NCERT Solutions for Class 11 Physics Chapter 13: Oscillations

Concept. A motion is periodic only if it repeats itself exactly, over and over, after equal intervals of time (the period). One-off or irregular motions are not periodic, even if they move back and forth.

Question. Which of the following examples represent periodic motion? (a) A swimmer completing one (return) trip from one bank of a river to the other and back. (b) A freely suspended bar magnet displaced from its N–S direction and released. (c) A hydrogen molecule rotating about its centre of mass. (d) An arrow released from a bow.

Analysis

What “periodic” demands

• The same state of motion (position and velocity) must recur after every fixed interval T.
• The motion should go on repeating indefinitely (apart from small losses): a single trip is not periodic.
• Periodic motion includes circular motion (c) and oscillation (b). All oscillations are periodic, but not all periodic motions are oscillatory.

Common trap in (a). The swimmer goes to the other bank and back, so the motion looks to-and-fro, but unless the swimmer keeps repeating the trip at constant speed and with constant intervals, we cannot call it periodic. A body is described as periodic only if its motion is repeated in identical fashion.

Examples to remember. Periodic: pendulum of a clock, a vibrating tuning fork, the Earth’s rotation and revolution, a planet’s orbit. Not periodic: a falling stone, the flight of a projectile, a ball thrown up and caught, the arrow of this question.

Extra depth. Strictly speaking even case (b) is periodic only in the idealisation of no damping. The magnetic needle of a compass slowly comes to rest because of friction at the pivot and air resistance: the amplitude decays (damped oscillation). For the purposes of the question we ignore small damping and treat the motion as periodic.

Conclusion: (a) not periodic; (b) periodic; (c) periodic; (d) not periodic.

Concept. SHM occurs only when the restoring force is proportional to the displacement (F = −kx), which means that the displacement is a single sine or cosine function of time. Periodic motions that are not of this simple form are periodic but not simple harmonic.

Question. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation of Earth about its axis; (b) motion of an oscillating mercury column in a U-tube; (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowest point; (d) general vibrations of a polyatomic molecule about its equilibrium position.

Analysis

How to decide

• Check for a restoring force F = −kx (or equivalently an acceleration a = −ω²x).
• A single sinusoidal x(t) means SHM; a sum of sinusoids with different frequencies is periodic (at best) but not SHM.
• Rotation is periodic but not oscillatory; SHM needs to-and-fro motion about a mean position.

Note on “nearly”. For (b) and (c) the motion is SHM only for small amplitudes. For large displacements the restoring force is no longer proportional to the displacement (for the bowl, sin θ differs from θ), and the motion is periodic but not SHM.

Extra depth. The superposition principle: a periodic function, however complicated, can be written as a sum of sines and cosines of frequencies ν, 2ν, 3ν, … (Fourier’s theorem). In a polyatomic molecule such as CO₂ or H₂O the normal modes are stretching and bending vibrations with different frequencies; each normal mode is SHM, but the general vibration is a sum of them and is not SHM.

Conclusion: (a) periodic, not SHM; (b) SHM; (c) SHM; (d) periodic, not SHM.

Concept. A plot is periodic only if its shape repeats exactly after a fixed time interval; the period is the smallest such interval.

Question. Fig. 13.18 depicts four x–t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?

Analysis of each plot

Why (c) is not periodic

The curve in (c) touches zero at the marked instants 1, 4, 7, 10 and 13 s, which is a regularity of the zero-crossings only. Between them the displacement has different peaks (the shape of every segment is different). For periodicity the entire function, not only the zeros, must repeat: x(t + T) = x(t) for all t. Here it does not, so the motion is not periodic.

Reading the period. Find the smallest time over which a feature (peak, trough or any given point) recurs with the same shape. In plot (d) successive peaks are 2 s apart, and in plot (b) the successive jumps at −3, −1, 1 and 3 s are also 2 s apart.

Not all periodic motions are SHM. Plot (d) looks sinusoidal, so it represents simple harmonic motion. The sawtooth-type plot (b) is periodic but not SHM because the shape is not a single sine or cosine.

Extra depth. A periodic function f(t) satisfies f(t + T) = f(t). If the motion is repeated every T, it is also repeated every 2T, 3T, …; the period is the least such time. The motion in (b) could be called “periodic with period 2 s, 4 s, 6 s …” but its period is the smallest, 2 s.

More to know. To tell whether a plot is periodic, shift the curve sideways by the proposed period T and see whether it falls exactly on itself. In plot (d) the displacement is a sine wave, so the motion is SHM with ω = 2π/T = π rad s⁻¹; the sawtooth (b) is periodic but not SHM, since it contains many harmonics. A curve like (c) with repeating zeros but changing shape between them is an example of how equal spacing of a few points can mislead: periodicity concerns the whole function x(t).

Conclusion: (a) not periodic; (b) periodic, T = 2 s; (c) not periodic; (d) periodic, T = 2 s.

