Oscillations & SHM

Introduction & Core Concept

*Assalam-o-Alaikum*, my dear students. My name is Dr. Amir Hussain, and for the next hour or so, I will be your guide into one of the most elegant and fundamental topics in all of physics: Oscillations.

Imagine you are in Lahore, sitting in the beautiful Bagh-e-Jinnah on a pleasant spring evening. You see a child on a *jhoola* (a swing). Her father pulls the swing back and releases it. It glides forward, reaches a peak, and glides back. Forward, back, forward, back. This rhythmic, repetitive motion is an oscillation. Now, think about the string of a *sitar* being plucked by a musician in a concert hall, the tremor of the earth during an earthquake near the northern areas, or even the alternating current (AC) that WAPDA supplies to our homes to power our lives. All of these are governed by the principles of oscillatory motion.

So, why do we dedicate so much time to studying this? Because oscillations are the very heart of waves. Understanding how a single particle oscillates is the first step to understanding how sound waves travel, how light illuminates our world, and how radio signals from a PTCL tower connect our phones. It is the language of vibrations, and vibrations build our physical universe.

The big-picture mental model for all oscillations is a constant tug-of-war. On one side, you have inertia – the tendency of an object to keep moving. On the other side, you have a restoring force – a force that always tries to pull the object back to a central, stable point called the equilibrium position.

Think of the *jhoola* again. As the child swings away from the lowest point (equilibrium), gravity acts as a restoring force, trying to pull her back down. At the lowest point, she is moving fastest (maximum inertia), which carries her up the other side. As she rises, the restoring force slows her down, stops her at the peak, and pulls her back again. This beautiful dance between inertia and a restoring force is the essence of oscillation. Our goal today is to describe this dance with the precise and powerful language of mathematics.

Theoretical Foundation

Let's build our understanding from the ground up, like constructing a solid building.

The Defining Equation of Simple Harmonic Motion (SHM)

While many things oscillate, physicists are particularly interested in the simplest, most fundamental type: Simple Harmonic Motion (SHM). An object performs SHM if it meets two strict conditions. Think of these as the two clauses in a legal contract for motion:

1. The object's acceleration is directly proportional to its displacement from the equilibrium position. We write this as `a ∝ x`.
2. The acceleration is always directed opposite to the displacement; that is, it's always directed towards the equilibrium position.

Combining these two conditions, we get the defining relationship of SHM:

`a ∝ -x`

This is the "identity card" of SHM. If you can prove that an object's motion follows this rule, you have proven it is executing SHM.

To turn this proportionality into an equation, we introduce a constant of proportionality. For reasons that will soon become clear, we call this constant `ω²` (omega-squared).

`a = -ω²x`

This is the most important equation in this topic. Every other formula for SHM is derived from this one. Here, `ω` is called the angular frequency. It's a measure of how rapidly the oscillation occurs, measured in radians per second (rad s⁻¹). A large `ω` means a fast, frenetic oscillation; a small `ω` means a slow, lazy one.

Derivation from a Mass-Spring System

Where does this equation come from? Let's consider the classic example: a mass `m` attached to a spring with spring constant `k`.

1. According to Newton's Second Law, the net force on the mass is `F = ma`.
2. According to Hooke's Law, the restoring force from the spring is `F = -kx`, where `x` is the displacement from the equilibrium position. The negative sign is crucial – it tells us the force opposes the displacement.
3. Since the spring force is the net force, we can equate them: `ma = -kx`.
4. Rearranging for acceleration `a`, we get: `a = -(k/m)x`.

Now, compare this to our defining equation for SHM, `a = -ω²x`. They are identical in form! By comparison, we can see that for a mass-spring system:

`ω² = k/m` or `ω = √(k/m)`

This is a powerful result. It tells us that the angular frequency of a mass-spring system depends *only* on the mass and the spring's stiffness, not on how far you pull it back (the amplitude).

