Solved Examples

How to Use This Problem Bank

Ten sections of theory, and now the part that actually earns marks. What follows is 44 worked problems covering the whole chapter — identifying which motions oscillate, reading a motion off a graph or an equation, reference circles, velocity and acceleration, springs alone and in combination, energy, the pendulum and its variants, damping, and driven oscillators at and away from resonance.

They are arranged easy first, hard last, in four parts that follow the order the chapter was taught in. Work them with a pen: cover the solution, try it, then compare — and read the takeaway at the end of each one, because that is where the marks usually leak away.

Notation for This Chapter

Key Point — THE THREE THAT GET CONFUSED: $$\omega = 2\pi\nu = \frac{2\pi}{T}$$

• $T$ is the period, in seconds — the time for one complete oscillation.
• $\nu$ is the frequency, in hertz — oscillations per second, $\nu = \frac{1}{T}$.
• $\omega$ is the angular frequency, in radians per second. It is $2\pi$ times the frequency, and it is not interchangeable with it.

A question that says "the frequency" wants $\nu$. A question that says "the angular frequency" wants $\omega$. Quoting one where the other was asked for is the single commonest wrong answer in this chapter, and it is worth zero marks even though the arithmetic was right.

Two more symbols that must not blur into each other. $\nu$ is the Greek letter nu, used only for frequency; $v$ is the italic vee of speed. And $k$ always means the spring constant in newtons per metre — never a wave number, never a Boltzmann constant.

Symbol	Meaning	Unit
$x$	displacement from the mean position	m
$A$	amplitude, the greatest value of $\lvert x \rvert$	m
$T$	period	s
$\nu$	frequency	Hz
$\omega$	angular frequency	rad/s
$(\omega t + \phi)$	phase at time $t$	rad
$\phi$	phase constant	rad
$k$	spring constant	N/m
$L$	pendulum length, support to centre of bob	m
$b$	damping constant	kg/s
$\omega_0$	natural angular frequency $\sqrt{\frac{k}{m}}$	rad/s
$\omega^{\,\prime}$	damped angular frequency	rad/s
$\omega_d$	driving angular frequency	rad/s

Two Conventions

1. Displacement is measured from the mean position. For a hanging mass on a vertical spring the mean position is the stretched equilibrium, a distance $x_0 = \frac{mg}{k}$ below the natural length. Measure from the natural length instead and every energy and every force comes out wrong. The gravity term cancels exactly about that new equilibrium, so a vertical spring has $T = 2\pi\sqrt{\frac{m}{k}}$ with no $g$ in it at all.
2. The standard form is the cosine one, $x = A\cos(\omega t + \phi)$, with the whole bracket called the phase and $\phi$ alone the phase constant. A sine start and the two-term form $A\cos\omega t + B\sin\omega t$ describe the same motion — $\phi$ absorbs the entire difference — and the conversions appear in Examples 8, 9 and 14. Angles inside $\sin$ and $\cos$ are in radians; a phase quoted as a bare number is in radians.

Key Point — every formula this section uses, in one place: $$x = A\cos(\omega t+\phi), \quad v = -\omega A\sin(\omega t+\phi), \quad a = -\omega^2 A\cos(\omega t+\phi) = -\omega^2 x$$ $$v_m = \omega A, \qquad a_m = \omega^2 A, \qquad v = \pm\omega\sqrt{A^2 - x^2}$$ $$F = -kx, \qquad k = m\omega^2, \qquad \omega = \sqrt{\frac{k}{m}}, \qquad T = 2\pi\sqrt{\frac{m}{k}}$$ $$k_{\text{eq}} = k_1 + k_2 \ \ \text{(parallel)}, \qquad \frac{1}{k_{\text{eq}}} = \frac{1}{k_1}+\frac{1}{k_2} \ \ \text{(series)}, \qquad k_{\text{piece}} = nk \ \ \text{(cut into } n)$$ $$K = \frac{1}{2}m\omega^2(A^2-x^2), \quad U = \frac{1}{2}kx^2, \quad E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2A^2$$ $$T_{\text{pendulum}} = 2\pi\sqrt{\frac{L}{g}}, \qquad T_{\text{lift}} = 2\pi\sqrt{\frac{L}{g \pm a}}, \qquad g_{\text{eff, car}} = \sqrt{g^2+a^2}$$ $$x = Ae^{-bt/2m}\cos(\omega^{\,\prime}t+\phi), \quad \omega^{\,\prime} = \sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}, \quad E(t) = \frac{1}{2}kA^2e^{-bt/m}$$ $$A_{\text{driven}} = \frac{F_0/m}{\sqrt{\left(\omega_0^2-\omega_d^2\right)^2 + \left(\frac{b\omega_d}{m}\right)^2}}$$

The series spring rule, the second's pendulum with its lift, car and liquid variants, damped oscillations and forced oscillations at resonance all sit outside the rationalised syllabus body text, yet a large fraction of the problems below cannot be done without them, and Boards, JEE and NEET ask about them every year. They are used freely here.

The constants used throughout

Quantity	Value
$g$ on the Earth	9.8 m/s²
$g$ on the Moon	1.7 m/s²
Density of water	1000 kg/m³
$\pi$	3.1416

Every solution also restates the constants it uses inside itself, so you never have to scroll back. Where a problem states a different value of $g$, that value is used and no other.

[Board Important] Every solution below writes the formula on its own line before any number goes into it, converts centimetres to metres on a line of its own, and names whether it is quoting $\omega$ or $\nu$. Do all three in the exam. A correct formula with an arithmetic slip still earns most of the marks; a $2\pi$ dropped between $\omega$ and $\nu$ loses them all.

Solved Examples

Part 1: Recognising the Motion, and Reading It Off

Eleven warm-ups. No springs and no pendulums yet — just the vocabulary, the three frequency-like quantities, and the business of turning a description, a graph or an equation into $x = A\cos(\omega t + \phi)$.

Constants for this part: none needed beyond $\pi = 3.1416$.

Example 1: Four motions, sorted into periodic and not

Which of these are periodic?

(a) A swimmer who crosses a river and comes back to the bank she started from, once. (b) A bar magnet hanging freely, turned away from the north-south line and released. (c) A hydrogen molecule spinning about its centre of mass. (d) An arrow shot from a bow.