Concept. A sum of sines or cosines of one frequency is a single SHM. A function with several frequencies (integer multiples of a basic frequency) is periodic but not SHM. A function that never repeats is non-periodic.

Question. Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give the period for each periodic case (ω is any positive constant): (a) sin ωt − cos ωt; (b) sin³ωt; (c) 3 cos(π/4 − 2ωt); (d) cos ωt + cos 3ωt + cos 5ωt; (e) exp(−ω²t²); (f) 1 + ωt + ω²t².

Solution

(i) sin ωt − cos ωt. Write as a single sinusoid: sin ωt − cos ωt = √2 sin(ωt − π/4). This is a sinusoidal function of time, so it is SHM, with period T = 2π/ω.

(ii) sin³ωt. Using sin 3θ = 3 sin θ − 4 sin³θ: sin³ωt = ¼(3 sin ωt − sin 3ωt). It is a sum of two SHMs of frequencies ω and 3ω, so it is periodic but not SHM; the period is 2π/ω (the longer of the two component periods).

(iii) 3 cos(π/4 − 2ωt). Since cos is even, 3 cos(2ωt − π/4): a single cosine of angular frequency 2ω, hence SHM with T = 2π/2ω = π/ω.

(iv) cos ωt + cos 3ωt + cos 5ωt. A sum of three SHMs with frequencies ω, 3ω, 5ω (integer multiples of ω). The sum is periodic but not SHM, with period 2π/ω (the fundamental period).

(v) exp(−ω²t²). A Gaussian that decays monotonically to zero and never repeats: non-periodic. (It is not even oscillatory.)

(vi) 1 + ωt + ω²t². A quadratic polynomial that increases without limit as t increases and never repeats: non-periodic.

Rule. A sum of harmonics ω, nω, mω (integers n, m) has the period 2π/ω of the lowest frequency. A sum of two SHMs with an irrational ratio of frequencies is not periodic at all.

Extra depth. All periodic functions can be written as a Fourier series, a sum of sines and cosines of frequencies ω, 2ω, 3ω, … In (iv) the function is already such a series with equal coefficients for the odd harmonics. The Gaussian in (v) is the shape of a pulse and the polynomial in (vi) is the displacement of a body under constant jerk: neither repeats.

Conclusion: (a) sin ωt − cos ωt: SHM, 2π/ω; sin³ωt: periodic, 2π/ω; 3 cos(π/4 − 2ωt): SHM, π/ω; cos ωt + cos 3ωt + cos 5ωt: periodic, 2π/ω; (c) exp(−ω²t²) and 1 + ωt + ω²t²: non-periodic.

Concept. In SHM about the mid-point O of AB, the acceleration and force are always directed towards O: a = −ω²x and F = −kx. The velocity is directed along the actual direction of motion and is zero at the extremes. Take A → B as the positive direction, so O is the origin, A is at x = −5 cm and B at x = +5 cm.

Question. A particle is in linear SHM between two points A and B, 10 cm apart. Take the direction from A to B as positive and give the signs of velocity, acceleration and force when it is (a) at the end A; (b) at the end B; (c) at the mid-point of AB going towards A; (d) at 2 cm away from B going towards A; (e) at 3 cm away from A going towards B; (f) at 4 cm away from B going towards A.

Set-up

Amplitude = 5 cm. Take the mid-point O as origin: A is at x = −5 cm, B at x = +5 cm. Then F and a have the sign opposite to x, and v is positive when moving towards B (positive direction) and negative when moving towards A.

Reasoning

• (a) At the extreme A the particle momentarily stops (v = 0) and the restoring force, directed towards the centre, points to B: F and a are positive.
• (b) At B: v = 0; F and a point towards A: negative.
• (c) At the mid-point the displacement is zero, so a = F = 0. The particle moves at maximum speed ωA towards A, so v is negative.
• (d) x = +3 cm (2 cm short of B); F, a = −kx, −ω²x are negative; moving towards A, v is negative.
• (e) x = −2 cm; a, F positive; moving towards B so v is positive.
• (f) x = +1 cm (4 cm from B); F and a are negative; v negative because it moves towards A.

Key point. The sign of the acceleration and force depends only on the position (side of the mean position), while the sign of the velocity depends on the direction of motion. At the same position the velocity can have either sign.

Numerical check. In (d) with ω = 2π/T and A = 5 cm, speed = ω√(5² − 3²) = 4ω cm s⁻¹; acceleration magnitude = 3ω² cm s⁻².

Extra depth. Cases (d) and (f) look similar (both going towards A, with F and a negative) but are different positions: x = +3 cm and x = +1 cm. The magnitudes differ: |a| = 3ω² and ω²; the speeds are 4ω and √24 ω = 4.9ω cm s⁻¹. Only the signs were asked, and the signs are the same in the two cases.

Conclusion: (a) v = 0, a and F positive; (b) v = 0, a and F negative; (c) v negative, a = F = 0; (d) all three negative; (e) all positive; (f) all negative.

