Periodic motion, simple harmonic motion, energy and the pendulum.
CHAPTER 13
Oscillations
Build SHM from its defining equation, then connect its motion, force, energy and pendulum applications.
13.1–13.2
Periodic and oscillatory motion
A motion can repeat without moving to and fro; oscillation is a narrower idea.
Periodic motion repeats after a least interval T; the state repeats at t+nT for integer n. Frequency ν=1/T is cycles per second (hertz), and angular frequency ω=2πν=2π/T.
Oscillatory motion is to-and-fro about an equilibrium position. Every oscillatory motion is periodic in the ideal case, but circular motion can be periodic without being oscillatory. Damping makes free oscillations die away; external periodic forcing can sustain them.
“Oscillation” and “vibration” are commonly used for lower- and higher-frequency to-and-fro motion; the distinction is conventional. A medium’s coupled oscillators transmit disturbances as waves.
Displacement may be position, angle, voltage, pressure or another variable that changes periodically. A periodic signal need not be a displacement in space.
NCERT Example 13.1
Heartbeat frequency
Question. A healthy human heart beats 75 times in a minute. What are its frequency and period?
Step 1. Frequency is beats per second: ν=75/60=1.25 Hz.
Step 2. The period is the reciprocal: T=1/ν.
Answer: ν=1.25 Hz and T=0.80 s.
13.2–13.3
Simple harmonic motion and phase
SHM is the special periodic motion with sinusoidal displacement or a linear restoring force.
Definition: x(t)=A cos(ωt+φ), where A≥0 is amplitude, ω is angular frequency and φ is the phase constant. The phase at time t is ωt+φ.
T=2π/ω and ν=ω/(2π). The phase sets the starting state; given ω, position and velocity at one instant determine A and φ.
Equivalent forms are x=A cos ωt+B sin ωt, x=A cos(ωt+α), or x=B sin(ωt+β). For A cosωt+B sinωt, amplitude is √(A²+B²); phase follows by matching coefficients.
A periodic function may be expanded as a sum of harmonics (Fourier series); the sum of two arbitrary SHMs is not necessarily periodic. Commensurate frequencies (their ratio rational) give a common repeat period.
Example 13.2: sinωt+cosωt=√2 sin(ωt+π/4) has period 2π/ω; sinωt+cos2ωt+sin4ωt repeats after 2π/ω. e^(−ωt) and log(ωt) are not periodic for t>0.
Example 13.3: sinωt−cosωt=√2 sin(ωt−π/4), so SHM with period 2π/ω. sin²ωt=(1−cos2ωt)/2 is periodic with period π/ω, but is not SHM about zero (its mean is 1/2).
NCERT Example 13.2
Testing periodic functions
Question. Which are periodic: (a) sinωt+cosωt, (b) sinωt+cos2ωt+sin4ωt, (c) e^(−ωt), (d) log(ωt)?
Step 1. (a) Both terms have angular frequency ω, so their sum repeats after 2π/ω.
Step 2. (b) The component periods are 2π/ω, π/ω and π/(2ω); the least common period is 2π/ω.
Step 3. (c) The exponential decays and does not repeat. (d) log(ωt) is undefined for t≤0 and is monotonic for t>0.
Answer: (a), (b) are periodic with T=2π/ω; (c), (d) are not.
NCERT Example 13.3
SHM or periodic motion?
Question. Classify (a) sinωt−cosωt and (b) sin²ωt; give each period.
Step 1. (a) Rewrite as √2 sin(ωt−π/4), a single sinusoid about zero.
Step 2. (b) Use sin²ωt=(1−cos2ωt)/2; it oscillates about 1/2, not zero.
Answer: (a) SHM, T=2π/ω. (b) Periodic but not SHM about zero, T=π/ω.
13.4
SHM as a projection of uniform circular motion
The shadow of a uniformly rotating point on a diameter follows SHM.
Let a point P rotate on a circle of radius A with angular speed ω and initial angle φ. Its coordinates are x=A cos(ωt+φ), y=A sin(ωt+φ); each projection is SHM, with the two projections π/2 out of phase.
This is a kinematic correspondence: the projected particle’s restoring force is not the centripetal force acting on P. In two dimensions the restoring law F=−mω²r does produce circular motion when the perpendicular components have equal amplitude and differ in phase by π/2.