Solution:

The test is the definition: periodic motion repeats itself at equal intervals of time, and it must keep doing so.

1. (a) Not periodic. One return trip is one event. Nothing repeats. If the swimmer kept going back and forth at a steady rate it would become periodic, but a single trip is not.
2. (b) Periodic — and oscillatory. The Earth's magnetic field supplies a restoring torque that always turns the magnet back towards the north-south line, so it swings to and fro about that line. Every oscillation is periodic.
3. (c) Periodic, not oscillatory. The molecule turns through the same angle in equal times and returns to the same configuration once per revolution, so the motion repeats. But it does not go to and fro about a mean position — it keeps going round the same way. Rotation is periodic without being an oscillation.
4. (d) Not periodic. The arrow flies once and stops in the target. There is no repetition at all.

Final Answer: (b) and (c) are periodic; (b) is also oscillatory. (a) and (d) are neither.

Takeaway: Every oscillation is periodic, but not every periodic motion is an oscillation. Rotation is the standard counterexample: it repeats without ever reversing. Ask two questions — does it repeat, and does it reverse about a mean position?

Example 2: Which of these are (nearly) simple harmonic?

For each of the following, say whether it is (nearly) simple harmonic or periodic but not simple harmonic.

(a) The Earth turning on its axis. (b) Mercury sloshing up and down in a U-tube. (c) A ball bearing released a little above the lowest point of a smooth curved bowl. (d) The general vibration of a polyatomic molecule about its equilibrium shape.

Solution:

The test for simple harmonic motion is a single sinusoid, equivalently a restoring force proportional to the displacement, equivalently $a = -\omega^2 x$.

1. (a) Periodic, not simple harmonic. The Earth spins steadily; nothing pulls it back towards a mean orientation. There is no restoring torque and no to-and-fro at all, so it cannot be SHM.
2. (b) Nearly simple harmonic. Push the mercury down on one side by $y$ and the level on the other side rises by $y$, so a column of excess height $2y$ is left unbalanced. The restoring force is proportional to $y$, which is exactly the condition for SHM. This one is worked in full in Example 30.
3. (c) Nearly simple harmonic. For a small displacement, the tangential restoring force on the bearing is proportional to the arc it has moved from the lowest point, again giving $a = -\omega^2 x$. Release it from far up the side of the bowl and the linearity fails, which is why the answer says "nearly".
4. (d) Periodic, not simple harmonic. A polyatomic molecule has several normal modes with different frequencies. A general vibration is a superposition of them, so it is periodic (if the frequencies are commensurate) but its displacement is a sum of sinusoids, not one.

Final Answer: (b) and (c) are nearly simple harmonic; (a) and (d) are periodic but not simple harmonic.

Takeaway: "Nearly" is doing real work in these answers. A U-tube and a bowl are simple harmonic only for small displacements; a superposition of several sinusoids is periodic but never simple harmonic, however many terms you add.

Example 3: Six functions of time, classified

For each function, say whether it represents simple harmonic motion, periodic motion that is not simple harmonic, or non-periodic motion. Give the period wherever there is one. Here $\omega$ is a positive constant.

(a) $\sin\omega t - \cos\omega t$ (b) $\sin^3\omega t$ (c) $3\cos\left(\frac{\pi}{4} - 2\omega t\right)$ (d) $\cos\omega t + \cos 3\omega t + \cos 5\omega t$ (e) $e^{-\omega^2t^2}$ (f) $1 + \omega t + \omega^2t^2$

Solution:

The test: a motion is simple harmonic only if it can be written as one term of the form $A\cos(\omega t + \phi)$.

1. (a) Take out $\sqrt{2}$: $$\sin\omega t - \cos\omega t = \sqrt{2}\left(\frac{1}{\sqrt{2}}\sin\omega t - \frac{1}{\sqrt{2}}\cos\omega t\right) = \sqrt{2}\sin\left(\omega t - \frac{\pi}{4}\right)$$ One sinusoid, so simple harmonic, with period $T = \frac{2\pi}{\omega}$.

2. (b) Use $\sin^3\theta = \frac{3\sin\theta - \sin 3\theta}{4}$: $$\sin^3\omega t = \frac{3\sin\omega t - \sin 3\omega t}{4}$$ Two sinusoids of different frequencies, so periodic but not simple harmonic. The longer of the two periods is $\frac{2\pi}{\omega}$, and the other, $\frac{2\pi}{3\omega}$, divides into it exactly three times, so the sum repeats after $T = \frac{2\pi}{\omega}$.

3. (c) Cosine is even, so $\cos\left(\frac{\pi}{4} - 2\omega t\right) = \cos\left(2\omega t - \frac{\pi}{4}\right)$: $$x = 3\cos\left(2\omega t - \frac{\pi}{4}\right)$$ One sinusoid, so simple harmonic — with angular frequency $2\omega$, and therefore period $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.

4. (d) Three sinusoids, so periodic but not simple harmonic. The individual periods are $\frac{2\pi}{\omega}$, $\frac{2\pi}{3\omega}$ and $\frac{2\pi}{5\omega}$; the smallest time containing a whole number of all three is $T = \frac{2\pi}{\omega}$.

5. (e) $e^{-\omega^2t^2}$ is a bell shape: it rises to 1 at $t = 0$ and dies away on both sides, never returning to a value it has already left. Non-periodic, and it is not even an oscillation.

6. (f) $1 + \omega t + \omega^2t^2$ grows without limit. Non-periodic.

Final Answer: SHM: (a) with $T = \frac{2\pi}{\omega}$, and (c) with $T = \frac{\pi}{\omega}$. Periodic but not SHM: (b) and (d), both with $T = \frac{2\pi}{\omega}$. Non-periodic: (e) and (f).

Takeaway: Count the sinusoids. One means simple harmonic; two or more of different frequencies means periodic but not simple harmonic; a growing or decaying function means neither. And watch the angular frequency inside the bracket in (c) — it is $2\omega$, so the period is half of what a careless reading gives.

Example 4: One measurement, three quantities

A metal strip clamped at one end is twanged and completes 40 full oscillations in 25 seconds. Find its period, its frequency and its angular frequency.

