Oscillations

What to remember

• In simple harmonic motion (SHM) the restoring force and the acceleration are proportional to the displacement and point towards the mean position: F = −kx, a = −ω²x, with ω = √(k/m).
• Displacement, velocity and acceleration are x = A cos(ωt + φ), v = −Aω sin(ωt + φ) and a = −Aω² cos(ωt + φ); the speed at displacement x is v = ω√(A² − x²).
• The total energy of SHM is constant: E = ½kA² = ½mω²A². Kinetic energy is ½k(A² − x²) and potential energy is ½kx².
• A loaded spring has T = 2π√(m/k) and a simple pendulum has T = 2π√(l/g). The spring period does not depend on g; the pendulum period does not depend on the mass of the bob.

Periodic and oscillatory motion

A motion that repeats itself after equal intervals of time is called periodic motion. The smallest interval after which the motion repeats is the period T (unit: second, s). The number of repetitions in one second is the frequency f = 1/T (unit: hertz, Hz = s⁻¹). The angular frequency is ω = 2πf = 2π/T (unit: rad s⁻¹).

An oscillatory motion is a periodic motion in which a body moves back and forth about a fixed point, called the mean (equilibrium) position. Every oscillatory motion is periodic, but not every periodic motion is oscillatory. For example, the motion of the Earth around the Sun is periodic but not oscillatory.

The displacement of an oscillating particle can be written as a function of time. The simplest form is a sine or cosine function of time. Any periodic function can be built from sine and cosine functions of different frequencies; this is why the sine and cosine functions are so important.

Simple harmonic motion and its equation

A particle is in SHM when its displacement from the mean position changes with time as

x(t) = A cos(ωt + φ)

Symbol	Name	Meaning	Unit
x	displacement	distance from the mean position at time t	m
A	amplitude	maximum magnitude of displacement	m
ω	angular frequency	2π/T = 2πf	rad s⁻¹
(ωt + φ)	phase	decides the state of motion at time t	rad
φ	phase constant (initial phase)	phase at t = 0	rad

The amplitude A and the phase constant φ are fixed by the initial position and initial velocity. The period is T = 2π/ω. The value of x repeats when the phase increases by 2π. Two particles with the same ω and different φ have a phase difference equal to the difference of their phase constants.

SHM and uniform circular motion

Take a particle moving on a circle of radius A with constant angular speed ω. Its projection (shadow) on any diameter performs SHM. If the radius vector makes an angle (ωt + φ) with the x-axis, the projection on the x-axis is x = A cos(ωt + φ). So the amplitude of the SHM is the radius of the circle and the angular frequency of the SHM is the angular speed of the circular motion. The speed of the circular motion is Aω, which is the maximum speed of the SHM.

Velocity and acceleration in SHM

Differentiating x = A cos(ωt + φ):

• Velocity: v = −Aω sin(ωt + φ).
• Acceleration: a = −Aω² cos(ωt + φ) = −ω²x.
• Speed at displacement x: v = ω√(A² − x²).

Position	x	Speed	Acceleration	Force
Mean position	0	maximum, Aω	zero	zero
Extreme position	±A	zero	maximum, Aω²	maximum, kA
Any x	x	ω√(A² − x²)	−ω²x	−kx

Velocity is ahead of displacement by a phase of π/2. Acceleration is ahead of velocity by π/2 and is opposite in phase (π) to displacement. The graph of acceleration against displacement is a straight line through the origin with a negative slope −ω².

Force law for SHM

From F = ma = −mω²x we get F = −kx, where k = mω² is the force constant (unit: N m⁻¹). The force is linear in x and always directed towards the mean position, so it is a restoring force. This gives ω = √(k/m) and T = 2π√(m/k). A body obeys SHM if and only if the restoring force is directly proportional to the displacement.

