Oscillations | ISC Class 11 Physics Notes

This note covers periodic motion, simple harmonic motion, amplitude and phase, the connection with uniform circular motion, displacement, velocity and acceleration, restoring forces, horizontal and vertical spring oscillations, energy changes, and the time period of a simple pendulum.

What distinguishes periodic motion from simple harmonic motion?

Periodic motion repeats itself at regular intervals of time. Oscillatory motion is to and fro motion about a mean position, the central position of the oscillation. Uniform circular motion is periodic, but it is not a to and fro motion along a line.

The equilibrium position is a position where the resultant force is zero. For the oscillators considered here, a small displacement produces a force tending to return the body towards equilibrium. A ball displaced slightly from the bottom of a bowl illustrates this behaviour.

Simple harmonic motion (SHM) is oscillatory motion in which acceleration is proportional to displacement from equilibrium and directed towards it. Its displacement varies sinusoidally with time, meaning that it follows a sine or cosine function with constant amplitude and frequency.

How are period and frequency related?

The time period T is the smallest time interval after which the motion repeats. The frequency f is the number of complete oscillations per unit time. The SI unit of time period is the second, symbol s. SI means the International System of Units.

f = 1/T

The SI unit of frequency is the hertz, symbol Hz. One hertz means one oscillation per second: 1 Hz = 1 s⁻¹. Frequency need not be an integer. A complete oscillation returns the particle to the same position with the same direction of motion.

Worked example 1. A heart beats, on average, 75 times in one minute. Find its frequency and period, taking one minute as 60 s.

Formula: f = number of beats/time; T = 1/f.

Substitute: f = 75/60; T = 1/1.25.

Answer: f = 1.25 Hz and T = 0.8 s. The fractional frequency is consistent with counting beats over a longer interval.

Periodicity alone does not establish SHM. A repeated motion can have a displacement graph that is not sinusoidal. To identify SHM, examine the displacement function or the relation between restoring acceleration and displacement, rather than simply checking whether the motion repeats.

How do amplitude, phase and periodic functions describe SHM?

Let y denote signed displacement from equilibrium and t denote elapsed time. A positive direction is chosen along the line of motion. Negative displacement means the particle lies on the other side of equilibrium; it does not mean a negative distance travelled.

y = A sin(ωt + φ₀)

Here A is amplitude, the magnitude of maximum displacement; ω is angular frequency, the rate at which phase increases; and φ₀ is the initial phase, also called the phase constant or epoch. The quantity ωt + φ₀ is the phase at time t.

The SI unit of displacement is the metre, symbol m; amplitude has the same unit. The SI unit of angular frequency is the radian per second, written rad s⁻¹. A radian is the angle subtended by an arc whose length equals its radius.

ω = 2πf

T = 2π/ω

The constant π is the ratio of a circle's circumference to its diameter. A full cycle advances phase by 2π radians. The functions sin and cos mean sine and cosine; angles in these equations are in radians.

How does the initial condition affect the equation?

For the sine form with φ₀ = 0, the particle starts at equilibrium and initially moves in the positive direction. A cosine form is equally valid: y = A cos(ωt + φ), where φ is the initial phase for that form.

The amplitude fixes the limits +A and −A. Phase distinguishes motions that have the same amplitude and frequency but begin at different stages of a cycle. Changing the phase constant shifts the displacement graph along the time axis without changing the period.

A periodic function repeats after its period. For example, sin ωt + cos ωt can be combined into √2 sin(ωt + π/4). The square-root sign √ denotes the positive square root. This sum has a single angular frequency and period 2π/ω.

By contrast, sin ωt + cos 2ωt + sin 4ωt repeats after 2π/ω but is not a single sinusoid. Its terms have different angular frequencies. This distinction separates a periodic displacement from the particular sinusoidal displacement required for SHM.

How is SHM related to uniform circular motion?

Uniform circular motion means motion around a circle at constant speed. Imagine a reference particle P travelling anticlockwise around a circle of radius A with constant angular speed ω. Let O be the centre and let the horizontal diameter define the x-axis.