Solution:

1. Period is time per oscillation: $$T = \frac{25}{40} = 0.625 \ \text{s}$$

2. Frequency is the reciprocal, in hertz: $$\nu = \frac{1}{T} = \frac{1}{0.625} = 1.6 \ \text{Hz}$$

3. Angular frequency is $2\pi$ times that, in radians per second: $$\omega = 2\pi\nu = 2 \times 3.1416 \times 1.6 = 10.05 \ \text{rad/s}$$

4. Check the two routes agree. Straight from the period, $$\omega = \frac{2\pi}{T} = \frac{6.2832}{0.625} = 10.05 \ \text{rad/s}$$

Final Answer: $T = 0.625$ s, $\nu = 1.6$ Hz, $\omega = 10.05$ rad/s.

Takeaway: Write the unit next to every one of the three. Hertz for $\nu$, radians per second for $\omega$, seconds for $T$. A bare number with no unit is where the $2\pi$ goes missing.

Example 5: Going both ways between $\nu$ and $\omega$

(a) A loudspeaker cone vibrates at 440 Hz. Find its angular frequency and its period. (b) A different body oscillates with an angular frequency of 31.4 rad/s. Find its frequency and its period.

Solution:

1. (a) Frequency to angular frequency — multiply by $2\pi$: $$\omega = 2\pi\nu = 2 \times 3.1416 \times 440 = 2765 \ \text{rad/s}$$ $$T = \frac{1}{\nu} = \frac{1}{440} = 2.27 \times 10^{-3} \ \text{s} = 2.27 \ \text{milliseconds}$$

2. (b) Angular frequency to frequency — divide by $2\pi$: $$\nu = \frac{\omega}{2\pi} = \frac{31.4}{6.2832} = 5.00 \ \text{Hz}$$ $$T = \frac{1}{\nu} = \frac{1}{5.00} = 0.200 \ \text{s}$$

3. Sense check on (b). $31.4$ rad/s is very nearly $10\pi$, and $\omega = 10\pi$ means $\nu = 5$ Hz exactly. That is the standard way this number is dressed up in a question.

Final Answer: (a) $\omega = 2765$ rad/s, $T = 2.27$ ms. (b) $\nu = 5.00$ Hz, $T = 0.200$ s.

Takeaway: Going from $\nu$ to $\omega$ you multiply by $2\pi$; going the other way you divide. Say which direction you are travelling out loud before you touch the calculator, and a factor of $6.28$ will never land upside down.

Example 6: Everything from one line of algebra

A particle moves as $$x = 6\cos\left(4\pi t + \frac{\pi}{3}\right)$$ with $x$ in centimetres and $t$ in seconds. Find the amplitude, the angular frequency, the frequency, the period and the phase constant. Then find the displacement, the velocity and the acceleration at $t = 0$.

Solution:

Convention: the motion is already in the standard form $x = A\cos(\omega t + \phi)$, with $x$ measured from the mean position, so the constants can be read straight off.

1. Read off the constants. $$A = 6 \ \text{cm}, \qquad \omega = 4\pi = 12.57 \ \text{rad/s}, \qquad \phi = \frac{\pi}{3} \ \text{rad} = 60°$$

2. Frequency and period, kept firmly apart from $\omega$: $$\nu = \frac{\omega}{2\pi} = \frac{4\pi}{2\pi} = 2 \ \text{Hz}, \qquad T = \frac{1}{\nu} = 0.5 \ \text{s}$$

3. Displacement at $t = 0$. The phase at that instant is just $\phi$: $$x(0) = 6\cos 60° = 6 \times 0.5 = 3 \ \text{cm}$$

4. Velocity at $t = 0$, by differentiating: $$v = -\omega A\sin(\omega t + \phi) \ \Rightarrow \ v(0) = -12.57 \times 6 \times \sin 60° = -65.3 \ \text{cm/s}$$ The minus sign says the particle is heading back towards the mean position.

5. Acceleration at $t = 0$, fastest from $a = -\omega^2 x$: $$a(0) = -(12.57)^2 \times 3 = -474 \ \text{cm/s}^2 = -4.74 \ \text{m/s}^2$$

Final Answer: $A = 6$ cm, $\omega = 12.57$ rad/s, $\nu = 2$ Hz, $T = 0.5$ s, $\phi = \frac{\pi}{3}$; at $t = 0$, $x = 3$ cm, $v = -65.3$ cm/s, $a = -4.74$ m/s².

Takeaway: Use $a = -\omega^2 x$ rather than differentiating twice. Once you have $x$ at that instant, the acceleration is one multiplication away — and the minus sign is automatic.

Example 7: Everything a displacement-time graph will tell you

The graph below is the record of a particle in simple harmonic motion. Its peaks reach 3 cm, the first at $t = 1$ s and the next at $t = 5$ s, and at $t = 0$ the particle is at the mean position and moving in the positive direction. Find $A$, $T$, $\nu$, $\omega$ and $\phi$, and write the equation of the motion.

Solution:

Convention: the standard form $x = A\cos(\omega t + \phi)$, with $x$ measured from the mean position.

1. Amplitude — the height of a peak above the mean line. $$A = 3 \ \text{cm}$$

2. Period — the gap between two successive maxima. $$T = 5 - 1 = 4 \ \text{s}$$

3. Frequency and angular frequency. $$\nu = \frac{1}{T} = 0.25 \ \text{Hz}, \qquad \omega = \frac{2\pi}{T} = \frac{6.2832}{4} = 1.571 \ \text{rad/s} = \frac{\pi}{2} \ \text{rad/s}$$

4. Phase constant. Two facts fix it. At $t = 0$ the displacement is zero: $$0 = A\cos\phi \ \Rightarrow \ \phi = \pm\frac{\pi}{2}$$ and the particle is moving in the $+x$ direction, so $v(0) > 0$: $$v(0) = -\omega A\sin\phi > 0 \ \Rightarrow \ \sin\phi < 0 \ \Rightarrow \ \phi = -\frac{\pi}{2}$$

5. The equation. $$x = 3\cos\left(\frac{\pi}{2}t - \frac{\pi}{2}\right) \ \text{cm} \qquad \text{equivalently} \qquad x = 3\sin\left(\frac{\pi}{2}t\right) \ \text{cm}$$

6. Check against the graph. Put $t = 1$: $x = 3\cos(0) = 3$ cm, the first maximum. Put $t = 3$: $x = 3\cos(\pi) = -3$ cm, the trough shown at $t = 3$.