Example 13.4(a): T=4 s and initial angle π/4 anticlockwise gives x=A cos(πt/2+π/4). (b) T=30 s from the top clockwise gives x=B sin(πt/15)=B cos(πt/15−π/2).
The rotating point P has constant angular speed; its x-projection moves back and forth as SHM.
Differentiate displacement to obtain the instantaneous velocity and acceleration.
For x=A cos(ωt+φ), v=dx/dt=−ωA sin(ωt+φ), so vmax=ωA. Speed is greatest at x=0; velocity is zero at x=±A.
a=dv/dt=−ω²A cos(ωt+φ)=−ω²x, so amax=ω²A. Acceleration always points toward equilibrium.
x, v and a have the same period T. Velocity is shifted by π/2 relative to displacement; acceleration is opposite in phase (π).
Example 13.5: x=5 cos(2πt+π/4) m at t=1.5 s gives x=−3.535 m; v=−10π sin(13π/4)=+22.2 m s⁻¹; a=−4π²x≈+140 m s⁻².
All three have period T; v is π/2 out of phase with x and a is π out of phase.
NCERT Example 13.5
Position, speed and acceleration
Question. For x=5 cos(2πt+π/4) m, find displacement, velocity and acceleration at t=1.5 s.
Step 1. The phase is 2π(1.5)+π/4=13π/4; cos(13π/4)=−√2/2.
Step 2. v=−10π sin(13π/4)=10π/√2.
Step 3. a=−(2π)²x.
Answer: x=−3.54 m; v≈+22.2 m s⁻¹ (speed 22.2 m s⁻¹); a≈+140 m s⁻².
13.6
Restoring force and spring oscillators
A force proportional to displacement and directed toward equilibrium produces SHM.
Newton’s second law gives F=ma=−mω²x. Writing k=mω² gives Hooke’s restoring law F=−kx; therefore ω=√(k/m) and T=2π√(m/k). This is a linear oscillator.
For two identical springs k attached on opposite sides of a block, each exerts −kx after displacement x; net force is −2kx. Thus ω=√(2k/m) and T=2π√(m/(2k)).
Example 13.6: the two-spring arrangement has combined restoring force −2kx; identify the effective spring constant 2k and apply the standard period formula.
Forces need not be exactly linear far from equilibrium. Expanding a restoring force may include higher powers (x², x³, …); if the displacement is small, the linear term often dominates.
NCERT Example 13.6
Block between two springs
Question. Two identical springs, each of force constant k, are fixed on opposite sides of a block of mass m. Find the period.
Step 1. A displacement x makes both spring forces point toward equilibrium; their sum is F=−kx−kx=−2kx.
Step 2. Thus k_eff=2k and mω²=2k.
Answer: T=2π√(m/(2k)).
13.7
Energy in SHM
Without friction, energy shifts between kinetic and potential forms while total mechanical energy stays fixed.
With x=A cos(ωt+φ), v=−ωA sin(ωt+φ), and k=mω²: U=½kx²=½kA²cos²(ωt+φ); K=½mv²=½kA²sin²(ωt+φ).
Adding and using sin²θ+cos²θ=1 proves E=K+U=½kA², independent of time. This is conservation of mechanical energy. K and U are non-negative and each repeats after T/2.
At equilibrium x=0, U=0 and K=E. At x=±A, v=0, K=0 and U=E. At any intermediate position, K=½k(A²−x²).
Example 13.7: m=1 kg, k=50 N m⁻¹, A=0.10 m and x=0.05 m. E=½kA²=0.25 J; U=½kx²=0.0625 J; K=E−U=0.1875 J.
Kinetic and potential energy exchange twice per cycle; their sum remains ½kA² without friction.
NCERT Example 13.7
Kinetic and potential energy
Question. A 1 kg oscillator has k=50 N m⁻¹ and amplitude 0.10 m. Find K and U at x=0.05 m.
Step 1. Total energy E=½kA²=½(50)(0.10)²=0.25 J.
Step 2. Potential energy U=½kx²=½(50)(0.05)²=0.0625 J.
Step 3. K=E−U.
Answer: K=0.1875 J≈0.19 J; U=0.0625 J; E=0.25 J.
13.8
The simple pendulum
A simple pendulum is approximately an angular harmonic oscillator only for small swings.
For a bob of mass m on a light inextensible string of length L, gravity’s tangential component gives restoring torque τ=−mgL sinθ. The radial tension/gravity components produce no torque about the pivot.