Energy in SHM

For a particle of mass m in SHM:

• Kinetic energy: K = ½mv² = ½mω²A² sin²(ωt + φ) = ½k(A² − x²).
• Potential energy: U = ½kx² = ½mω²A² cos²(ωt + φ).
• Total energy: E = K + U = ½kA², a constant that does not depend on time or position.

Graphs in words:

• U against x is an upward-opening parabola with its minimum (zero) at x = 0 and the value ½kA² at x = ±A.
• K against x is a downward-opening parabola with its maximum ½kA² at x = 0 and zero at x = ±A.
• E against x is a horizontal straight line at the height ½kA². The K and U curves cross it at the points where K + U = E.
• K and U both oscillate with frequency 2f, that is, twice the frequency of the displacement. The time average of each over one period is E/2.

Position	K	U
x = 0	E (maximum)	0
x = ±A/2	3E/4	E/4
x = ±A/√2	E/2	E/2
x = ±A	0	E (maximum)

Oscillations of a loaded spring

A mass m attached to a spring of force constant k, on a smooth horizontal surface, is pulled and released. The spring force F = −kx makes the motion SHM with

ω = √(k/m), T = 2π√(m/k), f = (1/2π)√(k/m).

A larger mass gives a longer period; a stiffer spring gives a shorter period. A mass hanging on a vertical spring also has T = 2π√(m/k), because gravity only shifts the equilibrium position. If the static extension under the weight is Δl, then k = mg/Δl and T = 2π√(Δl/g).

Combination of springs

Arrangement	Effective force constant	Result
Two springs in parallel (same extension)	k = k₁ + k₂	stiffer than either spring
Two springs in series (same force)	1/k = 1/k₁ + 1/k₂, so k = k₁k₂/(k₁ + k₂)	softer than either spring

For two identical springs of constant k: parallel gives 2k, and series gives k/2.

The simple pendulum

A simple pendulum is a point mass (bob) hung from a fixed point by a light, inextensible string of length l. For a small angle θ (in radian) from the vertical, the restoring tangential force is mg sin θ ≈ mgθ. The arc displacement is s = lθ, so the force is F = −(mg/l)s. This has the form F = −ks with k = mg/l, hence

ω = √(g/l), T = 2π√(l/g).

The period depends on the length l and on g. It does not depend on the mass of the bob, and for small amplitudes it does not depend on the amplitude. The length l is measured from the point of suspension to the centre of the bob. A seconds pendulum has a period of 2 s (one second for each swing). Its length at g = 9.8 m s⁻² is l = gT²/4π² = g/π², which is about 0.99 m, nearly 1 m. Since T ∝ √l, a small increase of length makes a pendulum clock run slow.

Damped and forced oscillations, resonance

In practice a free oscillation dies down because of friction and air drag. These are damping forces. If the damping force is F_d = −bv (b is the damping constant), the amplitude falls exponentially with time:

x(t) = A e^(−bt/2m) cos(ω′t + φ), with ω′ = √(k/m − b²/4m²).

The mechanical energy decreases with time because it is lost to the surroundings. The damped angular frequency ω′ is slightly less than ω = √(k/m) for small damping, so the period is slightly longer.

A free (natural) oscillation takes place at the natural angular frequency ω without any external periodic force. A forced oscillation is the oscillation of a system under an external periodic force of angular frequency ω_d. After the initial transient dies out, the system oscillates at the driving frequency ω_d, not at its natural frequency. The amplitude of the forced oscillation is large when ω_d is close to ω. Resonance occurs when ω_d = ω; the amplitude is then maximum. With damping the maximum amplitude is finite; the less the damping, the sharper and taller the resonance peak.