Final Answer: $A = 3$ cm, $T = 4$ s, $\nu = 0.25$ Hz, $\omega = \frac{\pi}{2}$ rad/s, $\phi = -\frac{\pi}{2}$, and $x = 3\cos\left(\frac{\pi}{2}t - \frac{\pi}{2}\right)$ cm.

Takeaway: A displacement value at $t = 0$ leaves two candidates for $\phi$; the direction of motion picks one. Never stop at $\cos\phi = 0$ — go on and check the sign of the velocity.

Example 8: A sine start rewritten in the standard form

A particle moves as $x = 4\sin\left(3t + \frac{\pi}{6}\right)$ cm. Write the same motion in the form $x = A\cos(\omega t + \phi)$, and state $A$, $\omega$, $\nu$, $T$ and $\phi$.

Solution:

1. The one identity that does the whole job. $$\sin\theta = \cos\left(\theta - \frac{\pi}{2}\right)$$

2. Apply it with $\theta = 3t + \frac{\pi}{6}$: $$x = 4\cos\left(3t + \frac{\pi}{6} - \frac{\pi}{2}\right) = 4\cos\left(3t - \frac{\pi}{3}\right) \ \text{cm}$$

3. Read off the constants. $$A = 4 \ \text{cm}, \qquad \omega = 3 \ \text{rad/s}, \qquad \phi = -\frac{\pi}{3} \ \text{rad} = -60°$$ $$\nu = \frac{\omega}{2\pi} = \frac{3}{6.2832} = 0.477 \ \text{Hz}, \qquad T = \frac{1}{\nu} = 2.09 \ \text{s}$$

4. Check at $t = 0$. From the sine form, $x(0) = 4\sin 30° = 2$ cm. From the cosine form, $x(0) = 4\cos(-60°) = 2$ cm. The same motion, two spellings.

Final Answer: $x = 4\cos\left(3t - \frac{\pi}{3}\right)$ cm, with $A = 4$ cm, $\omega = 3$ rad/s, $\nu = 0.477$ Hz, $T = 2.09$ s, $\phi = -\frac{\pi}{3}$.

Takeaway: Converting a sine to a cosine changes only the phase constant — it subtracts $\frac{\pi}{2}$. The amplitude, the angular frequency and the period are untouched, because it is the same motion timed from the same instant.

Example 9: A cosine and a sine added together

A particle moves as $x = 6\cos 5t + 8\sin 5t$ centimetres. Find its amplitude and its phase constant, and write the motion in the standard form.

Solution:

1. The general result. Any sum $x = P\cos\omega t + Q\sin\omega t$ with the same $\omega$ in both terms is a single simple harmonic motion: $$x = A\cos(\omega t + \phi), \qquad A = \sqrt{P^2+Q^2}, \qquad \tan\phi = -\frac{Q}{P}$$

2. Amplitude. Here $P = 6$ cm and $Q = 8$ cm: $$A = \sqrt{6^2 + 8^2} = \sqrt{36+64} = 10 \ \text{cm}$$

3. Phase constant. $$\tan\phi = -\frac{8}{6} = -1.333 \ \Rightarrow \ \phi = -53.13° = -0.927 \ \text{rad}$$ Take the value in the fourth quadrant, because $A\cos\phi = P = 6 > 0$ and $-A\sin\phi = Q = 8 > 0$ together need $\cos\phi > 0$ and $\sin\phi < 0$.

4. The motion. $$x = 10\cos(5t - 0.927) \ \text{cm}$$

5. Check at $t = 0$. The original gives $x(0) = 6$ cm; the standard form gives $10\cos(-0.927) = 10 \times 0.6 = 6$ cm. Check the velocity too: the original gives $v(0) = 8 \times 5 = 40$ cm/s, and the standard form gives $-10 \times 5 \times \sin(-0.927) = 50 \times 0.8 = 40$ cm/s.

Final Answer: $A = 10$ cm, $\phi = -0.927$ rad, and $x = 10\cos(5t - 0.927)$ cm.

Takeaway: Adding a cosine and a sine of the same frequency never makes a new kind of motion. It makes another simple harmonic motion, with amplitude $\sqrt{P^2+Q^2}$ — bigger than either term but not their sum.

Example 10: Amplitude and phase constant from a position and a velocity

A particle in simple harmonic motion is described by $x(t) = A\cos(\omega t + \phi)$ with $\omega = \pi$ rad/s. At $t = 0$ it is at $x = 1$ cm and moving with velocity $\omega$ cm/s in the positive direction. Find $A$ and $\phi$. Then find the amplitude $B$ and phase constant $\alpha$ if the same motion is written as $x = B\sin(\omega t + \alpha)$.

Solution:

Convention: displacement measured from the mean position, in centimetres, and all phases in radians.

1. Write down both conditions in the cosine form. $$x(0) = A\cos\phi = 1$$ $$v(0) = -\omega A\sin\phi = \omega \ \Rightarrow \ A\sin\phi = -1$$

2. Divide to get $\phi$, add squares to get $A$. $$\tan\phi = \frac{A\sin\phi}{A\cos\phi} = \frac{-1}{1} = -1$$ $$A^2 = (A\cos\phi)^2 + (A\sin\phi)^2 = 1 + 1 = 2 \ \Rightarrow \ A = \sqrt{2} = 1.41 \ \text{cm}$$

3. Choose the right root for $\phi$. $\tan\phi = -1$ allows $-\frac{\pi}{4}$ and $\frac{3\pi}{4}$. We need $\cos\phi > 0$, which only $-\frac{\pi}{4}$ gives: $$\phi = -\frac{\pi}{4} \ \text{rad} = -45°$$

4. Now the sine form. With $x = B\sin(\omega t + \alpha)$, $v = \omega B\cos(\omega t + \alpha)$: $$B\sin\alpha = 1, \qquad \omega B\cos\alpha = \omega \ \Rightarrow \ B\cos\alpha = 1$$ $$B = \sqrt{1+1} = \sqrt{2} = 1.41 \ \text{cm}, \qquad \tan\alpha = 1 \ \text{with both sine and cosine positive} \ \Rightarrow \ \alpha = \frac{\pi}{4}$$

5. The two answers agree, as they must. $\sqrt{2}\sin\left(\pi t + \frac{\pi}{4}\right) = \sqrt{2}\cos\left(\pi t + \frac{\pi}{4} - \frac{\pi}{2}\right) = \sqrt{2}\cos\left(\pi t - \frac{\pi}{4}\right)$.