The projection P′ is the foot of the perpendicular from P to that diameter. If OP initially makes an angle φ with the positive x-axis, its angle after time t is ωt + φ. The projected coordinate is x = A cos(ωt + φ).

Here x is displacement along the horizontal diameter. As P completes a revolution, P′ moves between +A and −A and completes an oscillation. Thus the radius gives the amplitude, while the period of revolution equals the period of the projected SHM.

The circle has centre O, horizontal x-axis and vertical y-axis. A radius reaches P in the upper-right part of the circle. A dotted perpendicular meets the horizontal axis at P′, and an arrow indicates anticlockwise rotation.

See Fig. 13.10 in your NCERT textbook

Why can both sine and cosine describe the motion?

Projection on the vertical diameter gives y = A sin(ωt + φ). The two projections have the same amplitude and angular frequency, with a phase difference of π/2. Either projection can serve as a geometrical representation of one-dimensional SHM.

The reference circle is a way to describe the motion; the oscillating body need not physically travel around that circle. The projected particle reverses its direction at the ends of the diameter, whereas the reference particle continues around the circumference.

The reference particle has a centripetal acceleration, directed towards the centre of its circular path. Its projection supplies the acceleration of the linear oscillator. This correspondence does not make the full circular force and the one-dimensional restoring force identical.

A complete geometrical description therefore specifies radius, angular speed, initial angle and sense of rotation. Omitting the initial angle or direction can leave the phase undetermined, even when the amplitude and period are already known.

How do displacement, velocity and acceleration vary during SHM?

Velocity v is the rate of change of signed displacement; acceleration a is the rate of change of velocity. The notation dy/dt means the derivative of displacement with respect to time. A second derivative, d²y/dt², gives acceleration.

Derivation: Velocity and acceleration from displacement

1. Begin with y = A sin(ωt + φ₀), keeping A, ω and φ₀ constant for the specified motion.
2. Differentiate once with respect to t: v = dy/dt = Aω cos(ωt + φ₀).
3. Differentiate again: a = dv/dt = −Aω² sin(ωt + φ₀). Replace the sine expression using the displacement equation.

a = −ω²y

The minus sign places acceleration opposite to displacement. For positive y, acceleration is negative; for negative y, it is positive. At equilibrium acceleration is zero, while speed is greatest. At either extreme, speed is zero but acceleration has its greatest magnitude.

v² = ω²(A² − y²)

This follows by adding the squared sine and cosine relations. The greatest speed is Aω, and the greatest acceleration magnitude is Aω². Squaring velocity removes its direction, so the speed relation alone cannot distinguish motion towards equilibrium from motion away from it.

Three vertically aligned plots share the time axis. Displacement starts at +A, velocity starts at zero and becomes negative, and acceleration starts at −ω²A. Each graph repeats after T.

See Fig. 13.13 in your NCERT textbook

These plotted curves use a cosine displacement with zero initial phase. Velocity differs in phase from displacement by π/2; acceleration differs by π. Each quantity has the same period, although their maximum magnitudes and physical units differ.

Worked example 2. A body has displacement x = 5 cos(2πt + π/4), with x in metres and t in seconds. Find displacement, speed and acceleration at t = 1.5 s.

Formula: v = −5(2π) sin(2πt + π/4); a = −(2π)²x.

Substitute: The phase is 3π + π/4; its cosine is approximately −0.707 and its sine is approximately −0.707.

Answer: x = −3.535 m, speed ≈ 22 m s⁻¹ and a ≈ +140 m s⁻². The positive acceleration points back towards equilibrium.

How does the restoring force produce the differential equation of SHM?

A restoring force tends to reduce displacement from equilibrium. For linear SHM, let F be the resultant restoring force, k the positive force constant and m the oscillating mass. The force constant measures the restoring force per unit displacement.

F = −ky

The SI unit of force is the newton, symbol N. The SI unit of force constant is the newton per metre, written N m⁻¹. The symbol m in a formula represents mass when defined that way; the upright unit symbol m after a value means metre.

Hooke's law describes a spring force proportional and opposite to its extension or compression, provided the spring remains within the range where that proportionality holds. The equation for SHM requires the resultant force about equilibrium, rather than one selected force acting on the body.