Formula summary

Quantity	Formula	Unit
Frequency, period	f = 1/T, ω = 2π/T	Hz, s, rad s⁻¹
Displacement	x = A cos(ωt + φ)	m
Velocity	v = −Aω sin(ωt + φ); v = ω√(A² − x²)	m s⁻¹
Maximum speed	v_max = Aω	m s⁻¹
Acceleration	a = −ω²x; a_max = Aω²	m s⁻²
Force law	F = −kx	N
Kinetic energy	½k(A² − x²)	J
Potential energy	½kx²	J
Total energy	½kA² = ½mω²A²	J
Spring-mass system	T = 2π√(m/k)	s
Springs	parallel k₁ + k₂; series k₁k₂/(k₁ + k₂)	N m⁻¹
Simple pendulum	T = 2π√(l/g)	s
Seconds pendulum	T = 2 s, l ≈ 0.99 m	s, m

Worked examples

Example 1. A block of mass 0.2 kg is attached to a spring of constant 80 N m⁻¹. Find ω and T.

ω = √(k/m) = √(80/0.2) = √400 = 20 rad s⁻¹. T = 2π/ω = 2π/20 = π/10 s, about 0.31 s.

Example 2. A particle has amplitude 0.05 m and ω = 10 rad s⁻¹. Find its speed at x = 0.03 m and its maximum acceleration.

v = ω√(A² − x²) = 10 × √(0.0025 − 0.0009) = 10 × 0.04 = 0.4 m s⁻¹. a_max = Aω² = 0.05 × 100 = 5 m s⁻².

Example 3. A simple pendulum of length 2.5 m oscillates where g = 10 m s⁻². Find its period.

T = 2π√(l/g) = 2π√(2.5/10) = 2π × 0.5 = π s, about 3.14 s.

Example 4. A spring has total energy 8 J. Find K and U when the displacement is half of the amplitude.

U = ½k(A/2)² = E/4 = 2 J. K = E − U = 6 J.

NEET traps

• 1. Every oscillatory motion is periodic, but not every periodic motion is SHM. SHM needs a force proportional to displacement.
• 2. Velocity is maximum at the mean position; acceleration is maximum at the extreme position. At x = 0 the acceleration is zero; at x = ±A the speed is zero.
• 3. Velocity of a particle in SHM is not always directed towards the mean position; only the acceleration is.
• 4. The period of a spring-mass system depends on m and k, not on g and not on the amplitude. The period of a simple pendulum does not depend on the mass.
• 5. Kinetic and potential energy each oscillate with frequency 2f, not f. The total energy does not oscillate at all.
• 6. Kinetic energy equals potential energy at x = ±A/√2, not at x = ±A/2. At x = A/2 the kinetic energy is 3E/4.
• 7. Springs in parallel add their constants; springs in series add the reciprocals. Series makes the system softer.
• 8. The effective length of a pendulum is measured up to the centre of the bob. T ∝ √l, so a length four times as large gives a period twice as large.
• 9. At resonance the amplitude is maximum but finite when damping is present; it is not infinite.
• 10. In the damped case the amplitude falls, but the period is slightly longer, not shorter, than the undamped period.

One-liners

• 1. Period T = 1/f and ω = 2π/T.
• 2. In x = A cos(ωt + φ), the quantity (ωt + φ) is the phase and φ is the phase constant.
• 3. The projection of uniform circular motion on a diameter is SHM.
• 4. The maximum speed in SHM is Aω; the maximum acceleration is Aω².
• 5. a = −ω²x: the acceleration is proportional to displacement and opposite to it.
• 6. The graph of a against x is a straight line through the origin with negative slope.
• 7. Total energy E = ½kA² = ½mω²A² is constant and proportional to A².
• 8. Kinetic energy is maximum at the mean position; potential energy is maximum at the extreme position.
• 9. K = U at x = ±A/√2.
• 10. Time-average kinetic energy equals time-average potential energy, each E/2.
• 11. Spring-mass period T = 2π√(m/k).
• 12. Parallel springs: k = k₁ + k₂. Series springs: 1/k = 1/k₁ + 1/k₂.
• 13. Simple pendulum period T = 2π√(l/g), valid for small angles.
• 14. A seconds pendulum has T = 2 s and length about 0.99 m.
• 15. Resonance happens when the driving frequency equals the natural frequency of the oscillator.