Final Answer: $A = \sqrt{2}$ cm with $\phi = -\frac{\pi}{4}$; in the sine form, $B = \sqrt{2}$ cm with $\alpha = +\frac{\pi}{4}$.

Takeaway: The amplitude never depends on which form you choose — only the phase constant does, and it shifts by exactly $\frac{\pi}{2}$. Getting $A$ and $B$ different is a sure sign of an algebra slip.

Example 11: Does starting it harder change anything?

A block on a spring is set going with an amplitude of 2 cm and found to have a period of 0.40 second. It is stopped and restarted, this time with an amplitude of 4 cm. What happens to the period, the frequency, the maximum speed and the maximum acceleration?

Solution:

1. Angular frequency first, since everything hangs off it: $$\omega = \frac{2\pi}{T} = \frac{6.2832}{0.40} = 15.71 \ \text{rad/s}, \qquad \nu = \frac{1}{T} = 2.5 \ \text{Hz}$$

2. Period and frequency: unchanged. $\omega = \sqrt{\frac{k}{m}}$ contains only the spring constant and the mass. The amplitude is set by how far you pulled the block, and does not appear. So $T$ stays at 0.40 second and $\nu$ stays at 2.5 Hz.

3. Maximum speed: doubled. $v_m = \omega A$ is proportional to $A$: $$v_m = 15.71 \times 0.02 = 0.314 \ \text{m/s} \ \longrightarrow \ 15.71 \times 0.04 = 0.628 \ \text{m/s}$$

4. Maximum acceleration: doubled. $a_m = \omega^2A$ is also proportional to $A$: $$a_m = (15.71)^2 \times 0.02 = 4.93 \ \text{m/s}^2 \ \longrightarrow \ 9.87 \ \text{m/s}^2$$

5. Why the period can stay put while the speed doubles. The block has twice as far to travel in each quarter cycle, but it travels the whole way twice as fast. The two changes cancel exactly, and only a linear restoring force makes them cancel exactly.

Final Answer: $T$ and $\nu$ unchanged at 0.40 s and 2.5 Hz; $v_m$ doubles from 0.314 to 0.628 m/s; $a_m$ doubles from 4.93 to 9.87 m/s².

Takeaway: The period of a simple harmonic oscillator is a property of the system, not of the start. Change the amplitude and the period will not move — which is exactly why a pendulum can keep time.

Part 2: The Reference Circle, Velocity and Acceleration

Eleven problems on the two tools that turn timing questions into geometry and position questions into one line of algebra: the reference circle, and $v = \pm\omega\sqrt{A^2-x^2}$.

Constants for this part: $\pi = 3.1416$.

Key Point: On the reference circle, a particle goes round a circle of radius $A$ with angular speed $\omega$, and its shadow on a diameter performs the SHM. Equal angles are swept in equal times, so a time interval is just $\frac{\text{angle swept}}{2\pi} \times T$. That is the whole method.

Example 12: How long does it take to get from here to there?

A particle executes simple harmonic motion of amplitude 6 cm with a period of 1.2 seconds. Find the time it takes to go (a) from $x = +3$ cm to the extreme at $x = +6$ cm, and (b) straight from $x = -3$ cm to $x = +3$ cm.

Solution:

Convention: $x = A\cos(\omega t + \phi)$ measured from the mean position; on the reference circle the phase angle is measured from the positive $x$-axis.

1. Turn each displacement into an angle on the circle using $x = A\cos\theta$: $$x = +6 \ \text{cm} \ \Rightarrow \ \cos\theta = 1 \ \Rightarrow \ \theta = 0°$$ $$x = +3 \ \text{cm} \ \Rightarrow \ \cos\theta = \frac{3}{6} = 0.5 \ \Rightarrow \ \theta = 60°$$ $$x = -3 \ \text{cm} \ \Rightarrow \ \cos\theta = -0.5 \ \Rightarrow \ \theta = 120°$$

2. (a) From $x = +3$ to $x = +6$ is a sweep from $60°$ to $0°$, an angle of $60°$: $$t = \frac{60°}{360°} \times T = \frac{T}{6} = \frac{1.2}{6} = 0.20 \ \text{s}$$

3. (b) From $x = -3$ to $x = +3$ is a sweep from $120°$ to $60°$, again $60°$: $$t = \frac{T}{6} = 0.20 \ \text{s}$$

4. Why the two are equal, and why that is surprising. The second journey covers 6 cm of ground, the first only 3 cm — but the second is made through the middle, where the particle is moving fastest, and the first near the extreme, where it is barely moving at all. The circle knows this automatically; a straight "distance over average speed" does not.

Final Answer: 0.20 second in both cases.

Takeaway: Convert displacements to angles, subtract, and scale by $\frac{T}{360°}$. No integration, no calculator gymnastics — and it never breaks for a displacement that is not a neat fraction of $A$.

Example 13: Two revolving particles, and the shadows they cast

Two particles move in uniform circular motion.

• Particle P moves on a circle of radius 3 cm with a period of 2 seconds. At $t = 0$ it is at the far left of the circle, on the negative $x$-axis, and it goes round anticlockwise.
• Particle Q moves on a circle of radius 2 m with a period of 4 seconds. At $t = 0$ it is at the top of the circle, on the positive $y$-axis, and it goes round clockwise.

Write the simple harmonic motion performed by the $x$-projection of each.

Solution:

Convention: angles measured anticlockwise from the positive $x$-axis; the projection is $x = A\cos\theta(t)$.

1. Particle P. Angular speed: $$\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \ \text{rad/s}$$ It starts at $\theta = 180°$ and the angle increases (anticlockwise), so $\theta(t) = \pi + \pi t$: $$x = 3\cos(\pi t + \pi) = -3\cos(\pi t) \ \text{cm}$$

2. Particle Q. Angular speed: $$\omega = \frac{2\pi}{4} = \frac{\pi}{2} \ \text{rad/s}$$ It starts at $\theta = 90°$ and the angle decreases (clockwise), so $\theta(t) = \frac{\pi}{2} - \frac{\pi}{2}t$: $$x = 2\cos\left(\frac{\pi}{2} - \frac{\pi}{2}t\right) = 2\cos\left(\frac{\pi}{2}t - \frac{\pi}{2}\right) = 2\sin\left(\frac{\pi}{2}t\right) \ \text{m}$$

3. Check both at $t = 0$. P: $x = -3$ cm, which is the far left of its circle. Q: $x = 0$, which is directly below the top of the circle. Both correct.