Derivation: Differential equation and angular frequency

1. Apply Newton's second law, F = ma, which relates resultant force to mass multiplied by acceleration.
2. Insert F = −ky and a = d²y/dt² to obtain m d²y/dt² = −ky.
3. Divide by m and rearrange: d²y/dt² + (k/m)y = 0. Compare this with a = −ω²y.

ω² = k/m

The differential equation, an equation containing derivatives, is d²y/dt² + ω²y = 0. Substitution verifies that y = A sin(ωt + φ₀) satisfies it: the second derivative is −ω² times the original displacement.

Two initial conditions, such as initial displacement and initial velocity, determine a particular motion when ω is given. The differential equation fixes the relation between acceleration and displacement; it does not by itself specify the amplitude or the starting phase.

Note: A force directed towards equilibrium is not sufficient by itself to establish SHM. Its magnitude must also be directly proportional to displacement, with a constant proportionality factor for the motion considered.

How does a horizontal mass and spring system oscillate?

Consider a block attached to a spring whose other end is fixed to a wall. The block moves on a frictionless horizontal surface, meaning that the surface offers no frictional resistance. Displacement is measured from the equilibrium position where the spring is unstretched.

The upward support force balances the downward weight. The spring supplies the horizontal restoring force −ky. When the spring stretches, it pulls the block back; when it is compressed, it pushes the block towards equilibrium.

T = 2π√(m/k)

f = (1/2π)√(k/m)

These expressions follow from ω² = k/m and T = 2π/ω. A greater mass gives a longer period for a fixed force constant. A stiffer spring, meaning a larger k, gives a shorter period for a fixed mass.

What fixes the amplitude?

If the block is displaced and released from rest, the release position is an extreme position. Its distance from equilibrium is therefore the amplitude. If a block instead starts with a non-zero velocity, initial displacement alone does not give the amplitude.

For an ideal linear spring oscillator, the period does not depend on amplitude or phase. This result assumes that the spring continues to obey the linear force law over the range of displacement used and that dissipative effects are neglected.

Worked example 3. A 3 kg block attached to a horizontal spring of force constant 1200 N m⁻¹ is displaced 2.0 cm and released from rest. Neglect friction. Find frequency, greatest speed and greatest acceleration magnitude.

Formula: ω = √(k/m); f = ω/(2π); greatest speed = Aω; greatest acceleration magnitude = Aω².

Substitute: A = 0.020 m and ω = √(1200/3) = 20 rad s⁻¹.

Answer: f = 10/π Hz ≈ 3.18 Hz; greatest speed = 0.40 m s⁻¹; greatest acceleration magnitude = 8.0 m s⁻².

For two identical springs attached on opposite sides of a block to fixed supports, a displacement stretches one spring and compresses the other. Both forces restore equilibrium. Their sum is −2ky, so the effective force constant is 2k and the period is 2π√(m/2k).

Why does a vertical spring have the same period formula?

For a vertical spring, gravity changes the equilibrium position. Let g be the local acceleration due to gravity, and let e be the extension from the spring's natural, unstretched length when the mass hangs at rest. Take downward displacement as positive.

At equilibrium the upward spring force balances the weight, the gravitational force on the mass. Thus ke = mg. Now let y be the additional downward displacement from this equilibrium position. The total spring extension is e + y.

Derivation: Vertical oscillations about equilibrium

1. The downward weight is mg, while the upward spring force has magnitude k(e + y).
2. The net downward force is F = mg − k(e + y).
3. Use the equilibrium relation mg = ke to cancel the constant terms, leaving F = −ky.
4. Consequently m d²y/dt² = −ky, giving ω² = k/m and the same period as the horizontal spring.

T = 2π√(m/k)

The mass oscillates around the loaded equilibrium position. Using extension from the natural length as though it were displacement from equilibrium would omit the constant gravitational contribution and give the wrong restoring-force equation.

Gravity determines e but does not appear explicitly in this period formula. The cancellation applies when g can be treated as constant during the motion, the spring's force remains proportional to extension and resistance is neglected. The period still depends on the mass and force constant.