The Equations of Motion for SHM

The equation `a = -ω²x` is a second-order differential equation. Solving it requires calculus, but you need to know the solutions. The solutions describe the displacement `x`, velocity `v`, and acceleration `a` as functions of time `t`. They are sinusoidal (i.e., they follow the pattern of a sine or cosine wave).

* Displacement (x):

If the object starts at the equilibrium position (`x=0` at `t=0`) and moves in the positive direction, the solution is:

`x = A sin(ωt)`

If the object starts at the maximum positive displacement (`x=A` at `t=0`), the solution is:

`x = A cos(ωt)`

Here, `A` is the amplitude, the maximum displacement.

* Velocity (v):

We can find velocity by differentiating displacement with respect to time (`v = dx/dt`).

If `x = A sin(ωt)`, then `v = Aω cos(ωt)`.

If `x = A cos(ωt)`, then `v = -Aω sin(ωt)`.

Notice that the maximum velocity, `v_max`, occurs when the cosine or sine term is 1. So, `v_max = Aω`. This happens when `x=0` (equilibrium position).

* Acceleration (a):

We find acceleration by differentiating velocity (`a = dv/dt`).

If `v = Aω cos(ωt)`, then `a = -Aω² sin(ωt)`.

Notice that `A sin(ωt)` is just `x`. So we can substitute back to get `a = -ω²x`, our original defining equation! This is a beautiful, self-consistent loop.

The maximum acceleration, `a_max`, occurs when the sine term is 1. So, `a_max = Aω²`. This happens when `x=±A` (the amplitude positions).

There is another incredibly useful formula for velocity that doesn't involve time. It relates velocity `v` to displacement `x`:

`v = ±ω√(A² - x²)`

This equation is perfect for finding the speed at a specific point in the oscillation. You can see that when `x=0` (equilibrium), `v = ±ω√(A²) = ±ωA` (which is `v_max`). And when `x=±A` (amplitude), `v = ±ω√(A² - A²) = 0`. This makes perfect physical sense.

Energy in SHM

Energy conservation provides another powerful lens through which to view SHM. In an ideal (undamped) system, the total mechanical energy is constant. This energy is continuously exchanged between two forms:

* Kinetic Energy (KE): The energy of motion. `KE = ½mv²`.

* Potential Energy (PE): The stored energy. For a mass-spring system, this is elastic potential energy, `PE = ½kx²`.

Let's analyse the energy at different points:

* At the amplitude (x = ±A): The mass is momentarily stationary (`v=0`), so `KE = 0`. The spring is maximally stretched or compressed, so PE is maximum: `PE_max = ½kA²`. The total energy is `E_total = PE_max = ½kA²`.

* At the equilibrium position (x = 0): The spring is at its natural length, so `PE = 0`. The mass is moving at its maximum speed (`v = v_max`), so KE is maximum: `KE_max = ½mv_max²`. The total energy is `E_total = KE_max = ½m(Aω)²`.

Since total energy is conserved, `½kA² = ½m(Aω)²`. If we simplify this, we get `k = mω²`, or `ω² = k/m`, which is the exact same relationship we found earlier! Physics is beautifully consistent.

At any point `x`, the total energy is the sum of the KE and PE at that point:

`E_total = KE + PE = ½mv² + ½kx² = ½kA²` (a constant value)

The Simple Pendulum

A simple pendulum (a point mass `m` on a massless string of length `L`) is another classic example of SHM.

For a pendulum, the restoring force is the component of gravity acting along the arc of the swing: `F_restoring = -mg sin(θ)`.

Using Newton's Second Law, `ma = -mg sin(θ)`. The displacement along the arc is `x = Lθ`. The acceleration is `a`, so we have `a = -g sin(θ)`.

This equation `a = -g sin(θ)` does *not* look like `a ∝ -x`. So, a pendulum is not, in general, a simple harmonic oscillator.

However, for small angles (typically less than 10° or about 0.17 radians), we can use the small-angle approximation: `sin(θ) ≈ θ` (where `θ` is in radians).