Concept. The defining property of SHM is that the acceleration is proportional to the displacement and directed opposite to it: a = −ω²x with ω² a positive constant. Any other dependence (a different power of x, or the wrong sign) is not SHM.

Question. Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? (a) a = 0.7x; (b) a = −200x²; (c) a = −10x; (d) a = 100x³.

Test

Details for the SHM case (c)

Comparing a = −10x with a = −ω²x gives ω² = 10 s⁻², so

ω = √10 = 3.16 rad s⁻¹, T = 2π/ω = 2π/3.16 = 1.99 s ≈ 2 s, ν = 0.50 Hz

Why the others fail

• (a) For a = +0.7x the force pushes the particle further away from the origin: the motion is exponential growth, not an oscillation (unstable equilibrium).
• (b) For a = −200x² the force is always negative, so the particle accelerates in the −x direction whichever side of the origin it is on; there is no oscillation.
• (d) For a = +100x³ the force is directed away from the origin and grows rapidly with distance: unstable. A restoring force proportional to x³ would be written a = −cx³ and gives a periodic but non-harmonic (anharmonic) oscillation.

Two conditions. (i) Linearity: a ∝ x (the first power). (ii) Restoring sign: the proportionality constant must be negative. Only (c) meets both.

Extra depth. Anharmonic oscillators are common in nature: for a diatomic molecule the potential energy curve is not exactly parabolic, so the force has terms in x² and x³ in addition to −kx; the motion is periodic but not SHM, and the period depends on the amplitude. For small amplitudes the −kx term dominates and the motion is approximately SHM.

Conclusion: Only (c) a = −10x is SHM (ω = √10 rad s⁻¹, T ≈ 2 s).

Concept. The initial position and velocity fix the two unknowns A and φ in x = A cos(ωt + φ), with v = −ωA sin(ωt + φ). The same motion can be described with a sine function using a different phase.

Question. The motion of a particle executing SHM is x(t) = A cos(ωt + φ). If at t = 0 the position is 1 cm and the velocity is ω cm s⁻¹, what are the amplitude and initial phase angle? The angular frequency is π s⁻¹. If instead of the cosine function we choose x = B sin(ωt + α), what are the amplitude and initial phase with the same initial conditions?

Given

• x(0) = 1 cm; v(0) = ω cm s⁻¹ (ω = π s⁻¹)
• x(t) = A cos(ωt + φ), v(t) = −ωA sin(ωt + φ)

Cosine form

At t = 0:

x(0) = A cos φ = 1 …(1)

v(0) = −ωA sin φ = ω ⟹ A sin φ = −1 …(2)

Square and add (1) and (2): A²(cos²φ + sin²φ) = 1 + 1, so A = √2 cm. Divide (2) by (1): tan φ = −1, so φ = −π/4 or 3π/4 or 7π/4. Since cos φ must be positive (A cos φ = 1) and sin φ negative (A sin φ = −1), φ is in the 4th quadrant:

A = √2 cm, φ = −π/4 (= 7π/4)

Thus x(t) = √2 cos(πt − π/4) cm. Check: x(0) = √2 cos(−π/4) = √2/√2 = 1 ✓; v(0) = −π√2 sin(−π/4) = π√2/√2 = π = ω ✓.

Sine form

x(t) = B sin(ωt + α), v(t) = ωB cos(ωt + α). At t = 0:

B sin α = 1 …(3)

ωB cos α = ω ⟹ B cos α = 1 …(4)

Squaring and adding gives B² = 2 and from (3)/(4) tan α = 1, with sin and cos both positive so α is in the 1st quadrant:

B = √2 cm, α = π/4

Thus x(t) = √2 sin(πt + π/4) cm. This is the same motion as √2 cos(πt − π/4), since sin(θ + π/4) = cos(θ + π/4 − π/2) = cos(θ − π/4).

Relation. α = φ + π/2 always: −π/4 + π/2 = π/4. The amplitude is the same in both forms because it describes the same oscillation.

Extra depth. Two initial conditions (position and velocity) are necessary and sufficient to fix A and φ. Geometrically, in the reference circle the particle starts at the angle φ = −π/4 with x = A cos φ = 1 cm and with a downward projection speed; it covers one revolution in T = 2π/ω = 2 s.

Conclusion: Cosine form: A = √2 cm, φ = −π/4; sine form: B = √2 cm, α = π/4.

Concept. A spring balance is a spring: the scale reading is proportional to the extension, so the spring constant is k = (full-scale load)/(length of the scale). A suspended body oscillates with T = 2π√(M/k), from which M and then the weight follow.