Release from rest a distance A below the loaded equilibrium position gives a downward extreme at the start. A convenient equation is y = A cos ωt. The initial acceleration is upward, consistent with a spring force then greater than the weight.

As with the horizontal system, the greatest speed occurs at equilibrium. Zero resultant force there means zero instantaneous acceleration; it does not imply zero velocity. The moving mass passes through equilibrium and continues because it already has velocity.

How is energy exchanged and conserved during SHM?

Kinetic energy K is the energy associated with motion. Potential energy U is energy associated with configuration under a conservative force, a force whose work between two positions is independent of the path. Total mechanical energy E is their sum.

K = ½mv²

U = ½ky²

The SI unit of energy is the joule, symbol J. Here potential energy is chosen to be zero at equilibrium. This choice gives non-negative potential energy for the oscillator; the zero of potential energy is a reference choice.

How does the total energy become independent of position?

Insert v² = ω²(A² − y²) into the kinetic-energy expression and use mω² = k. This gives K = ½k(A² − y²). Add U = ½ky²: the displacement-dependent terms cancel.

E = ½kA²

In the absence of friction and other dissipative effects, total mechanical energy remains constant. Energy moves between kinetic and potential forms as the particle travels. At equilibrium the energy is kinetic; at either extreme it is potential.

The upper graph shows alternating kinetic and potential energy peaks beneath a horizontal total-energy line. The lower graph shows potential energy rising on either side of zero displacement, kinetic energy peaking at zero, and constant total energy between −A and +A.

See Fig. 13.16 in your NCERT textbook

Against displacement, U forms an upward-opening parabola and K an inverted parabola over the allowed interval. Against time, both energies repeat after T/2 because squaring removes the sign of displacement or velocity. Each reaches its maximum twice in one complete oscillation.

Worked example 4. A 1 kg block on a frictionless horizontal surface is attached to a spring with k = 50 N m⁻¹. It is pulled 10 cm from equilibrium and released from rest. Find its energies at a displacement of 5 cm, taking U = 0 at equilibrium.

Formula: E = ½kA²; U = ½ky²; K = E − U.

Substitute: A = 0.10 m and y = 0.05 m. Thus E = ½ × 50 × 0.10² and U = ½ × 50 × 0.05².

Answer: E = 0.25 J, U = 0.0625 J and K ≈ 0.19 J. The kinetic-energy value is rounded; conservation refers to the unrounded sum.

Why is a simple pendulum approximately simple harmonic?

A simple pendulum consists of a small bob of mass m suspended from a rigid support by a massless, inextensible string. Inextensible means that the string does not stretch. Let L be its length and θ the string's angular displacement from the vertical.

The bob oscillates in a vertical plane. Its weight mg acts downward, and the tension, the pulling force exerted by the string, acts along the string towards the support. The component mg sin θ acts tangentially, along the direction of the circular arc.

The upper drawing labels the rigid support, string length L and bob mass m. The lower drawing marks θ from the vertical, tension T along the string, weight mg downward, and the components mg cos θ and mg sin θ.

See Fig. 13.17 in your NCERT textbook

In that figure, T labels tension; elsewhere in this note T denotes time period. The string's tension gives no turning effect about the support because its line of action passes through that point. The tangential component of weight supplies the restoring effect.

Derivation: Time period of a simple pendulum

Let τ be torque, the turning effect of force about the support; I the moment of inertia, measuring resistance to angular acceleration; and α the angular acceleration d²θ/dt².

1. The restoring torque is τ = −mgL sin θ. The negative sign indicates that it acts to reduce angular displacement.
2. For a small bob and massless string, I = mL². The rotational equation τ = Iα therefore gives α = −(g/L) sin θ.
3. For small θ measured in radians, use sin θ ≈ θ. Then α ≈ −(g/L)θ, which is the SHM form.
4. Identify ω² = g/L and substitute into T = 2π/ω to obtain the small-angle period.

T = 2π√(L/g)

The motion is approximately simple harmonic for small angular displacements. The small-angle approximation is essential; the exact restoring torque depends on sin θ. The period is independent of bob mass, and its independence of amplitude belongs to this small-angle approximation.

