Kinematics

This section explores two fundamental areas of mechanics: simple harmonic motion (SHM) and collisions. Learners will examine the principles of SHM, including displacement, velocity, acceleration, and energy transformation in oscillatory systems such as pendulums and springs. Practical activities and worked examples highlight how SHM can be used to determine physical quantities, such as acceleration due to gravity. The section also develops an understanding of momentum, impulse, and the laws of motion through the study of collisions.

Distinctions between elastic and inelastic collisions are drawn, supported by real- world examples and simulations, to show how momentum and energy behave in different situations. Together, these topics link earlier work on Newton’s laws of motion (Year 1) and circular motion (Year 2), providing a strong foundation for applying mechanics to oscillations and interactions in everyday systems.

KEY IDEAS

· Acceleration in SHM: Acceleration is always directed towards the equilibrium position and is proportional to displacement, leading to sinusoidal motion.

· Collisions: Momentum is conserved in all collisions, while kinetic energy may or may not be conserved, distinguishing elastic from inelastic collisions.

· Energy Transformation in SHM: Mechanical energy remains constant, with continuous interchange between potential and kinetic energy during oscillation.

· Impulse and Momentum Change: Impulse equals the change in momentum, providing a basis for Newton’s second law of motion.

· Newton’s Third Law Verification: Conservation of momentum demonstrates that forces between interacting bodies are equal in magnitude and opposite in direction.

· Simple Harmonic Motion (SHM): SHM is a periodic oscillation where the restoring force is directly proportional to displacement and opposite in direction.

In previous lessons, you have studied different types of motion such as rectilinear motion, circular motion, and oscillatory motion. A common example of oscillatory motion is seen when playing on a see-saw in the playground. When one side is pushed down, the other rises, and after a while the motion reverses, continuing back and forth around the central balance point. This repeated to-and-fro movement is an example of oscillatory motion.

Figure 9.1: Two learners enjoying a seesaw ride in their school compound Simple Harmonic Motion (SHM) Among oscillatory motions, there is a special type called Simple Harmonic Motion (SHM). This occurs when the force that brings the object back to its central position (equilibrium) is always directed towards that position and is directly proportional to how far the object has moved from it. A good

example of this is a pendulum bob; when pulled slightly to the side and released, a component of its weight pulls it back to its original position. When it passes this position the direction of this force reverses, once again pulling it back to the centre and causing oscillations back and forth.

In SHM, displacement, velocity, and acceleration are all linked. The displacement decides how strong the restoring force is, the velocity changes continuously as the object swings, and the acceleration always points back to the centre. This makes SHM important for understanding not only playground examples like swings, but also vibrations in musical instruments, clocks, and even some machines.

Simple harmonic motion is defined as the motion of a particle whose acceleration is always directed towards its equilibrium position and is directly proportional to its displacement from that position.

In a simple harmonic motion, the period of oscillation is independent of the amplitude and when displacement of particle is maximum, speed is zero.

Key terms associated with SHM · Restoring Force: The force that brings the object back to its equilibrium position. For example, for a mass on a spring this force is given by Hooke’s Law, F=−kx, where k is the spring constant and x is the displacement.

The negative sign indicates that the force is always directed opposite to the displacement.

· Period (T): The time it takes for one complete oscillation.

· Frequency (f): The number of oscillations per unit time, f=1/T.

· Amplitude (A): The maximum displacement from the equilibrium position.

· Displacement (x): The distance of the oscillating object from its equilibrium position at any instant.

· Angular Frequency/velocity (ω): Related to frequency by ω = 2πf = 2π/T.

Characteristics of SHM

1. Periodic: The motion repeats itself at regular intervals.

2. Oscillatory: The object moves back and forth about an equilibrium position.

3. Isochronous: The period (T) of oscillation is independent of the amplitude (for small amplitudes).

4. Sinusoidal Variation: Displacement, velocity, and acceleration all vary sinusoidally with time.

Key Conditions for SHM

1. Restoring Force: There must be a restoring force that always acts to bring the object back to its equilibrium position.

2. Proportionality to Displacement: The magnitude of this restoring force must be directly proportional to the magnitude of the displacement from the equilibrium position.

3. Direction: The restoring force must always be directed towards the equilibrium position (opposite to the displacement).