Question. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Given

• Maximum load 50 kg stretches the spring by 20 cm = 0.20 m
• Period of oscillation T = 0.6 s; g = 9.8 m s⁻²

Solution

Step 1: spring constant.

k = F/x = (50 × 9.8)/0.20 = 490/0.20 = 2450 N m⁻¹

Step 2: mass from the period. T = 2π√(M/k) ⟹ M = kT²/(4π²):

M = (2450 × 0.6²)/(4π²) = (2450 × 0.36)/39.48 = 882/39.48 = 22.3 kg

Step 3: weight.

W = Mg = 22.3 × 9.8 = 219 N ≈ 2.2 × 10² N

The weight of the body is about 2.2 × 10² N (a mass of about 22 kg).

Cross-check

With M = 22.3 kg the static extension would be Mg/k = 219/2450 = 0.0894 m = 8.9 cm, and the period T = 2π√(M/k) = 2π√(22.3/2450) = 2π × 0.0954 = 0.600 s ✓. The static extension relation also gives T = 2π√(x₀/g) = 2π√(0.0894/9.8) = 0.600 s, which shows that the period of a vertical spring oscillation is determined by the static stretch.

Note. Do not use the formula T = 2π√(M/k) with M = 50 kg (the full-scale mass). The scale range fixes only the spring constant.

Extra depth. The period of a body hanging on a spring depends on the mass but not on g: on the Moon the weight would be one-sixth, the static extension also one-sixth, but the period of oscillation would be unchanged (0.6 s), since T = 2π√(M/k). A spring balance reads weight, not mass, so it would read a different value on the Moon while a beam balance would not.

More to know. The “weight” shown on a spring balance is really the force that stretches the spring, so the reading changes if the balance is accelerated (in a lift). The period of oscillation depends on the mass and the spring constant only, so it is a clean way to measure the mass of astronauts in orbit: the astronaut sits in a chair attached to a spring, and the measured period gives the mass, because there is no apparent weight in orbit.

Conclusion: Weight of the body ≈ 2.2 × 10² N (M ≈ 22 kg).

Concept. For a mass–spring system on a frictionless horizontal table, ω = √(k/m). The amplitude is the initial displacement (the mass is released from rest), so a_max = ω²A and v_max = ωA.

Question. A spring with spring constant 1200 N m⁻¹ is mounted on a horizontal table (Fig. 13.19). A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Determine (i) the frequency of oscillations, (ii) the maximum acceleration of the mass, (iii) the maximum speed of the mass.

Given

• k = 1200 N m⁻¹, m = 3 kg
• Amplitude A = 2.0 cm = 0.020 m (released from rest at that displacement)

Solution

ω = √(k/m) = √(1200/3) = √400 = 20 rad s⁻¹

(i) Frequency:

ν = ω/2π = 20/(2 × 3.14) = 3.18 Hz (T = 0.314 s)

(ii) Maximum acceleration (at the extreme positions):

a_max = ω²A = 400 × 0.020 = 8.0 m s⁻²

Check by Newton’s law: a_max = kA/m = 1200 × 0.020/3 = 8.0 m s⁻² ✓.

(iii) Maximum speed (at the mean position):

v_max = ωA = 20 × 0.020 = 0.40 m s⁻¹

Check by energy: ½kA² = ½mv_max² gives v_max = A√(k/m) = 0.02 × 20 = 0.40 m s⁻¹ ✓.

Energy of the system. E = ½kA² = ½ × 1200 × (0.02)² = 0.24 J; at the mean position all of it is kinetic.

Note. The amplitude is 2.0 cm because the mass is pulled by 2.0 cm and released from rest; its initial speed is zero. If the initial speed were not zero, the amplitude would be larger than the initial displacement.

Extra depth. The period T = 2π√(m/k) = 0.314 s does not depend on the amplitude: pulling the mass by 5 cm instead of 2 cm would give the same frequency 3.18 Hz but a_max = 20 m s⁻² and v_max = 1.0 m s⁻¹. If the table were rough, friction would damp the oscillation and the equilibrium position would also shift slightly, so the horizontal frictionless table is an idealisation.

Conclusion: (i) 3.18 Hz; (ii) 8.0 m s⁻²; (iii) 0.40 m s⁻¹.

Concept. In x(t) = A cos(ωt + φ), the angular frequency ω = √(k/m) = 20 rad s⁻¹ and the amplitude A = 2.0 cm are the same in all three cases; only the initial condition (and hence the phase constant φ) differs.

Question. In Exercise 13.9, take the position of the mass when the spring is unstretched as x = 0 and the direction from left to right as the positive x-axis. Give x as a function of time t for the oscillating mass if at t = 0 the mass is (a) at the mean position, (b) at the maximum stretched position, (c) at the maximum compressed position. In what way do these functions differ from each other?

Common data

ω = 20 rad s⁻¹; A = 2.0 cm. The general form is x = A cos(ωt + φ), with A = 2 cm.

(a) At the mean position at t = 0

Here x(0) = 0, so cos φ = 0 and φ = ±π/2. If the mass is moving in the +x direction at t = 0 (the usual choice): x = 2 sin 20t cm, i.e. x = 2 cos(20t − π/2) with φ = −π/2.