Substituting this in, we get: `a ≈ -gθ`.

Since `x = Lθ`, we can write `θ = x/L`.

So, `a ≈ -(g/L)x`.

This equation is in the form `a = -ω²x`! Therefore, for small angles, a simple pendulum executes SHM. By comparing the two equations, we see that for a pendulum:

`ω² = g/L` or `ω = √(g/L)`

Damping & Resonance

In the real world, systems don't oscillate forever. Friction and air resistance cause the system to lose energy. This dissipation of energy from an oscillating system is called damping.

* Light Damping: The amplitude gradually decreases with each oscillation. The period remains almost constant. (e.g., the *jhoola* in Bagh-e-Jinnah slowly coming to a stop).

* Critical Damping: The system returns to equilibrium in the shortest possible time without oscillating at all. This is vital in systems like car suspensions, which need to absorb bumps without bouncing up and down.

* Heavy Damping: The system returns to equilibrium very slowly, without oscillating. (e.g., a piston moving in thick oil).

Resonance is a phenomenon that occurs when you apply a periodic driving force to an oscillating system. If the frequency of this driving force (`f_d`) matches the system's natural frequency (`f_n`), the system will absorb energy very efficiently, and its amplitude will increase dramatically.

Think of pushing the child on the *jhoola*. You can't just push randomly. You have to time your pushes to match the swing's natural rhythm (its natural frequency). When you do this, each push adds more energy, and the child swings higher and higher. This is resonance.

Resonance can be useful (e.g., tuning a radio to a specific station's frequency, or in medical MRI machines) or destructive (e.g., soldiers breaking step when crossing a bridge to avoid matching its natural frequency, the famous collapse of the Tacoma Narrows Bridge). Damping is crucial in limiting the amplitude at resonance.

Key Definitions & Formulae

Here is a summary of the key terms and equations you must know.

| Term | Symbol | Definition | SI Unit | Key Formulae |

| --------------------- | ------ | ------------------------------------------------------------------------------------------------------ | ------------ | ------------------------------------------------------------------------- |

| Displacement | `x` | The instantaneous position of the oscillating object relative to its equilibrium position. | m (meters) | `x = A sin(ωt)` or `x = A cos(ωt)` |

| Amplitude | `A` | The maximum displacement from the equilibrium position. A positive scalar quantity. | m (meters) | Appears in all `x, v, a` equations. `E_total = ½kA²` |

| Period | `T` | The time taken to complete one full oscillation. | s (seconds) | `T = 1/f`, `T = 2π/ω` |

| Frequency | `f` | The number of complete oscillations per unit time. | Hz (Hertz) | `f = 1/T`, `f = ω/2π` |

| Angular Frequency | `ω` | A measure of the rate of oscillation in radians per unit time. (`ω = 2πf`) | rad s⁻¹ | `a = -ω²x`, `ω = √(k/m)`, `ω = √(g/L)` |

| Phase Difference | `φ` | The difference in the stage of oscillation between two points or waves, measured in radians or degrees. | rad or ° | `x = A sin(ωt + φ)` |

| Spring Constant | `k` | A measure of the stiffness of a spring. | N m⁻¹ | `F = -kx`, `T = 2π√(m/k)` |

Dimensional Analysis Example: Period of a Pendulum

Let's check if the formula `T = 2π√(L/g)` makes sense in terms of units.

The constant `2π` is dimensionless.

`L` is length, with units of meters (m).

`g` is acceleration due to gravity, with units of meters per second squared (m s⁻²).

So, the units of `L/g` are `m / (m s⁻²) = m * (s² / m) = s²`.

The square root of `s²` is `s` (seconds).

The units match! The formula is dimensionally consistent. This is a great way to check your work in an exam.

Worked Examples

Example 1: The Rice Seller in Karachi

A shopkeeper in Saddar Bazaar, Karachi, is weighing rice. He hangs a 2.5 kg bag of rice from a vertical spring, which extends by 15 cm. He then pulls the bag down a further 10 cm and releases it, causing it to oscillate. (Assume `g = 9.81 m s⁻²`).