4. Inertia: The oscillating system must possess inertia (mass) so that it can overshoot the equilibrium position.

Mathematical Definition (Differential Equation)

The defining equation for SHM is:

a = − ω²x Where:

a is the acceleration of the oscillating body.

x is the displacement from the equilibrium position.

ω is the angular frequency (constant for a given SHM system), in radians per second (rad/s). The negative sign indicates that acceleration is always opposite to the displacement.

Principles of SHM

The core principle of SHM is the relationship between the restoring force and displacement. This can be understood through Newton’s Second Law, F = ma. By substituting the restoring force in a spring, we get: −kx = ma Since acceleration is the second derivative of displacement with respect to time (a = d²x___ dt²), the equation of motion for SHM is a second-order linear differential equation:

d²x_ dt²+ k/m x = 0 The general solution to this equation is: x = Asin (ωt + φ ) Where x is the displacement from the mean position A is the amplitude.

ω is the angular frequency, ω = k/m ϕ is the phase constant, determined by the initial conditions In simple terms, the solution given above indicates that the position of the oscillating object varies sinusoidally when plotted against time. A graph demonstrating this can be seen below.

Consider a particle executing SHM such that the path taken by the particle is on a straight line, then x is the displacement of particle from the mean position at that instant. The displacement of a particle executing SHM at time t is given by x = Asin (ωt + ϕ ) ………………………………. (1)

Note that this displacement from the mean position can also be derived by considering the vertical component of the motion of an object moving in a circle as shown in the image below. There are many parallels between SHM and circular motion.

Figure 9.2: SHM from circular motion Before reading on, have a go at using the formula for the position of an object undergoing simple harmonic motion to derive equations for the velocity and the acceleration of the same object. Remember, velocity is the first derivative of displacement and acceleration is the second! Sketch graphs of your equations when you are finished.

Velocity of Simple Harmonic Motion - Derivation The velocity v of a simple harmonic motion can be derived by differentiating the displacement x with respect to time t Thus v = dx__ dt Where x = Asin (ωt + ϕ ) v = d[Asin (ωt + ϕ)]____________ dt v = ωAcos(ωt + φ ) …………………………. (2) From eqn (1) x/A = sin(ωt + ϕ ) ………………….. (3) From trigonometry, sin²(ωt + ϕ) + cos²(ωt + ϕ) = 1 This implies that, cos(ωt + ϕ) = √______________ 1 − sin²(ωt + ϕ) ………………. (4) Putting eqn (3) into eqn (4) gives cos(ωt + ϕ) = √_______ 1 − ( ˣ__ A)²cos(ωt + ϕ) = √__________ 1 − x 2 _ A 2 cos(ωt + ϕ) = 1/A √A²− x²putting cos(ωt + ϕ) = 1/A √A²− x²into eqn (2) gives v = ω√A²− x²……………………. (5) Hence equation 2 and 5 represent the velocity of a simple harmonic motion Putting x = 0 into equations 2 and 5 will each give v = ωA ……………….. (6) This is the maximum velocity of SHM vₘₐₓwhich occurs at the equilibrium position.

Hence vₘₐₓ = ωA The minimum velocity of SHM occurs at the amplitude where x = ±A Acceleration of Simple Harmonic Motion - Derivation Acceleration a = dv/dt But from eqn 2, v = ωAcos(ωt + ϕ ) This implies that, a = d[ωAcos(ωt + ϕ )]______________ dt a = −ω²Asin(ωt + φ ) ……………….. (7) Hence the acceleration of SHM “a” is given as shown in the eqn (7) above When sin(ωt + ϕ ) = ± 1 a = ω²A Thus aₘₐₓ = ω²A This is the maximum acceleration of SHM and this occurs at the amplitude, thus when x = ±A The minimum acceleration occurs at the mean position where x = 0 Systems or bodies that can be caused to perform simple harmonic motion

1. A loaded spiral spring

2. A loaded test tube floating in a liquid

3. A cantilever

4. The balance wheel of a clock or watch

5. Simple pendulum

6. Tuning fork The Simple Pendulum – Deriving an Expression For The Time Period of Oscillations A simple pendulum consists of a point mass (bob) suspended from a fixed support by a light, inextensible string. When displaced from its equilibrium position (vertical) and released, gravity provides a restoring force that acts to bring the bob back. For small angular displacements (typically less than about 10-15 degrees), the restoring force component (-mg sinθ) is approximately proportional to the angular displacement θ (since sinθ ≈ θ for small angles).