Worked example 5. A seconds pendulum has a complete period of 2 s. Find its length where g = 9.8 m s⁻², assuming small oscillations.

Formula: T = 2π√(L/g); L = gT²/(4π²).

Substitute: L = 9.8 × 2²/(4π²).

Answer: L ≈ 1 m. The time for a complete oscillation is 2 s, although successive ticks are one second apart.

How should spring and pendulum results be applied and checked?

Start by identifying the equilibrium position and the displacement measured from it. A horizontal spring has its equilibrium at the unstretched position in the model used here. A vertical spring oscillates around its loaded equilibrium, while a simple pendulum oscillates about the downward vertical.

Which quantities determine the period?

For the spring, mass affects period. For the simple pendulum, bob mass cancels. The pendulum period increases with length and decreases with gravitational acceleration. These differences come from different restoring mechanisms even though both lead to an SHM equation under the stated assumptions.

Worked example 6. A simple pendulum has a period of 3.5 s on Earth. Find its period on the Moon, keeping its length unchanged and using g = 9.8 m s⁻² on Earth and 1.7 m s⁻² on the Moon. Assume small oscillations.

Formula: T = 2π√(L/g). For the same L, the Moon-to-Earth period ratio equals √(9.8/1.7).

Substitute: Moon period = 3.5√(9.8/1.7).

Answer: The period on the Moon is approximately 8.4 s. The smaller restoring acceleration makes the oscillation slower.

What checks prevent errors in a calculation?

1. Convert lengths into metres before combining them with force constants in newtons per metre or g in metres per second squared.
2. Distinguish amplitude from the full separation of extreme positions, which is twice the amplitude.
3. Distinguish frequency f from angular frequency ω using ω = 2πf, and retain units in the final answer.
4. Check force and acceleration signs against displacement, and check whether a requested quantity is speed or signed velocity.

Real oscillating bodies eventually come to rest because of friction and other dissipative causes. Damping is this loss of oscillatory motion through dissipation. Constant amplitude and conserved mechanical energy describe the ideal undamped models used in these calculations.

Glossary

• Periodic motion — Motion that repeats itself after equal time intervals, with the smallest repeat interval defining its period.
• Oscillatory motion — To and fro motion of a body about a mean position within its path.
• Equilibrium position — Position where the resultant force on the body is zero in the system considered.
• Simple harmonic motion — Oscillation whose acceleration is proportional to displacement from equilibrium and directed towards that equilibrium.
• Time period — Smallest time interval after which the motion repeats, corresponding to one complete oscillation.
• Frequency — Number of complete oscillations per unit time, equal to the reciprocal of the period.
• Amplitude — Magnitude of the greatest displacement from equilibrium reached by the particle during its oscillation.
• Angular frequency — Rate of increase of phase in SHM, equal to twice π multiplied by frequency.
• Phase — Quantity specifying the stage of an oscillation and determining displacement and velocity for a given amplitude and frequency.
• Epoch — Initial phase of an oscillation at the chosen zero of time, also called phase constant.
• Restoring force — Force directed towards equilibrium, tending to reduce a body's displacement from that position.
• Force constant — Positive proportionality constant relating the magnitude of a linear restoring force to displacement.
• Simple pendulum — Ideal system consisting of a small massive bob suspended by a massless, inextensible string from a rigid support.
• Damping — Reduction of oscillatory motion through friction or other causes that dissipate the system's mechanical energy.

Common errors and misconceptions

• Misconception: Every periodic motion is SHM. Correct: SHM additionally requires acceleration proportional and opposite to displacement from equilibrium.
• Misconception: Angular frequency and frequency are equal. Correct: ω = 2πf; their usual units are radians per second and hertz respectively.
• Misconception: Acceleration vanishes wherever velocity vanishes. Correct: At an SHM extreme, velocity is zero but acceleration has its greatest magnitude.
• Misconception: The restoring force opposes velocity throughout the motion. Correct: It opposes displacement; it acts along the velocity while the particle moves towards equilibrium.
• Misconception: A vertical spring oscillates about its natural length. Correct: It oscillates about the loaded equilibrium position where spring force balances weight.
• Misconception: Kinetic and potential energies each remain constant. Correct: They exchange continuously; their sum remains constant in ideal undamped SHM.
• Misconception: A pendulum is exactly simple harmonic at every amplitude. Correct: Its motion is approximately SHM for small angular displacements, using sin θ ≈ θ.
• Misconception: Every pendulum tick is a complete oscillation. Correct: A seconds pendulum has successive ticks one second apart and a complete period of two seconds.