(a) Calculate the spring constant `k` of the spring.

(b) Show that the subsequent motion is SHM and calculate the period `T` of the oscillations.

(c) Calculate the maximum speed of the bag of rice.

Solution:

(a) Finding the spring constant `k`

At equilibrium, the weight of the rice (`mg`) is balanced by the spring force (`kx`).

The extension `x` is 15 cm = 0.15 m.

`F = kx` => `mg = kx`

`k = mg / x`

`k = (2.5 kg * 9.81 m s⁻²) / 0.15 m`

`k = 24.525 / 0.15`

`k = 163.5 N m⁻¹` (Let's use 164 N m⁻¹ for subsequent calculations).

(b) Proving SHM and finding the Period `T`

When the mass is displaced by an additional distance `y` from its equilibrium position, the net restoring force is `F_net = -ky`.

From Newton's Second Law, `F_net = ma`.

So, `ma = -ky`, which gives `a = -(k/m)y`.

Since acceleration `a` is directly proportional to displacement `y` and directed opposite to it, the motion is Simple Harmonic Motion.

The period `T` for a mass-spring system is given by:

`T = 2π√(m/k)`

`T = 2π√(2.5 kg / 163.5 N m⁻¹)`

`T = 2π√(0.01529)`

`T = 2π * 0.12365`

`T = 0.777 s` (to 3 s.f.)

(c) Finding the maximum speed `v_max`

The bag was pulled down a further 10 cm, so this is the amplitude `A`.

`A = 10 cm = 0.10 m`.

Maximum speed is given by `v_max = Aω`.

First, we need `ω`. We know `T = 2π/ω`, so `ω = 2π/T`.

`ω = 2π / 0.777 s = 8.086 rad s⁻¹`.

Now, calculate `v_max`:

`v_max = (0.10 m) * (8.086 rad s⁻¹)`

`v_max = 0.809 m s⁻¹` (to 3 s.f.)

Example 2: The Cricketer's Pendulum

During a cricket match warm-up in Lahore's Gaddafi Stadium, a player, Bilal, swings his bat, which can be modelled as a simple pendulum of length 0.85 m. He swings it through a small angle such that its maximum horizontal displacement is 0.12 m.

(a) Calculate the period of the bat's swing.

(b) Calculate the maximum speed of the end of the bat.

Solution:

(a) Calculating the period `T`

Since it's modelled as a simple pendulum with a small angle, we can use the formula:

`T = 2π√(L/g)`

`T = 2π√(0.85 m / 9.81 m s⁻²)`

`T = 2π√(0.08665)`

`T = 2π * 0.2943`

`T = 1.85 s` (to 3 s.f.)

(b) Calculating the maximum speed `v_max`

The maximum horizontal displacement is the amplitude, `A = 0.12 m`.

The maximum speed is `v_max = Aω`.

We need `ω` first. `ω = 2π/T`.

`ω = 2π / 1.85 s = 3.396 rad s⁻¹`.

Now, calculate `v_max`:

`v_max = (0.12 m) * (3.396 rad s⁻¹)`

`v_max = 0.408 m s⁻¹` (to 3 s.f.)

Example 3: Energy in an Oscillator

A 0.50 kg mass is attached to a horizontal spring of spring constant `k = 50 N m⁻¹` on a frictionless surface. It is pulled back 8.0 cm from its equilibrium position and released.

(a) What is the total energy of the system?

(b) What is the potential energy of the system when the mass is 4.0 cm from equilibrium?

(c) What is the kinetic energy and speed of the mass at this point (4.0 cm from equilibrium)?

Solution:

(a) Total Energy `E_total`

The total energy is constant and equals the maximum potential energy, which occurs at the amplitude.

Amplitude `A = 8.0 cm = 0.080 m`.