Figure 9.3: Simple pendulum performing SHM From the diagram above the restoring force F = - mgsinθ and the displacement x Which is given as F = - m g θ But θ = x_ l substituting θ = x_ l into F = - m g θ gives F = - m g x_ l Also, F = ma Hence ma = - m g x_ l This implies that, a = - g x_ l But the acceleration of SHM is given as a = − ω²x Hence − ω²x = - g x_ l Thus ω² = [[OMML-EQ-45]] ω = √_ g/l …………………….. (8) But ω = 2π__ T 2π__ T =√_ g/l Hence T = 2π √_ l/g …………………………(9) The Helical Spring – Deriving An Expression For The Time Period Of Oscillations

Figure 9.4: A mass on helical spring performing SHM Consider a mass m suspended at rest from a helical spring and let the extension produced be “e” as shown in Figure 9.4. If the spring constant is k we have:

m g = ke The mass is then pulled down a small distance x and released. The mass will oscillate due to both the effect of the gravitational attraction (mg ) and the varying force in the spring [k (e + x)].

At any point distance x from the midpoint Restoring force = -[k (e + x) – mg ] But F = ma Hence ma = − ke − kx + mg But mg = ke ma = − ke − kx+ ke ma = − kx a = − k/m x Putting a = − ω²x into the equation gives ω² = k/m ω = √_ k/m But ω = 2π__ T 2π__ T = √_ k/m T = 2π √_ m/k ……………………. (10) From mg = ke m/k = e/g T = 2π √_ e/g ………………….(11)

Activity 9.1 Examples of Simple Harmonic Motion (SHM) in Daily Life Objective: To identify and understand examples of simple harmonic motion (SHM) in daily life by observing real systems, recognising the role of restoring forces, and summarising the common features that define SHM in order to build a clear and practical definition of the concept.

What you need

1. Access to the video: 5 Simple Harmonic Motion Examples In Physics & Daily Life

2. Paper or notebook and pen or pencil What to do

1. Open the video and watch carefully as it presents five real-life examples of simple harmonic motion.

2. For each example shown, pay attention to how the object moves back and forth around a stable central (equilibrium) position. Notice the smooth, repetitive oscillations and how a restoring force (gravity, tension, elasticity) acts to bring the object back toward equilibrium.

3. In your notebook, write down for each example

a. The type of object or system

b. The kind of motion observed

c. What acts as the restoring force; why is it being pulled back to where it came from?

d. Whether the motion looks smooth and repeating regularly

4. Based on the video and your notes, list the common characteristics that make these motions examples of simple harmonic motion

5. Using your observations, write your own definition of simple harmonic motion, including its key features. Ensure that you comment on direction and magnitude of the restoring force at different points in the cycle of oscillations.

Activity 9.2 Identifying Examples of Periodic Motion Which are SHM Objective: To distinguish simple harmonic motion (SHM) from other types of periodic motion by observing different examples, analysing their characteristics, and identifying the key features—such as restoring force, sinusoidal displacement, and linear motion—that define SHM.

What you need

1. Paper or notebook

2. Pen or pencil

3. Access to internet

4. Ruler or protractor (optional for sketching or measurement) What to do

1. Review a set of given examples showing different types of periodic motion.

Examples you might see include:

a. A mass attached to a spring oscillating back and forth

b. A simple pendulum swinging at small angles

c. The vibration of a tuning fork’s tines

d. A bouncing ball moving up and down repeatedly

e. Riders on a carousel going around in circles

f. A pendulum swinging with large angles

2. For each example, search online and watch the motion or view pictures/ videos, then sketch a graph of displacement versus time showing the motion’s pattern (have you best guess at this! Look for sudden changes in direction or velocity and make sure that you make these clear on your graph if they are present).

3. Use these criteria to decide if the motion is SHM:

a. Restoring force is directly proportional to the displacement from equilibrium and acts toward the equilibrium position (linear restoring force following Hooke›s law).

b. Displacement as a function of time follows a sinusoidal pattern (like a sine or cosine wave).

c. Motion is along a straight line and smooth, continuous, without sudden changes in direction or speed.

d. Amplitude remains constant over time (in ideal conditions without ‘damping’, which is a term meaning resistance causing a loss of amplitude).