Rotational Newton’s law Iα=τ gives α=−(mgL/I)sinθ. For small θ in radians, sinθ≈θ, so α=−(mgL/I)θ: this is angular SHM with ω=√(mgL/I).
For a point bob, I=mL²; hence ω=√(g/L), T=2π√(L/g). The period is independent of bob mass and (approximately) amplitude. The approximation is good for small angles; at large amplitudes the motion is not exactly SHM.
Example 13.8: a seconds-ticking pendulum has T=2 s. L=gT²/(4π²)=9.8×4/(4π²)≈1.0 m.
For small θ in radians, sin θ≈θ, so angular acceleration is proportional and opposite to θ.
NCERT Example 13.8
Length of a seconds pendulum
Question. What is the length of a simple pendulum that ticks seconds?
Step 1. “Ticks seconds” means a complete oscillation takes T=2 s.
Step 2. From T=2π√(L/g), L=gT²/(4π²).
Step 3. With g=9.8 m s⁻², L=9.8×4/(4π²).
Answer: L≈0.993 m, about 1.0 m.
13
Points to ponder
The least repeating time is T; motion repeats after integer multiples nT.
Not every periodic motion is SHM: SHM requires restoring force proportional to displacement and opposite in direction.
A two-dimensional force −mω²r produces uniform circular motion for equal perpendicular amplitudes and a π/2 phase difference.
For a fixed ω, two initial conditions determine SHM: x and v, or amplitude and phase, or energy and phase.
Given amplitude/energy, the initial position and velocity determine the phase.
Two SHMs with arbitrary amplitudes and phases need not sum to periodic motion; a common period exists when frequency ratios are rational. A periodic function can be represented as a Fourier sum.
The ideal SHM period does not depend on amplitude, energy or phase.
A simple pendulum is SHM only at small angular displacement.
x=A cosωt+B sinωt, x=A cos(ωt+α), and x=B sin(ωt+β) are equivalent forms. Damped motion is only approximately SHM for intervals much shorter than 2m/b.
MODEL
Mass on a spring
The moving block, spring force and live x, v and a values show the restoring-force rule. Change amplitude and frequency to compare displacement with acceleration.
x = 0.00 A · v = 0.00 ωA · a = 0.00 ω²A
30 QUESTIONS · ALL SECTIONS
Quiz
Choose an answer to see why it is right.
0 / 30 AnsweredCorrect: 0 (0%)
1. The least repeat time of periodic motion is called:
The period T is the smallest positive repeat interval.
2. A periodic motion has frequency 4 Hz. Its period is:
T=1/ν.
3. Which statement is correct?
Oscillatory motion repeats to and fro; periodic motion can also be circular.
4. For x=A cos(ωt+φ), the phase at time t is:
The argument of the cosine is the phase.
5. The relation between period and angular frequency is:
ω=2π/T.
6. A cosωt+B sinωt has amplitude:
Combine as one sinusoid by matching coefficients.
7. For fixed ω, complete initial data for SHM can be:
Two independent initial conditions set amplitude and phase.
8. Which sum is guaranteed periodic?
Rational ratios allow a common repeat time.
9. The x-projection of a point on a circle of radius A is:
Projection produces the SHM displacement equation.
10. The x and y projections of uniform circular motion differ in phase by:
Cosine and sine components are π/2 apart.
11. A circular point begins at φ=π/3 and rotates anticlockwise with ω=2 rad s⁻¹. Its x projection is:
The phase advances as φ+ωt.
12. For x=A cos(ωt+φ), velocity is:
Differentiate x with respect to time.
13. Maximum speed in SHM is:
|v| reaches ωA at equilibrium.
14. At x=0 in SHM, acceleration is:
a=−ω²x.
15. Acceleration and displacement are:
a=−ω²x.
16. The defining linear acceleration relation for SHM is:
It is proportional to x and opposite in direction.
17. For a mass–spring oscillator, angular frequency is:
Set mω²=k.
18. Two identical springs act in parallel on a mass. Effective spring constant is:
Their restoring forces add.
19. If a restoring force contains dominant −kx near equilibrium, small oscillations are:
Higher-order force terms are small sufficiently near equilibrium.
20. Total mechanical energy of an ideal spring oscillator is:
At an endpoint all energy is potential: E=½kA².
21. At the mean position, an ideal oscillator has:
x=0, so U=0 and K=E.
22. The period of K and U separately is:
Squaring sine/cosine halves the period.