Exam-style questions with model answers

Q1. Define simple harmonic motion and state its acceleration-displacement relation. Define the symbols used. [2 marks]

1. SHM is oscillation in which acceleration is directly proportional to displacement from equilibrium and directed towards equilibrium.
2. The relation is a = −ω²y, where a is acceleration, y is signed displacement from equilibrium and ω is constant angular frequency.

Q2. A heart beats 75 times in 60 s. Calculate its frequency and time period. [2 marks]

1. Frequency is the number of beats divided by the elapsed time: f = 75/60 = 1.25 Hz.
2. The time period is the reciprocal of frequency: T = 1/1.25 = 0.8 s for each beat.

Q3. A particle has displacement y = A sin(ωt + φ₀), where A is constant amplitude, ω is constant angular frequency, t is time and φ₀ is initial phase. Derive velocity and acceleration and state their values at equilibrium. [3 marks]

1. Velocity v is the time derivative of displacement. Differentiating gives v = Aω cos(ωt + φ₀), so the velocity varies sinusoidally with the same period as displacement.
2. Acceleration a is the derivative of velocity: a = −Aω² sin(ωt + φ₀) = −ω²y. The negative sign shows its direction towards equilibrium.
3. At equilibrium y = 0, acceleration is zero and velocity has magnitude Aω, its greatest value. Its sign depends on the direction of passage.

Q4. A 3 kg block on a frictionless horizontal surface is attached to an ideal spring of force constant 1200 N m⁻¹. It is displaced 2.0 cm from equilibrium and released from rest. Calculate angular frequency, frequency, greatest speed and greatest acceleration magnitude. [4 marks]

1. Using k for force constant and m for mass, angular frequency is ω = √(k/m) = √(1200/3) = 20 rad s⁻¹.
2. Frequency is f = ω/(2π) = 10/π Hz ≈ 3.18 Hz. It is the number of complete oscillations per second.
3. The release displacement is amplitude A = 0.020 m because the block starts at rest. Greatest speed is Aω = 0.40 m s⁻¹.
4. Greatest acceleration magnitude is Aω² = 0.020 × 20² = 8.0 m s⁻², reached at the extreme positions.

Q5. A mass m hangs from an ideal vertical spring of force constant k in uniform gravitational acceleration g. Neglect resistance. Using downward displacement y from the loaded equilibrium, derive the differential equation and period, explaining why g cancels. [5 marks]

1. Let e be the spring's extension at rest. Equilibrium requires ke = mg because the upward spring force balances the downward weight.
2. After an additional downward displacement y, the total extension is e + y. The upward spring force is therefore k(e + y).
3. The resultant downward force is F = mg − k(e + y). Substitution of mg = ke reduces this expression to F = −ky.
4. Newton's second law gives m d²y/dt² = −ky, or d²y/dt² + (k/m)y = 0. Thus the motion about equilibrium is SHM.
5. Angular frequency ω satisfies ω² = k/m, giving period T = 2π√(m/k). Gravity sets the equilibrium extension; its constant contribution cancels when displacement is measured from that equilibrium.

Q6. An ideal simple pendulum has a small bob of mass m and a massless, inextensible string of length L in gravitational acceleration g. Neglect resistance. Derive its small-angle period, defining any additional symbols, and state the amplitude restriction. [5 marks]

1. Let θ be angular displacement from the downward vertical. The tangential component of weight acts towards equilibrium and has magnitude mg sin θ.
2. The restoring torque, denoted τ, about the support is τ = −mgL sin θ. Tension acts through the support and contributes no torque about it.
3. The moment of inertia I about the support is mL². Using angular acceleration α = d²θ/dt², the equation τ = Iα gives α = −(g/L) sin θ.
4. For small θ measured in radians, sin θ ≈ θ. Therefore α ≈ −(g/L)θ, the SHM relation with angular frequency satisfying ω² = g/L.
5. The period is T = 2π/ω = 2π√(L/g). It is independent of bob mass. This amplitude-independent expression is an approximation valid for small angular oscillations.