4. For each motion that you identify as SHM or not SHM, write a short explanation justifying your reasoning

5. Make two lists: one for motions that are SHM and one for motions that are periodic but not SHM.

6. Write a brief conclusion stating what key factors distinguish SHM from other periodic motions based on your investigation.

Activity 9.3 The Period of a Simple Pendulum for Different Lengths Objective: To investigate how the period of a simple pendulum depends on its length, and to confirm the relationship T∝√l by measuring, recording, and analysing experimental data.

What you need

1. A small dense bob (metal ball or small weight) attached to a string

2. A sturdy support stand or place to hang the pendulum so it can swing freely

3. A stopwatch or timer (smartphone timer is fine)

4. A ruler or measuring tape (to measure length of the pendulum)

5. Paper or notebook

6. Pen or pencil

7. Graph paper Or, in the absence of the appropriate practical material, you could perform this experiment using the PHeT simulation Pendulum Lab What to do

1. Attach the bob securely to one end of the string. Tie or fix the other end of the string to a support so the bob can swing freely like a pendulum.

Measure the length l of the pendulum from the point of suspension to the centre of the bob. Record this length in your notebook.

2. Perform the Timing for Oscillations

a. Pull the bob gently to a small angle (less than 15°) and release it to start swinging.

b. Use the stopwatch to measure the time Ttotal it takes to complete a fixed number of oscillations, for example, 10 full swings (back and forth counts as 1 oscillation).

Note

To improve accuracy, timing multiple oscillations reduces reaction time error.

3. Calculate the period T (time for one oscillation) by dividing the total time by the number of oscillations. Record the period value for that length l.

4. Repeat for Different Lengths - e.g., try 20 cm, 40 cm, 60 cm, 80 cm, 100 cm).

a. For each new length, repeat steps 2 and 3 to measure and calculate the period.

b. Record all values of length and corresponding period in a table.

5. On graph paper or your notebook, plot a graph of the period T (vertical axis) against the square root of the length l (horizontal axis).

a. Calculate the square root of each length value before plotting.

b. Label the axes clearly with units (e.g., T in seconds and l in metres1/2).

6. Analyse the Graph: Observe that the graph should be a straight line, showing a linear relationship between the period T and the square root of the length l.

7. Summarise your findings by describing how the period changed with length

Activity 9.4 Investigating Oscillations of a Mass on a Spring Objective: To investigate the oscillations of a mass on a spring in order to understand how displacement, velocity, and acceleration vary during simple harmonic motion.

What you need

1. Spring

2. Set of masses (known weights)

3. Stand and clamp to suspend the spring vertically

4. Stopwatch or timer

5. Meter ruler or measuring tape

6. Calculator

7. Paper and pencil for recording data and calculations What to do

1. Set Up the Apparatus

a. Attach the spring securely to the stand and clamp so it hangs vertically.

b. Suspend a known mass from the spring’s lower end, allowing it to hang freely.

2. Measure Displacement

a. Pull the mass down gently to stretch the spring.

b. Measure the maximum displacement (amplitude) from the rest position using the ruler.

c. Record this value as the amplitude A.

3. Measure Oscillation Time

a. Release the mass to start oscillations.

b. Use the stopwatch to time how long it takes the mass to complete 10 full oscillations.

c. Repeat this timing two or three times for accuracy and record the results.

d. Take a moment to consider the accuracy of this experiment; is measuring the time challenging? Why? What effect may this have on your results?

4. Calculate Period T and Angular Frequency ω

a. Calculate the period T (time for one oscillation) by dividing the total time for 10 oscillations by 10.

b. Calculate the angular frequency using the formula:

ω = 2 π_ T

5. Calculate the Spring Constant k

a. Measure the mass m suspended on the spring in kilograms.

b. Using the relationship between the period T, mass m, and spring constant k, calculate the spring constant k = 4π²m/T²6. Calculate Velocity and Acceleration at Various Points

a. Pick several points during the oscillation (e.g., at certain displacements x from the rest position).

b. Use the displacement and angular frequency to calculate velocity v at each point using:

v = ω √A²− x²c. Calculate acceleration a at each point using:

a = −ω²x

d. Record the values of x, v, and a in a table.

7. Analyse and Reflect

a. Compare how velocity changes with displacement — it’s highest at the equilibrium point and zero at the maximum displacement.

b. Note acceleration is zero at equilibrium and highest (in magnitude) at maximum displacement, and it’s directed opposite to the displacement.

c. Sketch graphs of how the displacement, velocity and acceleration are changing with time.

d. Write a few sentences describing what these results show about oscillatory motion.