4. Check the direction of travel just after $t = 0$. P is at the leftmost point and moving anticlockwise, so it is heading downwards and its shadow starts moving in the $+x$ direction: $v = 3\pi\sin(\pi t)$, which is positive for small $t$. Q is at the top moving clockwise, so it heads to the right: $v = \pi\cos\left(\frac{\pi}{2}t\right)$, positive at $t = 0$.

Final Answer: P: $x = -3\cos(\pi t)$ cm. Q: $x = 2\sin\left(\frac{\pi}{2}t\right)$ m.

Takeaway: The only thing the sense of rotation changes is the sign in front of $\omega t$. Anticlockwise gives $\theta_0 + \omega t$, clockwise gives $\theta_0 - \omega t$. Everything else follows from $x = A\cos\theta$.

Example 14: Turning three equations into circle diagrams

For each motion below ($x$ in centimetres, $t$ in seconds), find the radius of the reference circle, the angular speed of the reference particle, and the angle at which it starts, taking the rotation to be anticlockwise throughout.

(a) $x = -2\sin\left(3t + \frac{\pi}{3}\right)$ (b) $x = \cos\left(\frac{\pi}{6} - t\right)$ (c) $x = 3\sin\left(2\pi t + \frac{\pi}{4}\right)$

Solution:

Convention: get every motion into $x = A\cos(\omega t + \phi)$ with $A > 0$; then the radius is $A$, the angular speed is $\omega$, and the starting angle is $\phi$.

1. (a) Use $-\sin\theta = \cos\left(\theta + \frac{\pi}{2}\right)$: $$x = 2\cos\left(3t + \frac{\pi}{3} + \frac{\pi}{2}\right) = 2\cos\left(3t + \frac{5\pi}{6}\right)$$ Radius 2 cm, angular speed 3 rad/s, starting angle $\frac{5\pi}{6} = 150°$. The period is $T = \frac{2\pi}{3} = 2.09$ s, and at $t = 0$ the particle sits at $x = 2\cos 150° = -1.73$ cm.

2. (b) Cosine is even, so the minus sign inside can be flipped: $$x = \cos\left(t - \frac{\pi}{6}\right)$$ Radius 1 cm, angular speed 1 rad/s, starting angle $-\frac{\pi}{6} = -30°$. Period $T = 2\pi = 6.28$ s, and $x(0) = \cos(-30°) = 0.87$ cm.

3. (c) Use $\sin\theta = \cos\left(\theta - \frac{\pi}{2}\right)$: $$x = 3\cos\left(2\pi t + \frac{\pi}{4} - \frac{\pi}{2}\right) = 3\cos\left(2\pi t - \frac{\pi}{4}\right)$$ Radius 3 cm, angular speed $2\pi = 6.28$ rad/s, starting angle $-\frac{\pi}{4} = -45°$. Here $\nu = 1$ Hz and $T = 1$ s, and $x(0) = 3\cos(-45°) = 2.12$ cm.

Final Answer: (a) radius 2 cm, $\omega = 3$ rad/s, start at $150°$. (b) radius 1 cm, $\omega = 1$ rad/s, start at $-30°$. (c) radius 3 cm, $\omega = 2\pi$ rad/s, start at $-45°$.

Takeaway: A minus sign in front of the amplitude is not a negative amplitude — it is a phase shift of $\pi$. Absorb every sign and every sine into $\phi$ before you draw anything, and the diagram comes out right first time.

Example 15: Position, velocity and acceleration at a stated instant

A particle moves as $x = 5\cos\left(2t + \frac{\pi}{6}\right)$ centimetres. Find its displacement, velocity and acceleration at $t = 1.0$ s, and say which way it is moving and which way it is being pushed.

Solution:

1. Read off the constants. $$A = 5 \ \text{cm}, \qquad \omega = 2 \ \text{rad/s}, \qquad \phi = \frac{\pi}{6} = 0.5236 \ \text{rad}$$

2. The phase at that instant, in radians: $$\omega t + \phi = 2(1.0) + 0.5236 = 2.5236 \ \text{rad} = 144.6°$$

3. Displacement. $$x = 5\cos(2.5236) = 5 \times (-0.8150) = -4.08 \ \text{cm}$$

4. Velocity. $$v = -\omega A\sin(\omega t + \phi) = -(2)(5)\sin(2.5236) = -10 \times 0.5794 = -5.79 \ \text{cm/s}$$

5. Acceleration, straight from $a = -\omega^2 x$: $$a = -(2)^2 \times (-4.08) = +16.3 \ \text{cm/s}^2$$

6. Read the signs. The particle is on the negative side, still travelling further negative ($v < 0$), and the acceleration points back towards the mean position ($a > 0$) and is slowing it down. It is on its way to the extreme at $x = -5$ cm.

Final Answer: $x = -4.08$ cm, $v = -5.79$ cm/s, $a = +16.3$ cm/s².

Takeaway: The acceleration always points back towards the mean position, so its sign is always opposite to $x$. If your $a$ and your $x$ come out with the same sign, you have made an arithmetic error, not discovered a new kind of motion.

Example 16: Speed at three displacements, without touching the clock

A particle executes simple harmonic motion of amplitude 5 cm with a period of 0.40 second. Find its speed at the mean position, at $x = 3$ cm and at $x = 4$ cm.

Solution:

1. Angular frequency. $$\omega = \frac{2\pi}{T} = \frac{6.2832}{0.40} = 15.71 \ \text{rad/s}$$

2. The relation that skips the time entirely. $$v = \pm\omega\sqrt{A^2 - x^2}$$

3. At the mean position, $x = 0$: $$\lvert v \rvert = 15.71 \times \sqrt{25 - 0} = 15.71 \times 5 = 78.5 \ \text{cm/s}$$ which is also $v_m = \omega A$, as it must be.