Practice questions

1. A particle executes SHM with a frequency of 5 Hz. What is its time period?

   1. 5 s
   2. 0.5 s
   3. 10π s
   4. 0.2 s
   Answer

   D. 0.2 s

   T = 1/f = 1/5 = 0.2 s.

2. The displacement of a particle in SHM is x = 0.1 cos(4πt + π/3), with x in metre and t in second. What is its time period?

   1. 0.5 s
   2. 4π s
   3. 2 s
   4. 0.25 s
   Answer

   A. 0.5 s

   ω = 4π rad s⁻¹, so T = 2π/ω = 2π/4π = 0.5 s. (Check: f = 2 Hz, T = 1/2 s.)

3. A block of mass 0.5 kg is attached to a horizontal spring of force constant 200 N m⁻¹ and oscillates on a smooth surface. What is its time period?

   1. π/20 s
   2. π/10 s
   3. π/5 s
   4. 10π s
   Answer

   B. π/10 s

   T = 2π√(m/k) = 2π√(0.5/200) = 2π × 0.05 = π/10 s. Check: ω = √(200/0.5) = 20 rad s⁻¹ and T = 2π/20.

4. A particle in SHM has amplitude 0.2 m and angular frequency 5 rad s⁻¹. What is the magnitude of its maximum acceleration?

   1. 0.04 m s⁻²
   2. 1 m s⁻²
   3. 5 m s⁻²
   4. 25 m s⁻²
   Answer

   C. 5 m s⁻²

   a_max = Aω² = 0.2 × 25 = 5 m s⁻². Check: v_max = Aω = 1 m s⁻¹ and a_max = v_max × ω = 5.

5. A particle in SHM has amplitude 0.13 m and angular frequency 10 rad s⁻¹. What is its speed when the displacement is 0.05 m?

   1. 1.3 m s⁻¹
   2. 1.8 m s⁻¹
   3. 1.2 m s⁻¹
   4. 0.5 m s⁻¹
   Answer

   C. 1.2 m s⁻¹

   v = ω√(A² − x²) = 10 × √(0.0169 − 0.0025) = 10 × 0.12 = 1.2 m s⁻¹.

6. Two identical springs, each of force constant 100 N m⁻¹, are connected in series. A mass of 2 kg is attached to the combination. What is the time period of vertical or horizontal oscillation?

   1. 4π/5 s
   2. π/10 s
   3. π/5 s
   4. 2π/5 s
   Answer

   D. 2π/5 s

   Series: k = 100 × 100/200 = 50 N m⁻¹. T = 2π√(2/50) = 2π/5 s. (Parallel, k = 200, would give π/5 s.)

7. What is the time period of a seconds pendulum?

   1. 2 s
   2. π s
   3. 1 s
   4. 0.5 s
   Answer

   A. 2 s

   A seconds pendulum takes 1 s for each swing, so its full period is 2 s.

8. A simple pendulum of length 1 m oscillates at a place where g = π² m s⁻². What is its time period?

   1. 2π s
   2. 2 s
   3. 1 s
   4. π s
   Answer

   B. 2 s

   T = 2π√(l/g) = 2π√(1/π²) = 2 s.

9. In SHM the velocity of the particle is zero at

   1. the mean position
   2. the extreme positions
   3. a displacement of A/2 only
   4. all positions
   Answer

   B. the extreme positions

   v = ω√(A² − x²) is zero when x = ±A, the extreme positions; it is maximum at x = 0.

10. A particle executes SHM with frequency f. The frequency with which its kinetic energy oscillates is

    1. f/2
    2. f
    3. 4f
    4. 2f
    Answer

    D. 2f

    K = ½mω²A² sin²(ωt + φ) repeats every half period, so its frequency is 2f.