Activity 9.5 Calculating Motion Parameters in SHM and Pendulums Study the worked examples carefully before attempting the sample questions that follow.

Worked Example 1

Determine the acceleration due to gravity in a region where a simple pendulum having a length 75.0 cm has a period of 1.7357s

Solution

The length l of a simple pendulum = 0.75m The period T = 1.7357s From T = 2π √_ l/g 1.7357 = 2π √___ 0.75/g 1.7357/2π = √___ 0.75/g 0.2762²= 0.75/g g = 0.75/0.076311 g = 9.8 m/s²Worked Example 2 A particle is executing SHM with an amplitude of A and an angular frequency of ω. At a certain time t, its displacement is x = A/2. Find the magnitude of its velocity and acceleration at this instant, in terms of ω and a.

Solution

We use the general formula for velocity in SHM: v = ±ω √A²− x²Substitute the given values: x=A/2.

v = ± ω √________ A²− (A/2 ) 2 v = +ω √_____ A²– A²__ 2²v = ± ω √______ A²− A/4 2 v = ± ω √__ 3A/4 2 v = ±√3/2 ωA The magnitude of the velocity is √3/2 ωA The formula for acceleration in SHM is a = −ω²x Substitute the given value for displacement: x=A/2.

a = −ω²A/2 a = −1/2 ω²A The magnitude of the acceleration is 1/2 ω²A Practice Problems

1. A load hanged from a vertical spring causes the spring to stretch by 30cm and oscillates when slightly displaced and released, calculate the period of oscillation (g = 10m/s2).

The total mechanical energy in a simple harmonic motion remains constant (if there is no damping). This energy is continuously converted between potential energy and kinetic energy.

Kinetic Energy (KE)

Kinetic energy is the energy possessed by a body due to its motion.

KE = 1/2 m v²Substitute v = Aωcos(ωt + ϕ) into the kinetic energy equation KE = 1/2 m[ Aωcos(ωt + ϕ)]²KE = 1/2 m ω²A²cos²(ωt + ϕ) From trigonometry, sin²(ωt + ϕ) + cos²(ωt + ϕ) = 1 Also, x = Asin(ωt + ϕ) Therefore KE = 1/2 m ω²( A²− x²) Maximum Kinetic Energy (KEₘₐₓ) Occurs when velocity is maximum, i.e., at the equilibrium position (x = 0).

KEₘₐₓ = 1/2 m ω²A² Minimum Kinetic Energy Kinetic energy is zero at the extreme positions (x = ±A), where the velocity is momentarily zero.

Potential Energy (PE)

Potential energy in SHM is the energy stored due to the displacement from the equilibrium position (e.g., elastic potential energy in a spring).

PE = 1/2 kx² (for a spring-mass system) The restoring force on the object can be found using Hooke’s Law: F = ma = − kx But a = ω²x , so:

m ω²x = kx k = m ω²Substituting into the PE equation above:

PE = m ω²x²______ 2 PE = m ω²x²______ 2 PE = m ω²(Asin(ωt + ϕ))²_______________ 2 Therefore, PE = m ω²A²______ 2 sin²(ωt + ϕ) Total Mechanical Energy (E) The total mechanical energy in an ideal (undamped) SHM system is conserved and is the sum of kinetic and potential energy.

E = KE + PE E = 1/2 m ω²( A²− x²) + m ω²x²______ 2 E = 1/2 m ω²A²The Energy Transformation Cycle The energy in an SHM system constantly cycles back and forth between these two forms:

1. At Maximum Displacement (x=±A): The object is momentarily at rest, so its velocity is zero. At this point, all the energy is stored as potential energy and the kinetic energy is zero (K.E=0).

2. Moving Towards Equilibrium (x=0): As the restoring force pulls the object back toward the equilibrium position, its speed increases. The potential energy is converted into kinetic energy.

3. At Equilibrium Position (x=0): The object is moving at its maximum speed (v=vₘₐₓ). All the energy is now kinetic energy and the potential energy is zero (P.E=0).

4. Moving Away from Equilibrium (x=±A): The object moves past the equilibrium position, and the restoring force now acts against its motion, causing it to slow down.

The kinetic energy is converted back into potential energy.