23. The restoring torque on a simple pendulum bob is:
29. In a U-tube with total moving liquid length L, a level displacement y gives acceleration:
The two arms create a 2y height difference.
30. In a car turning at speed v on radius R, effective gravity magnitude is:
Gravity and horizontal centripetal acceleration are perpendicular.
NCERT · EXERCISES 13.1–13.18
Exercises
Try each textbook question before opening its worked solution.
Ex 13.1NCERT
13.1Which 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.
Step 1. One return trip alone is not a repeating motion.
Step 2. A released bar magnet oscillates about the N-S direction (approximately periodic, with damping).
Step 3. Steady molecular rotation repeats after each revolution.
Step 4. An arrow’s flight does not repeat.
Answer: (b) and (c) are periodic in the idealised descriptions; (a) and (d) are not.
Ex 13.2NCERT
13.2Which examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) rotation of Earth about its axis; (b) an oscillating mercury column in a U-tube; (c) a ball bearing in a smooth curved bowl released slightly above its lowest point; (d) general vibrations of a polyatomic molecule about equilibrium.
Step 1. Earth’s rotation is periodic, not to-and-fro SHM.
Step 2. For a small level difference, hydrostatic restoring force is proportional to displacement, so the U-tube oscillation is nearly SHM.
Step 3. Near the bowl’s bottom, the restoring force is approximately linear in displacement, so motion is nearly SHM.
Step 4. General polyatomic vibrations can combine modes and need not be a single SHM; near equilibrium, normal modes are harmonic in the small-amplitude approximation.
Answer: (a) periodic, not SHM; (b), (c) nearly SHM; (d) generally a combination of normal-mode oscillations (periodic only for commensurate mode frequencies).
Ex 13.3NCERT
13.3Fig. 13.18 depicts four x-t plots for linear motion of a particle. Which plots represent periodic motion? What is the period in each periodic case?
Step 1. A motion is periodic only if its full x–t pattern repeats after the same least time. Bounded motion alone is not enough.
Step 2. In Fig. 13.18, panels (b) and (d) repeat after 2 s; panels (a) and (c) do not repeat.
Answer: (b) and (d) are periodic, each with T=2 s. (a) and (c) are non-periodic.
Ex 13.4NCERT
13.4Which functions represent (a) SHM, (b) periodic but not SHM, and (c) non-periodic motion? Give each period (ω>0): (a) sinωt−cosωt; (b) sin³ωt; (c) 3cos(π/4−2ωt); (d) cosωt+cos3ωt+cos5ωt; (e) exp(−ω²t²); (f) 1+ωt+ω²t².
Step 1. (a) Combine into √2 sin(ωt−π/4), so it is SHM.
Step 2. (b) sin³u=(3sin u−sin3u)/4: periodic but not a single sinusoid. Its least period is 2π/ω.
Step 3. (c) One sinusoid with angular frequency 2ω.
Step 4. (d) Harmonics ω, 3ω, 5ω share least period 2π/ω but the sum is not a single SHM.
Step 5. (e) A decaying Gaussian and (f) a polynomial do not repeat.
Answer: SHM: (a), T=2π/ω; (c), T=π/ω. Periodic but not SHM: (b), T=2π/ω; (d), T=2π/ω. Non-periodic: (e), (f).
Ex 13.5NCERT
13.5A particle is in linear SHM between A and B, 10 cm apart. Take A→B positive. Give signs of v, a and F: (a) at A; (b) at B; (c) midpoint going towards A; (d) 2 cm from B going towards A; (e) 3 cm from A going towards B; (f) 4 cm from B going towards A.
Step 1. SHM requires a=−ω²x with a positive constant ω².
Step 2. Only a relationship linear in x and with a negative coefficient has this form.
Answer: Only (c): ω=√10 in the corresponding units.
Ex 13.7NCERT
13.7For x(t)=A cos(ωt+φ), initial position is 1 cm and initial velocity is ω cm s⁻¹. Find amplitude and initial phase. Here ω=π s⁻¹. Repeat for x=B sin(ωt+α).
Step 1. The 50 kg mark corresponds to a force 50g over extension 0.20 m, so k=50g/0.20=2450 N m⁻¹.
Step 2. From T=2π√(m/k), m=kT²/(4π²).
Step 3. Weight W=mg.
Answer: m≈22.4 kg, so W≈219 N (about 22.4 kg-wt).