(b) At the maximum stretched position at t = 0

At the maximum stretch x = +A, so x(0) = A and φ = 0:

x = 2 cos 20t (cm)

(c) At the maximum compressed position at t = 0

At the maximum compression x = −A, so φ = π:

x = −2 cos 20t = 2 cos(20t + π) (cm)

How do they differ?

All three have the same amplitude (2 cm) and the same frequency (20 rad s⁻¹, ν = 3.18 Hz). They differ only in the initial phase: φ = −π/2, 0 and π respectively. Frequency and amplitude depend on the system (k, m) and on the energy; the phase depends on the instant when t = 0 is chosen.

Remark. In (a) if the mass is moving in the −x direction at t = 0, then x = −2 sin 20t, φ = +π/2. The statement of the problem leaves the direction open, so a short note on the sign is useful.

Extra depth. The phase constant tells where in the cycle the oscillation starts. Shifting the stopwatch start by a quarter period (T/4 = 0.0785 s) changes φ by π/2: (b) becomes (a) etc. The function sin 20t is the same as cos(20t − π/2), so all three are the same sinusoid shifted along the time axis.

Conclusion: (a) x = 2 sin 20t cm; (b) x = 2 cos 20t cm; (c) x = −2 cos 20t cm. They differ only in the initial phase.

Concept. The x-projection of the radius vector of a particle moving uniformly on a circle is SHM: x = A cos(θ), with θ = θ₀ ± ωt (plus for anticlockwise, minus for clockwise), A = radius and ω = 2π/T.

Question. Fig. 13.20 shows two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution (clockwise or anticlockwise) are indicated. Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P in each case.

Reading the figure

Case (a)

ω = 2π/T = 2π/2 = π rad s⁻¹. The angle of OP with the +x axis at time t, for clockwise motion, decreases with time: θ(t) = θ₀ − ωt = −π/2 − πt. The x-projection is

x(t) = A cos θ = 3 cos(−π/2 − πt) = 3 cos(πt + π/2) = −3 sin(πt) (cm)

Check at t = 0: x = 0 ✓ (P is on the y-axis). A moment later P moves clockwise from the bottom towards the left, so x becomes negative ✓ (−3 sin πt is negative for small t).

Case (b)

ω = 2π/4 = π/2 rad s⁻¹; θ(t) = π − (π/2)t. Then

x(t) = 2 cos(π − πt/2) = −2 cos(πt/2) (m)

Check: at t = 0, x = −2 m ✓ (P on the negative x-axis). At t = 1 s, θ = π/2: P is on top, x = 0 and −2 cos(π/2) = 0 ✓.

Rule for the sense of rotation

For anticlockwise rotation the argument is θ₀ + ωt, giving x = A cos(ωt + θ₀). For clockwise rotation the argument is θ₀ − ωt; since cos is even, x = A cos(ωt − θ₀): the phase constant is −θ₀. Case (a): θ₀ = −π/2, clockwise ⟹ φ = +π/2. Case (b): θ₀ = π, so φ = −π ≡ π: same cosine since cos is even, so the sense of rotation does not matter if P starts on the x-axis.

Extra depth. The sense of rotation affects the x-projection only when the starting point is off the x-axis. For (b) the particle starts on the x-axis, so the clockwise and anticlockwise cases give the same x(t); the y-projection would differ in sign. Any SHM can be modelled by a rotating vector (phasor), an idea that is used throughout alternating current theory and wave optics.

Conclusion: (a) x = −3 sin(πt) cm; (b) x = −2 cos(πt/2) m.

Concept. Convert each SHM to the standard form x = A cos(ωt + φ). Then the reference circle has radius A (amplitude), the particle rotates anticlockwise with angular speed ω, and at t = 0 the radius vector makes the angle φ with the +x-axis.

Question. Plot the reference circle for each of the following SHMs. Indicate the initial (t = 0) position of the particle, the radius of the circle and the angular speed of the rotating particle (anticlockwise sense, x in cm and t in s): (a) x = −2 sin(3t + π/3); (b) x = cos(π/6 − t); (c) x = 3 sin(2πt + π/4); (d) x = 2 cos πt.

Conversion to the cosine form

Describing each reference circle

• (a) Circle of radius 2 cm; the particle starts at 150° from the +x axis (in the second quadrant) and rotates anticlockwise at 3 rad s⁻¹ (T = 2π/3 = 2.09 s).
• (b) Circle of radius 1 cm; the particle starts at −30° (in the fourth quadrant) and rotates anticlockwise at 1 rad s⁻¹ (T = 2π s = 6.28 s).
• (c) Circle of radius 3 cm; the particle starts at −45° (fourth quadrant) and rotates anticlockwise at 2π rad s⁻¹ (T = 1 s).
• (d) Circle of radius 2 cm; the particle starts on the +x axis (angle 0) and rotates anticlockwise at π rad s⁻¹ (T = 2 s).