11. The graph of acceleration against displacement for a particle in SHM is

    1. a straight line through the origin with negative slope
    2. a horizontal straight line
    3. a parabola opening upward
    4. a straight line through the origin with positive slope
    Answer

    A. a straight line through the origin with negative slope

    a = −ω²x is linear in x, passes through the origin and has slope −ω².

12. At what displacement from the mean position are the kinetic energy and the potential energy of a particle in SHM (amplitude A) equal?

    1. A/2
    2. A/4
    3. A/√2
    4. A/3
    Answer

    C. A/√2

    ½k(A² − x²) = ½kx² gives x² = A²/2, so x = A/√2.

13. The length of a simple pendulum is made four times. Its time period becomes

    1. 2 times the original
    2. 4 times the original
    3. 1/2 of the original
    4. 16 times the original
    Answer

    A. 2 times the original

    T ∝ √l, so T′/T = √4 = 2.

14. In a damped oscillation where the damping force is proportional to velocity, the amplitude of the oscillation

    1. decreases linearly with time
    2. increases exponentially with time
    3. decreases exponentially with time
    4. remains constant
    Answer

    C. decreases exponentially with time

    The amplitude is A e^(−bt/2m), an exponential decay.

15. Resonance occurs in a forced oscillation when

    1. the driving frequency is double the natural frequency
    2. the driving force is zero
    3. the damping is maximum
    4. the driving frequency equals the natural frequency
    Answer

    D. the driving frequency equals the natural frequency

    The amplitude is maximum when the angular frequency of the driving force equals the natural angular frequency.

16. A particle starts from the mean position and executes SHM of time period 12 s. What is the time taken to reach half of its amplitude?

    1. 1.5 s
    2. 1 s
    3. 3 s
    4. 2 s
    Answer

    B. 1 s

    x = A sin ωt = A/2 gives ωt = π/6, so t = T/12 = 1 s. Check: phase π/6 is 1/12 of 2π.

17. Given below are two statements: Statement I: In SHM the acceleration of the particle is always directed towards the mean position. Statement II: In SHM the velocity of the particle is always directed towards the mean position.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    a = −ω²x always points towards x = 0. The velocity points away from the mean position during the outward half of the motion.

18. Given below are two statements: Statement I: The total energy of a particle in SHM does not depend on its displacement. Statement II: The total energy of a particle in SHM is proportional to the square of the amplitude.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    E = ½kA² is constant and proportional to A².

19. Given below are two statements: Statement I: The time period of a simple pendulum, for small oscillations, depends on the mass of its bob. Statement II: The time period of a mass oscillating on a vertical spring does not depend on the value of g.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    T = 2π√(l/g) has no mass; T = 2π√(m/k) has no g. Statement I is wrong, Statement II is right.

20. Which of the following statements are correct for a particle in SHM? 1. The acceleration is zero at the mean position. 2. The speed is maximum at the extreme position. 3. The potential energy is maximum at the extreme position.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    B. 1 and 3 only

    a = −ω²x is zero at x = 0 (1 true). Speed is zero at the extremes (2 false). U = ½kx² is maximum at x = ±A (3 true).

21. Given below are two statements: Statement I: When two springs of constants k₁ and k₂ are connected in parallel, the effective force constant is k₁ + k₂. Statement II: When two springs are connected in series, the effective force constant is greater than the larger of the two constants.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Parallel gives k₁ + k₂. Series gives k₁k₂/(k₁ + k₂), which is smaller than either constant.

22. Given below are two statements: Statement I: At resonance the amplitude of a forced oscillation is maximum. Statement II: At resonance the amplitude becomes infinite whatever the damping.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Resonance gives the maximum amplitude, but damping keeps it finite.

23. Given below are two statements: Statement I: Every periodic motion is simple harmonic motion. Statement II: Every simple harmonic motion is a periodic motion.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    SHM repeats after a fixed period, so it is periodic. A periodic motion need not obey F = −kx, so it need not be SHM.