Ex 13.9NCERT
13.9A spring with force constant 1200 N m⁻¹ is mounted on a horizontal table and carries a 3 kg mass. It is pulled sideways by 2.0 cm and released. Find (i) frequency, (ii) maximum acceleration, and (iii) maximum speed.
Answer: (i) f≈3.18 Hz; (ii) amax=8.0 m s⁻²; (iii) vmax=0.40 m s⁻¹.
Ex 13.10NCERT
13.10In Exercise 13.9, take the unstretched spring position as x=0 and left-to-right as positive. Give x(t) if at t=0 the mass is (a) at the mean position, (b) at maximum stretch, (c) at maximum compression. How do frequency, amplitude and phase differ?
Step 1. For Exercise 13.9, A=0.020 m and ω=20 rad s⁻¹.
Step 2. At the mean position, x(0)=0; without the direction of initial velocity, both signs are possible.
Step 3. At maximum stretch x(0)=+A; at maximum compression x(0)=−A.
Answer: (a) x=±0.020 sin(20t), with sign set by initial direction; (b) x=0.020 cos(20t); (c) x=−0.020 cos(20t). All have the same amplitude and frequency; only phase (and for (a), direction) changes.
Ex 13.11NCERT
13.11Figures 13.20 show two circular motions with indicated radius, period, initial position and sense of revolution. Obtain the corresponding SHM of the x-projection of P in each case.
Step 1. For (a), R=3 cm, T=2 s, the point starts at the bottom (φ=−π/2) and rotates clockwise, so ω=π rad s⁻¹ and x=3cos(πt−π/2).
Step 2. For (b), R=2 m, T=4 s, the point starts at the left (φ=π) and rotates anticlockwise, so x=2cos(πt/2+π).
Answer: (a) x=−3 sin(πt) cm. (b) x=−2 cos(πt/2) m.
Ex 13.12NCERT
13.12Plot the reference circle for each SHM; indicate initial position, radius and angular speed (x in cm, t in s; take anticlockwise rotation): (a) x=−2sin(3t+π/3); (b) x=cos(π/6−t); (c) x=3sin(2πt+π/4); (d) x=2cosπt.
Step 1. Write each as R cos(ωt+φ), with R≥0. Then the rotating point begins at angle φ, radius R; anticlockwise angular speed is ω.
Step 2. (a) −sin u=cos(u+π/2); (b) cos(π/6−t)=cos(t−π/6); (c) sin u=cos(u−π/2).
Answer: (a) R=2 cm, φ=5π/6, ω=3 s⁻¹; (b) R=1 cm, φ=−π/6, ω=1 s⁻¹; (c) R=3 cm, φ=−π/4, ω=2π s⁻¹; (d) R=2 cm, φ=0, ω=π s⁻¹.
Ex 13.13NCERT
13.13A spring of force constant k is clamped at one end and carries mass m at the other. A force F stretches it. The same spring is then free at both ends with equal masses m and each end pulled by F. (a) Find maximum extension in both cases. (b) Find oscillation period after release.
13.16A pendulum of length l and bob mass M is in a car moving at speed v around a circular track of radius R. It makes small radial oscillations about equilibrium. Find its period.
Step 1. In the car frame, the effective gravity combines downward g and outward centrifugal acceleration v²/R.
Step 2. Its magnitude is geff=√(g²+v⁴/R²). Small angular oscillations obey ω=√(geff/l).
Answer: T=2π√(l/√(g²+v⁴/R²)). The bob mass cancels.
Ex 13.17NCERT
13.17A cylindrical cork of density ρ, base area A and height h floats in a liquid of density ρl. It is depressed slightly and released. Show it oscillates vertically in SHM and find the period (ignore viscosity).
Step 1. For a small downward displacement x, extra displaced liquid volume is Ax, so the buoyancy increase is ρlAgx upward: F=−ρlAgx.
Step 2. The cork mass is m=ρAh. Newton’s law gives x¨=−(ρl g/(ρh))x.
Step 3. Compare with x¨=−ω²x, then use T=2π/ω.
Answer: SHM with ω=√(ρl g/(ρh)) and T=2π√(ρh/(ρl g)).
Ex 13.18NCERT
13.18One end of a U-tube containing mercury is connected to a suction pump and the other to atmosphere. A small pressure difference is maintained between the columns. Show that, when the pump is removed, the mercury column executes SHM.