4. At $x = 3$ cm: $$\lvert v \rvert = 15.71\sqrt{25-9} = 15.71 \times 4 = 62.8 \ \text{cm/s}$$

5. At $x = 4$ cm: $$\lvert v \rvert = 15.71\sqrt{25-16} = 15.71 \times 3 = 47.1 \ \text{cm/s}$$

6. Notice the pattern. The particle is 60% of the way out at $x = 3$ cm but still moving at 80% of top speed; at 80% of the way out it is still at 60%. Speed falls off slowly at first and then collapses near the ends.

Final Answer: 78.5 cm/s, 62.8 cm/s and 47.1 cm/s respectively.

Takeaway: $v = \pm\omega\sqrt{A^2-x^2}$ answers "how fast is it moving when it is here?" in one line. The sign is the only thing it cannot tell you — that depends on which way the particle happens to be going.

Example 17: Where is the speed half its maximum, and where is the acceleration?

A particle executes simple harmonic motion of amplitude 8 cm. Find the displacements at which (a) its speed is half its maximum speed, and (b) the magnitude of its acceleration is half its maximum.

Solution:

1. (a) Speed. Set $\lvert v \rvert = \frac{v_m}{2} = \frac{\omega A}{2}$ in $\lvert v \rvert = \omega\sqrt{A^2-x^2}$: $$\omega\sqrt{A^2-x^2} = \frac{\omega A}{2} \ \Rightarrow \ A^2 - x^2 = \frac{A^2}{4} \ \Rightarrow \ x^2 = \frac{3A^2}{4}$$ $$x = \pm\frac{\sqrt{3}}{2}A = \pm 0.866 \times 8 = \pm 6.93 \ \text{cm}$$

2. (b) Acceleration. Here $\lvert a \rvert = \omega^2\lvert x \rvert$ is directly proportional to the displacement, so half the maximum acceleration means half the amplitude: $$\omega^2\lvert x \rvert = \frac{\omega^2A}{2} \ \Rightarrow \ x = \pm\frac{A}{2} = \pm 4.0 \ \text{cm}$$

3. Why the two answers are so different. Acceleration is linear in $x$, so halving it halves the displacement. Speed depends on $\sqrt{A^2-x^2}$, so halving it barely moves the particle in from the extreme — it is already at $86.6\%$ of the amplitude.

Final Answer: (a) $x = \pm 6.93$ cm. (b) $x = \pm 4.0$ cm.

Takeaway: "Half the maximum" means two completely different places for $v$ and for $a$. $\lvert a \rvert$ scales with $x$; $\lvert v \rvert$ scales with $\sqrt{A^2-x^2}$. Never carry an answer from one to the other.

Example 18: The piston in a locomotive cylinder

The piston in the cylinder head of a locomotive has a stroke — that is, twice the amplitude — of 1.0 m, and it moves in simple harmonic motion with an angular frequency of 200 rad/min. Find its maximum speed, and also its frequency, its period and its maximum acceleration.

Solution:

1. Amplitude from the stroke. The stroke is the full travel from one extreme to the other, which is $2A$: $$A = \frac{1.0}{2} = 0.50 \ \text{m}$$

2. Convert the angular frequency to SI. It was given per minute, and every formula below wants per second: $$\omega = \frac{200}{60} = 3.333 \ \text{rad/s}$$

3. Maximum speed. $$v_m = \omega A = 3.333 \times 0.50 = 1.67 \ \text{m/s}$$

4. Frequency and period, keeping them distinct from $\omega$: $$\nu = \frac{\omega}{2\pi} = \frac{3.333}{6.2832} = 0.531 \ \text{Hz}, \qquad T = \frac{1}{\nu} = 1.88 \ \text{s}$$

5. Maximum acceleration. $$a_m = \omega^2A = (3.333)^2 \times 0.50 = 5.56 \ \text{m/s}^2$$

Final Answer: $v_m = 1.67$ m/s, with $\nu = 0.531$ Hz, $T = 1.88$ s and $a_m = 5.56$ m/s².

Takeaway: The stroke is $2A$, and radians per minute are not radians per second. Two unit traps in one short question, and each of them alone is enough to lose the mark.

Example 19: The signs of velocity, acceleration and force, point by point

A particle is in linear simple harmonic motion between two points $A$ and $B$ which are 10 cm apart. Take the direction from $A$ to $B$ as positive. Give the signs of the velocity, the acceleration and the force on the particle when it is:

(a) at the end $A$; (b) at the end $B$; (c) at the mid-point of $AB$, going towards $A$; (d) 2 cm from $B$, going towards $A$; (e) 3 cm from $A$, going towards $B$; (f) 4 cm from $B$, going towards $A$.

Solution:

Convention: the mean position is the mid-point of $AB$, and displacement is measured from there. With $AB = 10$ cm, the amplitude is 5 cm, so $A$ sits at $x = -5$ cm and $B$ at $x = +5$ cm.

1. The two rules that decide everything. The velocity sign is simply the direction of travel: towards $B$ is positive, towards $A$ is negative. The acceleration is $a = -\omega^2 x$ and the force is $F = ma$, so both always have the sign opposite to $x$ — and both are zero at the mean position.

2. Locate each case on the $x$-axis, then apply the rules.

Case	Position $x$	Moving	$v$	$a$	$F$
(a) at $A$	$-5$ cm	at rest, turning	$0$	$+$	$+$
(b) at $B$	$+5$ cm	at rest, turning	$0$	$-$	$-$
(c) mid-point, towards $A$	$0$	towards $A$	$-$	$0$	$0$
(d) 2 cm from $B$, towards $A$	$+3$ cm	towards $A$	$-$	$-$	$-$
(e) 3 cm from $A$, towards $B$	$-2$ cm	towards $B$	$+$	$+$	$+$
(f) 4 cm from $B$, towards $A$	$+1$ cm	towards $A$	$-$	$-$	$-$

3. Reading the table. In cases (d), (e) and (f) the velocity and the force come out with the same sign. That is no accident: in each of those three the particle is heading towards the mean position, so the restoring force pulls it the way it is already going and the speed is rising. Whenever a particle is heading away from the mean position instead, the two signs disagree and it is slowing down.

Final Answer: as tabulated: (a) $0, +, +$; (b) $0, -, -$; (c) $-, 0, 0$; (d) $-, -, -$; (e) $+, +, +$; (f) $-, -, -$.