`E_total = ½kA²`

`E_total = ½ * (50 N m⁻¹) * (0.080 m)²`

`E_total = 25 * 0.0064`

`E_total = 0.16 J`

(b) Potential Energy at `x = 4.0 cm`

Displacement `x = 4.0 cm = 0.040 m`.

`PE = ½kx²`

`PE = ½ * (50 N m⁻¹) * (0.040 m)²`

`PE = 25 * 0.0016`

`PE = 0.04 J`

(c) Kinetic Energy and Speed at `x = 4.0 cm`

By the principle of conservation of energy, `E_total = PE + KE`.

`KE = E_total - PE`

`KE = 0.16 J - 0.04 J`

`KE = 0.12 J`

Now, to find the speed `v`, we use the kinetic energy formula:

`KE = ½mv²`

`v = √(2 * KE / m)`

`v = √(2 * 0.12 J / 0.50 kg)`

`v = √(0.24 / 0.50)`

`v = √(0.48)`

`v = 0.693 m s⁻¹` (to 3 s.f.)

Visual Mental Models

To truly master SHM, you need to see it in your mind's eye.

1. The Reference Circle (The King of Mental Models)

Imagine a point P moving in a circle of radius `A` at a constant angular velocity `ω`. Now, imagine a light shining from the side, casting a shadow of point P onto a wall. Let's call the shadow Q.

(P) •

/|\

/ | \

/ | \

/ | \ <-- Radius A

/ |y \

/ | \

<------o-----(Q)---->

-A x=0 +A

As P goes around the circle, its shadow Q moves back and forth along the diameter. The motion of this shadow Q is perfect SHM.

* The radius of the circle is the amplitude `A`.

* The angular velocity of P (`ω`) is the angular frequency of the shadow Q.

* This model beautifully explains why the motion is sinusoidal. The vertical position of P is `y = A sin(θ)`. Since `θ = ωt`, we get `y = A sin(ωt)`, which is our equation for displacement!

2. SHM Graphs

Graphs are a physicist's best friend. You must be able to sketch and interpret these three graphs for an object starting at `x=0` (`sin` form) or `x=A` (`cos` form). Let's assume it starts at `x=A`.

* Displacement-Time (x-t): A cosine curve. Starts at `+A`, goes to 0 at `T/4`, `-A` at `T/2`, 0 at `3T/4`, and back to `+A` at `T`.

x

^ +A

| o

| / \

--|/---\---t-->

|\ /

| \ /

| o -A

+-----------

* Velocity-Time (v-t): A negative sine curve. Starts at 0 (since `x=A`), becomes most negative at `T/4` (as it passes through equilibrium), back to 0 at `T/2` (at `x=-A`), most positive at `3T/4`, and back to 0 at `T`. It is π/2 (90°) out of phase with displacement.

v

^

| / \

--|--/---\--t-->

| / \

|/ \

o---------o

* Acceleration-Time (a-t): A negative cosine curve. Starts at its most negative value (`-Aω²`), goes to 0 at `T/4`, becomes most positive at `T/2`, back to 0 at `3T/4`, and ends at its most negative value. It is π (180°) out of phase with displacement (i.e., it's an inverted version of the x-t graph).

a

^

|

--|---------t-->

|\ /

| \ /

| \ /

o---o---o -a_max

3. The Energy-Displacement Graph

This graph shows how energy changes with position.

* The Potential Energy (`PE = ½kx²`) is a parabola opening upwards, zero at the center (`x=0`) and maximum at the ends (`x=±A`).

* The Kinetic Energy (`KE = E_total - PE`) is an inverted parabola, maximum at the center and zero at the ends.

* The Total Energy (`E_total = ½kA²`) is a constant horizontal line across the top.

Energy

^

| /-------------------\ <-- Total Energy (E_total)

| / KE \

|/ \ / \ / \|

o---\---/-----\---/---o <-- Potential Energy (PE)

+----|---------|----|----> Displacement (x)

-A 0 +A

Common Mistakes & Misconceptions

1. Mixing up Period (T) and Frequency (f): Students often calculate one when asked for the other. Correction: Always remember `f = 1/T`. Frequency is "how many per second", Period is "how many seconds per one". They are reciprocals. Double-check the question.