Check by evaluating x(0). (a) x(0) = −2 sin(π/3) = −1.73 cm and 2 cos(5π/6) = −1.73 ✓. (b) x(0) = cos(π/6) = 0.866 = cos(−π/6) ✓. (c) x(0) = 3 sin(π/4) = 2.12 and 3 cos(−π/4) = 2.12 ✓. (d) x(0) = 2 cm ✓.

The initial projection on the x-axis is the initial displacement: P′ is the foot of the perpendicular from P(0) on the x-axis, in the figure above shown as a violet dot.

Extra depth. To plot a reference circle by hand: draw a circle of radius A; draw the radius at the angle φ from the +x axis (measured anticlockwise, negative φ clockwise); mark P(0) at its end; drop a perpendicular to the x-axis to get the initial displacement; indicate anticlockwise rotation at ω. The period is T = 2π/ω, so (c) completes one revolution per second.

Conclusion: (a) r = 2 cm, ω = 3 rad/s, φ = 150°; (b) r = 1 cm, ω = 1 rad/s, φ = −30°; (c) r = 3 cm, ω = 2π rad/s, φ = −45°; (d) r = 2 cm, ω = π rad/s, φ = 0.

Concept. A spring stretched by a force F at each end has a tension F, so its extension is F/k in both arrangements. In the two-mass case the centre of the spring does not move (by symmetry), so each half acts as a spring of constant 2k, attached to one mass.

Question. Fig. 13.21(a) shows a spring of force constant k clamped rigidly at one end with a mass m at its free end; a force F applied at the free end stretches the spring. Fig. 13.21(b) shows the same spring with both ends free and a mass m at either end; each end is stretched by the same force F. (a) What is the maximum extension of the spring in the two cases? (b) If the masses are released, what is the period of oscillation in each case?

(a) Maximum extension

Case (a). The wall holds one end, and the applied force F stretches the spring. The tension in the spring is F, so by Hooke’s law F = kx:

x_max = F/k

Case (b). Each end is pulled outward by a force F, and the spring is in equilibrium (the net force on it is zero). The tension everywhere in the spring is F, so

x_max = F/k

The extension is again F/k, the same as in (a): it depends on the tension, which is F in both cases (the wall in (a) supplies the force F on the other end).

(b) Periods

Case (a). A single mass m on a spring of constant k: T = 2π√(m/k).

Case (b). The two masses oscillate in opposite directions with equal amplitudes (the centre of mass of the system stays fixed and the midpoint of the spring is a node). Each mass is therefore attached to a half-spring of length L/2. A spring of half the length has twice the spring constant (k₁ = 2k, as the same force F gives half the extension). Each mass sees k_eff = 2k:

T = 2π√(m/2k) = (2π/√2)√(m/k)

Alternative route: reduced mass. For two masses m₁ and m₂ joined by a spring, T = 2π√(μ/k) with μ = m₁m₂/(m₁ + m₂) = m/2. So T = 2π√(m/2k), the same result. The period in (b) is shorter by the factor 1/√2 = 0.707.

Application. The same formula with reduced mass describes the vibration of a diatomic molecule: the atoms oscillate as two masses joined by the bond (spring), with ν = (1/2π)√(k/μ).

Extra depth. The case (b) is an example of the motion of a two-body system: the relative coordinate x = x₂ − x₁ obeys μ d²x/dt² = −kx. Heavier atoms (larger μ) vibrate at lower frequency: the isotope effect on molecular spectra (H₂ vs D₂) is explained by this: ν ∝ 1/√μ.

Conclusion: (a) F/k in both cases; (b) T = 2π√(m/k) in (a) and T = 2π√(m/2k) in (b).

Concept. For SHM the maximum speed is v_max = ωA. The stroke is the total distance travelled between the extreme positions, which is twice the amplitude.

Question. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?

Given

• Stroke = 2A = 1.0 m, so amplitude A = 0.50 m
• ω = 200 rad min⁻¹ = 200/60 = 3.33 rad s⁻¹

Solution

v_max = ωA = 200 rad min⁻¹ × 0.50 m = 100 m min⁻¹

v_max = 100/60 = 1.67 m s⁻¹

The maximum speed of the piston is about 1.7 m s⁻¹ (100 m min⁻¹), reached when the piston passes through the mid-point of its stroke.

Related quantities

Units. Radians are dimensionless, so rad min⁻¹ × m = m min⁻¹. Convert minutes to seconds at the end: 100 m min⁻¹ ÷ 60 = 1.67 m s⁻¹. Alternatively convert ω to s⁻¹ first and then multiply.

Common mistake. Using A = 1.0 m (the stroke) would double the answer. The amplitude is the distance from the mid-point to one end, half the stroke.

Extra depth. The motion of a real locomotive piston is approximately SHM only if the connecting rod is long compared with the crank radius; the exact motion has a small second harmonic. For a short connecting rod the piston is faster in one half of the stroke than the other, which matters in engine balancing.