Q7. A 1 kg block attached to an ideal spring with force constant 50 N m⁻¹ moves on a frictionless horizontal surface. It is released from rest 10 cm from equilibrium. Calculate total, potential and kinetic energies at 5 cm from equilibrium. Take potential energy as zero at equilibrium. [3 marks]

1. The amplitude is A = 0.10 m. With k = 50 N m⁻¹, total mechanical energy is E = ½kA² = ½ × 50 × 0.10² = 0.25 J.
2. At displacement y = 0.05 m, potential energy is U = ½ky² = ½ × 50 × 0.05² = 0.0625 J.
3. Conservation gives kinetic energy K = E − U = 0.25 − 0.0625 ≈ 0.19 J. The last value is rounded; the total energy remains 0.25 J.

Q8. A pendulum has period 3.5 s on Earth, where gravitational acceleration is 9.8 m s⁻². Its length is unchanged on the Moon, where gravitational acceleration is 1.7 m s⁻². Assuming small oscillations and negligible resistance, calculate its lunar period and explain the change. [3 marks]

1. For length L and local gravitational acceleration g, the small-angle period is T = 2π√(L/g). Thus, at fixed length, period varies inversely as the square root of g.
2. Dividing the lunar period by the terrestrial period cancels the common length and 2π. The ratio is √(9.8/1.7).
3. The lunar period is therefore 3.5√(9.8/1.7) ≈ 8.4 s. It is longer because the smaller gravitational acceleration provides a weaker restoring effect.

Key takeaways

• SHM requires acceleration proportional and opposite to displacement from equilibrium, giving sinusoidal displacement with time.
• Period is the smallest repeat time; frequency is its reciprocal, and angular frequency equals 2π times frequency.
• Amplitude fixes the displacement limits, while initial phase specifies the starting stage of a given oscillation.
• The projection of uniform circular motion on a diameter executes SHM with amplitude equal to the circle's radius.
• At equilibrium, speed is greatest and acceleration zero; at either extreme, speed is zero and acceleration magnitude greatest.
• Horizontal and vertical ideal spring oscillators have period 2π√(m/k), with displacement measured from their respective equilibrium positions.
• Kinetic and potential energies exchange during ideal SHM, while their sum remains equal to ½kA².
• The simple pendulum has period approximately 2π√(L/g) for small angular oscillations; its bob mass cancels.

Test yourself

Why is repeated motion not sufficient evidence of SHM?

Periodicity establishes repetition, but SHM additionally requires acceleration proportional and opposite to displacement from equilibrium.

What is the difference between amplitude and displacement?

Displacement is the signed position relative to equilibrium at an instant; amplitude is its greatest magnitude during the oscillation.

Can a particle have zero acceleration and non-zero velocity during SHM?

Yes. At equilibrium its acceleration is zero, but its speed is greatest and generally non-zero.

What does the minus sign in F = −ky mean, where F is force, k is a positive force constant and y is displacement?

It means the restoring force points opposite to displacement, towards the equilibrium position.

Why must a vertical spring's displacement be measured from its loaded equilibrium?

At that position the constant spring force balances weight; additional displacement then gives the resultant restoring force directly.

If SHM has period T, why do its kinetic and potential energies repeat after T/2?

Each depends on a squared sinusoidal quantity, so reversing the sign of velocity or displacement leaves the corresponding energy unchanged.

Which approximation makes the pendulum equation simple harmonic?

For small angular displacement θ in radians, sin θ ≈ θ, making angular acceleration approximately proportional and opposite to θ.

Does a seconds pendulum have a period of one second?

No. Its complete oscillation takes two seconds; the one-second interval is between successive ticks.