At x = 0 , PE = 0 and the energy is purely kinetic i.e., E = KEₘₐₓ = 1/2 m ω²A²At extreme points or turning points of the SHM, kinetic energy is zero and the energy is purely potential i.e., E = PEₘₐₓ = 1/2 m ω²A²Thus, the total energy remains constant for the system no matter where the particle is.

Figure 9.5: Energy transformation in SHM

Activity 9.6 Exploring Energy Variation in Simple Harmonic Motion Using Conceptual Energy vs. Displacement Graphs Objective: To understand how energy changes during simple harmonic motion (SHM) by using energy–displacement graphs to show the transformation between kinetic and potential energy, and to recognise that the total mechanical energy remains constant, demonstrating conservation of energy.

What you need

1. Paper or notebook

2. Pen or pencil

3. Access to internet to see simulation: oPhysics

4. Printed or digital copy of the energy graph diagram (see below) What to do

1. Open the Simulation and Select ‘Run’. Look at the graphs being generated on the right, which show the position, velocity, acceleration, potential and kinetic energies of the mass on a spring. Take time to look at how these are related; which graphs show maxima at the same time? Why? Which graphs are completely out of phase? Why?

2. Examine the graph that illustrates how kinetic energy, potential energy, and total energy change as the oscillator moves between its extreme positions (-A to +A).

3. Write down answers addressing

a. How does kinetic energy change as the displacement moves from -A to +A?

b. How does potential energy vary with displacement in that range?

c. What does the total energy curve indicate about the overall energy of the system?

4. Consider and note

a. At the equilibrium position (displacement 0), what do you observe about kinetic and potential energies?

b. At maximum displacement points (-A and +A), what happens to each form of energy?

c. Are there points where kinetic energy and potential energy are equal? How many such points are visible?

5. Using the graph shapes, describe in your own words how energy transforms between kinetic and potential forms during oscillation.

6. Reflect on why the total mechanical energy remains constant regardless of displacement and what this implies about energy conservation in the ideal system.

7. Reinforce your learning by drawing a simplified version of the graph, labelling key points such as maximum displacement, equilibrium, and energy equality points.

Activity 9.7 Calculating Total Energy, Potential Energy, and Kinetic Energy Objective: To calculate and analyse the distribution of total, potential, and kinetic energy during oscillations of a mass on a spring, and to verify the conservation of mechanical energy in simple harmonic motion.

What you need

1. Access to your notes from Activity 9.4.

2. Calculator

3. Paper and pencil for calculations and notes What to do

1. Review Activity 9.4

a. Write down the mass m (in kilograms) of the object and the spring constant k (in Newtons per metre).

b. Look at your table from Activity 9.4 which contains values of displacement (x) and velocity (v) recorded during oscillation.

2. Calculate Total Energy E

a. Recall that total mechanical energy E in a simple harmonic oscillator is constant and is the sum of potential energy and kinetic energy.

b. Calculate total energy using the maximum displacement A (amplitude) with this formula:

E = 1__ 2k A²

c. Use the known spring constant k and amplitude A from your data.

3. For each displacement x in your table, calculate the potential energy stored in the spring using:

PE = 1/2 kx² Write the results in a new column next to your table.

4. Using corresponding velocity v values, calculate kinetic energy at each point with:

KE = 1/2 m v²Add these values in another column next to P.E.

5. Verify Energy Conservation

a. For each row, add the calculated P.E. and K.E. values. This sum should be close to the total energy E you calculated earlier.

b. Note any small differences and consider reasons (such as experimental error or rounding).

6. Analyse the Energy Changes

a. Observe how potential energy is maximum at maximum displacement and zero at equilibrium.

b. Notice kinetic energy is zero at maximum displacement and maximum at equilibrium.

c. Write a short paragraph explaining these energy changes during oscillations.

Activity 9.8 Calculating Energy in Simple Harmonic Motion

Study the worked examples carefully before attempting the sample questions that follow.

Worked Example 1

A 2 kg mass is attached to a spring with a spring constant of 50 N/m. The system oscillates with an amplitude of 0.3 m. Calculate the total energy of the system and the kinetic energy of the mass when its displacement is 0.1 m.

Solution

The total energy (Eₜₒₜₐₗ) in SHM is constant and can be calculated using the maximum potential energy, which occurs at the maximum displacement (the amplitude, A).