More to know. In an exam always convert the units carefully: ω = 200 rad/min = 3.33 rad/s. The motion of the piston, if the crank has radius r and rotates at angular speed ω, is approximately x = r cos ωt: the piston is the projection of the crank pin moving on a circle, the physical realisation of the reference circle. The maximum speed ωr is attained when the crank is at right angles to the cylinder axis.

Conclusion: Maximum speed = 100 m/min ≈ 1.7 m/s.

Concept. For a simple pendulum T = 2π√(L/g), so for the same length the period is inversely proportional to √g. Therefore T₂/T₁ = √(g₁/g₂).

Question. The acceleration due to gravity on the surface of the Moon is 1.7 m s⁻². What is the time period of a simple pendulum on the surface of the Moon if its time period on the surface of the Earth is 3.5 s? (g on the Earth = 9.8 m s⁻²)

Given

• g_E = 9.8 m s⁻², T_E = 3.5 s
• g_M = 1.7 m s⁻²; same length L

Solution

T_E = 2π√(L/g_E), T_M = 2π√(L/g_M)

T_M/T_E = √(g_E/g_M) = √(9.8/1.7) = √5.765 = 2.40

T_M = 2.40 × 3.5 = 8.4 s

The period of the pendulum on the Moon is about 8.4 s.

Check by finding the length

L = g_E T_E²/4π² = 9.8 × 12.25/39.48 = 3.04 m. On the Moon: T_M = 2π√(3.04/1.7) = 2π × 1.337 = 8.40 s ✓.

Why the pendulum is slower on the Moon. The restoring force mg sin θ is smaller by the factor g_M/g_E = 0.173, so the bob accelerates back more slowly and the oscillation takes 2.4 times longer. A pendulum clock calibrated on the Earth would run slow by this factor on the Moon: it would show about 1 h while 2.4 h of real time have passed.

Note. The period does not depend on the mass of the bob, so a heavier or lighter bob makes no difference.

Extra depth. Because T depends on g, a pendulum can measure g: g = 4π²L/T². A seconds pendulum (T = 2 s) has a length of 0.994 m at the Earth’s surface, 0.172 m on the Moon. Geologists have used sensitive pendulums to detect variations of g with altitude and underground density (gravity surveys).

More to know. The period on any planet scales as 1/√g: for Jupiter (g = 24.8 m s⁻²) the same pendulum would have T = 3.5 × √(9.8/24.8) = 2.2 s. The Moon’s weaker gravity (1/6 of the Earth’s) explains why astronauts’ pendulum-like walking is slower and why lunar jumps are high. Always check which quantity stays the same between the two places: here it is the length of the pendulum, not the period or the mass.

Conclusion: T on the Moon = 8.4 s.

Concept. In the car’s frame the bob experiences gravity mg (downward) and a pseudo (centrifugal) force Mv²/R directed radially outward. These two forces are perpendicular, and their resultant defines the effective acceleration g_eff. The pendulum oscillates about the direction of g_eff with T = 2π√(l/g_eff).

Question. A simple pendulum of length l and bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

Forces in the frame of the car

• Weight Mg, vertically downward.
• Centrifugal (pseudo) force Mv²/R, horizontally, radially outward from the centre of the track (v²/R is the centripetal acceleration of the car).
• Tension in the string: it balances the resultant of these two at equilibrium.

The two forces are at right angles, so the effective weight per unit mass has magnitude

g_eff = √(g² + (v²/R)²) = √(g² + v⁴/R²)

The pendulum hangs at an angle θ₀ to the vertical, with tan θ₀ = (v²/R)/g = v²/(gR), outward. For small oscillations in the radial direction about this new equilibrium line, the restoring force is M g_eff sin(θ − θ₀) ≈ M g_eff (θ − θ₀), exactly as for an ordinary pendulum in a gravitational field g_eff. Hence

T = 2π √(l/g_eff) = 2π √( l / √(g² + v⁴/R²) )

Limiting cases

Interpretation. Because of the extra outward acceleration, the pendulum behaves as if gravity were stronger, so the period is shorter than 2π√(l/g). The pendulum’s equilibrium line is tilted outwards from the vertical by θ₀ = tan⁻¹(v²/gR). It is like a pendulum in a lift that accelerates upwards (g_eff = g + a), but here the extra acceleration is horizontal, at 90° to g.

Example. v = 20 m s⁻¹ on a track of R = 100 m: v²/R = 4 m s⁻², g_eff = √(9.8² + 4²) = 10.6 m s⁻²; for l = 1 m, T = 2π√(1/10.6) = 1.93 s against 2.01 s for a pendulum at rest.

Extra depth. An observer on the ground sees the car accelerating towards the centre with a = v²/R. In the inertial frame the bob’s equilibrium requires a tension with a horizontal component M v²/R (providing the centripetal force) and vertical component Mg, giving the same g_eff for the relative motion. The result is general: in an accelerated frame g_eff = g − a (vector difference).