Takeaway: Force and acceleration never depend on which way the particle is going — only on where it is. Fix the origin at the mean position, read the sign of $x$, and flip it. The velocity is the only entry that needs the direction of travel.

Example 20: Which acceleration laws give simple harmonic motion?

Which of these relations between acceleration $a$ and displacement $x$ (SI units throughout) describe simple harmonic motion? For any that do, find $\omega$, $T$ and $\nu$.

(a) $a = 0.7x$ (b) $a = -200x^2$ (c) $a = -10x$ (d) $a = 100x^3$

Solution:

1. The test. Simple harmonic motion means $a = -\omega^2x$: proportional to $x$, with a minus sign, and $\omega^2$ positive.

2. (a) $a = 0.7x$. Proportional, but the sign is wrong. The acceleration points away from the mean position, so any small displacement grows. This is unstable equilibrium — a ball balanced on top of a dome — not oscillation.

3. (b) $a = -200x^2$. The sign in front is negative, but $x^2$ is never negative, so $a$ is negative on both sides of the origin. That is not a restoring acceleration, and in any case it is not proportional to $x$. Not SHM.

4. (c) $a = -10x$. Proportional and restoring. Simple harmonic, with $$\omega^2 = 10 \ \Rightarrow \ \omega = 3.162 \ \text{rad/s}$$ $$T = \frac{2\pi}{\omega} = \frac{6.2832}{3.162} = 1.99 \ \text{s}, \qquad \nu = \frac{1}{T} = 0.503 \ \text{Hz}$$

5. (d) $a = 100x^3$. Not proportional to $x$, and the sign is wrong as well. Not SHM.

Final Answer: only (c), with $\omega = 3.16$ rad/s, $T = 1.99$ s and $\nu = 0.503$ Hz.

Takeaway: Two conditions, both compulsory: first power of $x$, and a minus sign. Fail either and it is not simple harmonic — and $\omega$ comes from the magnitude of the coefficient, with $\omega = \sqrt{\lvert \text{coefficient} \rvert}$.

Example 21: A spring on a table, from equation to graphs

A spring of spring constant 1200 N/m is clamped at one end on a horizontal table, and a 3 kg block on a frictionless surface is attached to its free end. The block is pulled sideways 2.0 cm and released. Find (i) the frequency of the oscillation, (ii) the maximum acceleration of the block and (iii) its maximum speed.

Solution:

1. Angular frequency from the spring and the mass. $$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{1200}{3}} = \sqrt{400} = 20 \ \text{rad/s}$$

2. (i) Frequency — the question asked for $\nu$, not $\omega$: $$\nu = \frac{\omega}{2\pi} = \frac{20}{6.2832} = 3.18 \ \text{Hz}, \qquad T = \frac{1}{\nu} = 0.314 \ \text{s}$$

3. Amplitude. The block was released from rest 2.0 cm from the mean position, so that is the amplitude: $$A = 2.0 \ \text{cm} = 0.020 \ \text{m}$$

4. (ii) Maximum acceleration, which happens at the extremes: $$a_m = \omega^2A = 400 \times 0.020 = 8.0 \ \text{m/s}^2$$

5. (iii) Maximum speed, which happens at the mean position: $$v_m = \omega A = 20 \times 0.020 = 0.40 \ \text{m/s}$$

6. The three curves together. Displacement peaks at the extremes, velocity a quarter of a period later at the mean position, and acceleration back in step with displacement but reversed in sign — the figure at the top of this section is drawn for exactly this oscillator.

Final Answer: $\nu = 3.18$ Hz, $a_m = 8.0$ m/s², $v_m = 0.40$ m/s.

Takeaway: Release from rest means the release point is the amplitude. And when a question says "frequency", divide $\omega$ by $2\pi$ before you write the answer down — quoting 20 here instead of 3.18 is the standard way to lose this mark.

Example 22: The same oscillator, three different stopwatch starts

Take the oscillator of the previous example — 1200 N/m, 3 kg, amplitude 2.0 cm — and measure $x$ from the mean position, with the positive direction to the right. Write $x$ as a function of time if, at the instant the stopwatch is started, the block is (a) at the mean position and moving to the right, (b) at its right-hand extreme, and (c) at its left-hand extreme. In what respect do the three answers differ?

Solution:

Convention: the standard form $x = A\cos(\omega t + \phi)$, with $x$ measured from the mean position. From the previous example, $A = 2.0$ cm and $\omega = 20$ rad/s.

1. (a) Starting at the mean position, moving right. We need $x(0) = 0$ with $v(0) > 0$: $$0 = A\cos\phi \ \Rightarrow \ \phi = \pm\frac{\pi}{2}, \qquad v(0) = -\omega A\sin\phi > 0 \ \Rightarrow \ \sin\phi < 0 \ \Rightarrow \ \phi = -\frac{\pi}{2}$$ $$x = 2\cos\left(20t - \frac{\pi}{2}\right) = 2\sin(20t) \ \text{cm}$$

2. (b) Starting at the right-hand extreme, $x(0) = +A$: $$A = A\cos\phi \ \Rightarrow \ \cos\phi = 1 \ \Rightarrow \ \phi = 0$$ $$x = 2\cos(20t) \ \text{cm}$$

3. (c) Starting at the left-hand extreme, $x(0) = -A$: $$-A = A\cos\phi \ \Rightarrow \ \cos\phi = -1 \ \Rightarrow \ \phi = \pi$$ $$x = 2\cos(20t + \pi) = -2\cos(20t) \ \text{cm}$$

4. Compare the three. All three have amplitude 2.0 cm and angular frequency 20 rad/s, hence the same frequency 3.18 Hz and the same period 0.314 s. The only difference is the phase constant — $-\frac{\pi}{2}$, $0$ and $\pi$ respectively. Starting the stopwatch at a different moment cannot change how stiff the spring is or how heavy the block is.

Final Answer: (a) $x = 2\sin(20t)$ cm; (b) $x = 2\cos(20t)$ cm; (c) $x = -2\cos(20t)$ cm. They differ only in the phase constant, not in amplitude or frequency.

Takeaway: The phase constant carries all the information about when you started watching. It is set by the initial conditions and by nothing else — which is why changing it never changes the period, and why an exam answer that also changes $A$ or $\omega$ has gone wrong somewhere.