velocity that leads by $\dfrac\pi2$.

(D) Both average $\tfrac E2$ over a period ✓.

Answer (A), (B), (D).

Exam tip Write $x$, $v$, $a$ all as cosines before comparing phases; mixing sines and cosines is where the $\dfrac\pi2$ errors come from.

E6. NEET A body of mass $1.0\ \text{kg}$ executes SHM given by $x=6.0\cos\left(100t+\dfrac\pi4\right)$, with $x$ in centimetres and $t$ in seconds. Its maximum kinetic energy is (a) $18\ \text{J}$ (b) $9.0\ \text{J}$ (c) $36\ \text{J}$ (d) $1.8\times10^5\ \text{J}$

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Idea $v_\text{max}=\omega A=100\times0.060=6.0\ \text{m/s}$.

Answer $K_\text{max}=\tfrac12(1.0)(6.0)^2=18\ \text{J}$: (a).

Exam tip (d) forgets to convert centimetres to metres; (c) drops the $\tfrac12$; (b) is the average kinetic energy. The phase constant $\dfrac\pi4$ is irrelevant to energies.

🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. Can a phone's accelerometer measure the amplitude of an oscillation it cannot see — using only $a_\text{max}=\omega^2A$?

Why it matters. Vibration engineers monitor bridges, wind turbines, aircraft engines and factory machines with accelerometers, not rulers: a sensor glued to the structure records $a(t)$, and the displacement is inferred from $A=\dfrac{a_\text{max}}{\omega^2}$. Earthquake engineers specify ground motion the same way, as a peak acceleration in units of $g$, and international standards limit the vibration that workers' hands and bodies may receive.

What to do. Use a free sensor app (for example phyphox) that records acceleration against time.

Tape the phone firmly to a mass hanging from a long spring or a loop of strong elastic, or lay it flat on a swing seat. Pull it aside and release.
Record the acceleration along the direction of motion for at least $10$ cycles. From the graph, find the period $T$ (time $10$ cycles) and the peak acceleration $a_\text{max}$.
Calculate $A=\dfrac{a_\text{max}T^2}{4\pi^2}$. Measure the actual amplitude with a ruler or a video and compare.
Repeat with three different amplitudes. Is $a_\text{max}$ proportional to $A$? Does $T$ stay the same?
Check the phase relation: on the recorded graph, is the acceleration largest when the phone is at the end of its swing?

Tools: Smartphone with a sensor app, tape, spring or elastic, ruler, spreadsheet.

Hand in: Your $a(t)$ graph; a table of measured and predicted amplitudes; a plot of $a_\text{max}$ against $A$; and a paragraph on the main sources of disagreement.

Going further. Estimate the peak acceleration of your own head while walking and running (phone in a headband). Find out what peak ground acceleration a building in your region is designed to survive.

Instructor note: a phone on a spring with $T=1.0\ \text{s}$ and $A=5.0\ \text{cm}$ gives $a_\text{max}=4\pi^2\times0.050=2.0\ \text{m/s}^2$, easily resolved. For a vertical spring the app may show $a$ offset by $g$ (it reports specific force) — the oscillation is the variation about that offset. Agreement within $10$–$15\%$ is typical; differences come from the sensor's position relative to the mass's centre, damping, and side-to-side swinging. On a swing (a pendulum, Section 13.3) the tangential acceleration is the one to use.

🧪 Self-check quiz

0/5 correct

Q1. In SHM the speed of the particle is greatest

Q2. The acceleration of a particle in SHM is

Q3. The phase difference between the displacement and the acceleration of a particle in SHM is

Q4. The amplitude of an oscillator is doubled, the mass and force constant being unchanged. Its total energy becomes

Q5. A body of mass $0.50\ \text{kg}$ moves under the net force $F=-50x$ (SI units). Its period of oscillation is about

← 13.1 Periodic Motion and Simple Harmonic Motion 13.3 The Spring–Block Oscillator and the Simple Pendulum →