Conclusion: T = 2π √( l / √(g² + v⁴/R²) ).

Concept. When the floating cork is pushed down, the buoyant force exceeds its weight and the net force is a restoring force proportional to the additional depth y: this is the force law of SHM. The constant of proportionality gives ω.

Question. A cylindrical piece of cork of density ρ, base area A and height h floats in a liquid of density ρ_l. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with period T = 2π√(hρ/(ρ_l g)). (Ignore damping due to viscosity of the liquid.)

Equilibrium

Mass of cork: m = (volume) × ρ = A h ρ. Let h′ be the depth submerged at equilibrium. Weight = buoyant force:

mg = A h ρ g = ρ_l A h′ g ⟹ h′ = h ρ/ρ_l

Displacement from equilibrium

Push the cork down by an additional small distance y. The submerged depth becomes h′ + y and the buoyant force becomes ρ_l A (h′ + y) g. The net upward force on the cork is

F = ρ_l A (h′ + y) g − m g = ρ_l A g y (upward)

(using ρ_l A h′ g = mg). Taking the downward displacement y as positive, the force is opposite to the displacement:

F = −ρ_l A g y = −k y, k = ρ_l A g

This is the SHM force law (restoring force proportional to the displacement). The cork therefore executes simple harmonic motion with

ω = √(k/m) = √(ρ_l A g/(A h ρ)) = √(ρ_l g/(h ρ))

T = 2π/ω = 2π √( h ρ/(ρ_l g) )

Result obtained. This is the required formula. It is valid for small displacements (while the cork stays partly in the liquid and is neither fully submerged nor lifted out).

Observations

• T is independent of the base area A.
• T increases with the height h and the density ρ of the cork (more inertia) and decreases for a denser liquid (stronger restoring force).
• In terms of the submerged depth, h′ = hρ/ρ_l, so T = 2π√(h′/g): the cork oscillates like a simple pendulum of length equal to the submerged depth.

Numerical illustration. A cork of height 4 cm and density 0.25 g cm⁻³ floating on water (ρ_l = 1 g cm⁻³): h′ = 1 cm and T = 2π√(0.01/9.8) = 0.20 s: it bobs about 5 times a second.

Extra depth. The same analysis applies to any floating body (a ship rolling or heaving, a hydrometer, a buoy). The restoring force ρ_l A g y is the change of buoyancy, called the “hydrostatic stiffness”. Damping by the viscosity of the liquid and the waves radiated by the oscillating body make the real oscillation die out, which is why the question asks to ignore damping.

Conclusion: T = 2π√(hρ/(ρ_l g)), SHM for small depression.

Concept. In the U-tube, when the mercury is displaced by y in one limb, the levels differ by 2y; the pressure difference 2ρgy acting on the cross-section A gives a restoring force. The force is proportional to the displacement, hence SHM.

Question. One end of a U-tube containing mercury is connected to a suction pump and the other end to the atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Set-up

Let the uniform tube have cross-section A and let the total length of the mercury column be h, with mass m = A h ρ (ρ = density of mercury). With the pump on, the pressure difference holds the mercury with one column higher than the other. When the pump is removed (both ends at atmospheric pressure), the mercury is displaced from the equilibrium position where both levels are equal; the mercury then oscillates.

Restoring force

Suppose the level in one limb is a distance y above the equilibrium level; then in the other limb it is y below it. The difference in the heights of the two columns is 2y, which gives a difference in pressure at the same horizontal level:

ΔP = 2yρg

This pressure difference acts over the cross-section A and tends to restore the equilibrium: the net force on the mercury is

F = −ΔP · A = −2Aρg y

F is directed opposite to the displacement y and is proportional to it, which is the condition for SHM: F = −ky with k = 2Aρg. The mercury column therefore executes SHM.

Period

ω = √(k/m) = √(2Aρg/(Aρh)) = √(2g/h)

T = 2π √(h/2g)

Here h is the total length of the mercury column (the sum of the lengths in both limbs and the bottom). For h = 0.20 m, T = 2π√(0.20/19.6) = 0.63 s.

Remarks. The result is independent of the density of the liquid and of the cross-section of the tube (so long as it is uniform). The same analysis applies to any liquid (water, oil) in a U-tube. The motion is damped by viscosity, which we have ignored; for narrow tubes the damping is large and the liquid returns to equilibrium without oscillating.

Extra depth. This system is a one-dimensional analogue of a liquid column oscillating in a manometer. Also note the energy: when the level of one limb is displaced by y the potential energy increases by ½(2Aρg)y² = Aρg y², which is the energy of raising a volume Ay of liquid through a height y (centre of mass shift). Check with E = ½kA₀² for amplitude A₀.

Conclusion: The restoring force F = −2Aρg y ∝ −y, so the mercury executes SHM with T = 2π√(h/2g).