1. Calculator in Degrees Mode: The argument of `sin` or `cos` in SHM equations (`ωt`) is an angle in radians. Using degrees will give completely wrong answers. Correction: Before any SHM calculation, switch your calculator to RAD mode. Write "RAD MODE!" at the top of your exam paper.

1. Assuming the Pendulum Formula is Always Valid: Students use `T = 2π√(L/g)` for large angles of swing. Correction: This formula is an approximation derived using `sin(θ) ≈ θ`, which is only true for small angles. If an exam question mentions a "large angle," this formula is no longer accurate, and the period will actually be slightly longer.

1. Thinking Velocity is Maximum at the Ends: It's intuitive to think the object is "trying hardest" at the ends, but this is wrong. Correction: At the amplitude positions (`x=±A`), the restoring force is maximum, but the object momentarily stops to change direction, so `v=0`. The velocity is maximum at the equilibrium position (`x=0`) where the net force is zero, but its inertia is at its peak.

1. Forgetting the Negative Sign in `a = -ω²x`: The negative sign is not just a mathematical formality; it represents the physics. Correction: The negative sign means the acceleration vector always points in the opposite direction to the displacement vector. It's the "restoring" nature of the motion. Omitting it in a definition will lose you marks.

1. Confusing Damping with a Change in Period: Students often think damping slows down the oscillation, significantly increasing the period. Correction: For the light damping studied at A-Level, the primary effect is a decrease in amplitude. The change in the period is negligible and can be ignored unless stated otherwise.

Exam Technique & Mark Scheme Tips

Cambridge examiners are precise. You must be too.

* Command Words are Law:

* State: Give a concise term or sentence. No explanation needed. E.g., "State the two conditions for SHM."

* Describe: Say what you see. For a graph, describe its shape, intercepts, and turning points. For damping, describe how the amplitude changes over time.

* Explain: Give the reason *why*. This requires linking physical principles. E.g., "Explain why the velocity is zero at the amplitude." (Because all the kinetic energy has been converted to potential energy).

* Calculate: Show your working clearly. A mark scheme will typically have one mark for the formula, one for correct substitution, and one for the final answer with the correct unit and significant figures (the "ECF" or Error Carried Forward rule often applies here).

* Derivations are a Story: When asked to derive a formula like `T = 2π√(L/g)`, don't just write down equations. Explain each step. "Start with F=ma... For a pendulum, the restoring force is... Using the small angle approximation...". Each logical step earns a mark.

* Significant Figures Matter: Look at the data given in the question. If it's given to 2 or 3 significant figures, your final answer should also be to 2 or 3 s.f. Don't write down a long string of numbers from your calculator. It looks unprofessional and can lose you a mark.

* Graphing Skills: Use a sharp pencil. Label your axes with the quantity and the unit (e.g., "Displacement / m"). Choose a sensible scale. Plot points as small, neat crosses (x). Draw a smooth best-fit curve, not a dot-to-dot mess.

* Common Examiner Tricks:

* Giving the total distance of an oscillation (from -A to +A) and students using it as the amplitude. Remember, amplitude is from the center to one extreme.

* Asking for the time taken to travel from x=0 to x=A. This is exactly one-quarter of a period (`T/4`).

* Giving frequency `f` when the formula requires angular frequency `ω`. Always remember to convert: `ω = 2πf`.