The formula for total energy is: Eₜₒₜₐₗ = 1__ 2k A²Substitute the given values k = 50 N/m, A = 0.3 m Eₜₒₜₐₗ = 1/2(50) (0.3)²Eₜₒₜₐₗ = 25(0.09) Eₜₒₜₐₗ = 2.25J The total energy of the system is 2.25J.

To find the kinetic energy (K. E) at a displacement of x=0.1 m, we can use the conservation of energy principle. The total energy is the sum of the kinetic energy and potential energy (P.E).

Eₜₒₜₐₗ= K . E + P . E K.E = Eₜₒₜₐₗ– P.E but P . E = 1/2 k x²P . E = 1/2(50) (0.1)²P . E = 0.25J K.E = 2.25 – 0.25 = 2.0 J The kinetic energy of the mass at a displacement of 0.1 m is 2.0 J.

Practice Problems

1. A block of mass m=0.5 kg is attached to a spring and executes simple harmonic motion (SHM). Its velocity is v=0.2 m/s when its displacement is x=0.1 m. The maximum displacement (amplitude) of the block is A=0.2 m.

Calculate the spring constant (k).

2. The bob of a simple pendulum of mass 0.6kg, performs a simple harmonic motion with amplitude 7.0cm and period 2.0s. If the motion of the bob is undamped, calculate the maximum kinetic energy of the bob.

Activity 9.9 Energy Transformations in SHM and Real-Life Systems Objective: To compare energy transformations between potential and kinetic energy in real-life oscillating systems and relate them to the mass-spring simple harmonic motion (SHM) model, thereby understanding similarities and differences in energy behaviour across systems.

What you need

1. Paper and pencils or pens

2. Reference materials or internet access (for researching examples if needed)

3. Coloured markers for diagrams (optional)

4. Presentation materials (poster paper, whiteboard, or digital presentation tools) What to do

1. Join with some classmates to work as a team; no more than 5 people in total.

2. Each group picks one of the following oscillating systems:

a. Child on a swing

b. Simple pendulum (a mass on a string)

c. Car suspension system

3. Research and Discuss Energy Transformation

a. Discuss how energy changes between potential energy (P.E.) and kinetic energy (K.E.) during the oscillation in your chosen system.

b. Identify when the potential energy is at its maximum (e.g., highest point or maximum displacement) and when the kinetic energy is at its maximum (e.g., lowest or equilibrium point).

c. Make connections to the mass-spring system you studied previously where:

i. P.E. is maximum at maximum displacement (amplitude)

ii. K.E. is maximum at the equilibrium position (midpoint)

4. Create simple diagrams showing the position of the system at maximum P.E. and maximum K.E. Label the points where energy transforms occur during the oscillation cycle.

5. Summarise your group’s discussion into a short presentation or poster with diagrams and key points. Explain how the energy transformation in your real-life example is similar to that of a mass-spring SHM system.

6. Share your findings and diagrams with the other groups or the whole class.

Listen to and learn how energy transformations occur in the other systems being presented.

1. A particle undergoes SHM with amplitude 10cm and period 2s. Calculate the time taken to move from the equilibrium position to a point 5cm away.

2. A 0.5kg mass is attached to a spring with spring constant k = 200N/m.

Find the total energy of the system if the amplitude is 0.3m.

3. A mass on a spring executes SHM. If the kinetic energy is half the total energy, find the displacement from equilibrium.

4. A simple pendulum has a length of 1.5m. calculate the change in period if taken to a place where the acceleration due to gravity decreases by 4%.

5. A block attached to a spring oscillates with angular frequency ω. Calculate the displacement at which the acceleration equals half of its maximum value.

6. The position of a particle is given by the equation x(t)=0.4cos(2t−π/6), where x is in metres and t is in seconds. Determine the time at which the particle first reaches its maximum speed after t=0.

7. A particle of mass m moving with an initial velocity of 10 m/s collides elastically with an identical stationary particle. After the collision, the first particle is deflected by an angle of 30∘ from its original path. Determine the final speeds of both particles and the direction of the second particle.

8. A golf ball with a mass of 46 g is at rest on a tee. A golf club strikes the ball, and the ball leaves the club face with a velocity of 60 m/s. The collision lasts for 0.0005 seconds.

a. Calculate the impulse delivered to the golf ball.

b. Calculate the average force exerted on the ball by the club.