Memory Tricks & Mnemonics

1. SHM's ID Card: The equation `a = -ω²x` is the official identification for SHM. If a system has this ID, it's in the SHM club.
2. V-Max in the Middle: Where is velocity maximum? At `x=0`, the middle. Think of the traffic on Shahrah-e-Faisal in Karachi; it's fastest in the middle lanes, and slowest (stopped) at the edges.
3. Radians Rule! The `ωt` in `sin(ωt)` is an angle. What kind? Omega Requires Radians. Don't use degrees.
4. Energy Exchange: Think of your mobile phone balance. At the end of the month (amplitude), your data (KE) is zero, but your potential to recharge (PE) is high. In the middle of the month (equilibrium), you're streaming videos like crazy (max KE), but your potential to recharge is low. The total value (money paid to PTCL/Ufone) remains the same.

Pakistan & Everyday Connections

1. The Indus River and Hydroelectric Power: The generators at Tarbela Dam are turned by massive turbines. These turbines rotate, and the generators use this rotation to produce the alternating current (AC) that powers our cities. The voltage in this AC supply oscillates sinusoidally, exactly like an object in SHM, with a frequency of 50 Hz.
2. Earthquake-Resistant Buildings: In cities like Islamabad and Karachi, which are near fault lines, modern high-rise buildings are designed with damping systems. They are essentially massive, inverted pendulums or mass-spring systems. These dampers are designed to have a natural frequency different from that of typical earthquakes, and they absorb the vibrational energy to prevent the building from resonating and collapsing.
3. The Tabla and Sitar: When a musician strikes a *tabla* or plucks a *sitar* string, the membrane or string vibrates. This vibration is a complex form of oscillation. The tension, mass, and length of the vibrating object determine its natural frequency, which in turn determines the musical note we hear. A tighter, lighter, or shorter string vibrates faster (higher frequency), producing a higher-pitched note.

Practice Problems

1. Definition: State the defining equation for simple harmonic motion and define each symbol used.

* *Answer Outline:* State `a = -ω²x`. Define `a` as acceleration, `ω` as angular frequency, and `x` as displacement from the equilibrium position. Explain that the negative sign indicates acceleration is always directed towards the equilibrium position.

1. Calculation (Mass-Spring): A 1.2 kg fish is hung from a spring scale in a market in Empress Market, Karachi. The spring extends by 6.0 cm. The fish is then pulled down slightly and released. Calculate the frequency of its oscillation.

* *Answer Outline:* First find `k` using `k = mg/x`. Then use `f = (1/2π)√(k/m)` to find the frequency. Remember to convert cm to m.

1. Application (Pendulum): A historic grandfather clock in the Lahore Museum has a pendulum that is 1.20 m long. On a hot day, the pendulum rod expands by 0.05%. By what percentage will the clock's timekeeping be inaccurate per day? Will it run fast or slow?

* *Answer Outline:* Calculate the original period `T_1`. Calculate the new length `L_2`. Calculate the new period `T_2`. Find the percentage change in period `((T_2 - T_1) / T_1) * 100`. Since the period increases, the clock will run slow. Calculate the total seconds in a day and apply the percentage error to find the time lost.

1. Graphing & Energy: An object undergoes SHM with amplitude `A`.

(a) Sketch a graph of velocity `v` against displacement `x`.

(b) Explain how the graph shows the maximum velocity and where it occurs.

* *Answer Outline:* (a) The graph is an ellipse, defined by `v² = ω²(A² - x²)`. It will be a circle if the axes are scaled appropriately. It intercepts the v-axis at `±v_max` and the x-axis at `±A`. (b) The maximum velocity (`v_max = Aω`) occurs at the intercepts with the v-axis, which corresponds to a displacement `x=0` (the equilibrium position).

1. Resonance & Damping: A child's swing has a natural period of 3.0 s.

(a) What is the frequency of the pushes a parent should apply to make the swing go as high as possible?

(b) Later, the child's older, heavier brother uses the swing. Will the required pushing frequency increase, decrease, or stay the same? Explain your reasoning.

* *Answer Outline:* (a) To cause resonance, the driving frequency must equal the natural frequency. `f = 1/T = 1/3.0 = 0.33 Hz`. (b) For a pendulum, the period `T = 2π√(L/g)` is independent of mass. Therefore, the natural period and frequency remain the same. The parent should push at the same frequency.