24. Which of the following statements are correct about a seconds pendulum? 1. Its time period is 2 s. 2. Its length is about 0.99 m where g = 9.8 m s⁻². 3. Its frequency is 2 Hz.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    T = 2 s, so f = 0.5 Hz (3 false). l = gT²/4π² = g/π² ≈ 0.99 m (2 true).

25. Given below are two statements: Statement I: In a damped oscillation the mechanical energy of the oscillator decreases with time. Statement II: For small damping, the time period of the damped oscillation is slightly longer than that of the undamped oscillation.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Energy is lost to the surroundings. ω′ = √(k/m − b²/4m²) is slightly less than ω, so T′ is slightly longer.

26. Which of the following statements are correct? 1. The projection of uniform circular motion on a diameter is SHM. 2. The angular frequency of the SHM equals the angular speed of the circular motion. 3. The amplitude of the SHM equals the radius of the circle.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    D. 1, 2 and 3

    All three are correct: the shadow on a diameter is SHM, with ω equal to the angular speed and A equal to the radius.

27. Assertion (A): The acceleration of a particle in SHM is zero at the mean position. Reason (R): The acceleration in SHM is directly proportional to the displacement from the mean position.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    a = −ω²x vanishes at x = 0 because it is proportional to x.

28. Assertion (A): The total energy of a particle in SHM changes with its position. Reason (R): During SHM the kinetic energy and potential energy are converted into each other.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    The total energy is constant. R is true: K and U exchange but their sum stays ½kA².

29. Assertion (A): A block oscillating on a spring has the same time period on the Moon as on the Earth. Reason (R): The value of g on the Moon is different from that on the Earth.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    T = 2π√(m/k) does not contain g, so A is true. R is also true but does not explain A (g is simply not in the formula).

30. Assertion (A): If the mass of a block on a spring is made four times, its time period doubles. Reason (R): The time period of a spring-mass system is directly proportional to the mass.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    T ∝ √m, so four times the mass gives twice the period (A true). R is false because T is not proportional to m.

31. Assertion (A): The kinetic energy of a particle in SHM oscillates at twice the frequency of its displacement. Reason (R): The total energy of a particle in SHM is constant.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    K ∝ sin²(ωt + φ), which repeats at 2f. R is true but does not explain the doubling of frequency.

32. Assertion (A): The amplitude of a damped oscillator decreases exponentially with time. Reason (R): The damping force, proportional to velocity, takes mechanical energy out of the oscillator.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    x = A e^(−bt/2m) cos(ω′t + φ); the damping force dissipates energy, which reduces the amplitude.

33. Match List I with List II. List I: a. Angular frequency of a spring-mass system b. Angular frequency of a simple pendulum c. Maximum speed in SHM d. Maximum acceleration in SHM. List II: i. √(g/l) ii. √(k/m) iii. Aω iv. Aω².

    1. a-ii, b-i, c-iv, d-iii
    2. a-ii, b-i, c-iii, d-iv
    3. a-i, b-ii, c-iii, d-iv
    4. a-iii, b-iv, c-i, d-ii
    Answer

    B. a-ii, b-i, c-iii, d-iv

    ω = √(k/m) for a spring, √(g/l) for a pendulum; v_max = Aω; a_max = Aω².

34. Match List I with List II. List I: a. Mean position b. Extreme position c. Displacement x = A/√2 d. Displacement x = A/2. List II: i. Kinetic energy equals potential energy ii. Speed zero and potential energy maximum iii. Kinetic energy maximum iv. Kinetic energy is 3/4 of the total energy.

    1. a-iii, b-i, c-ii, d-iv
    2. a-iii, b-ii, c-i, d-iv
    3. a-iii, b-ii, c-iv, d-i
    4. a-ii, b-iii, c-i, d-iv
    Answer

    B. a-iii, b-ii, c-i, d-iv

    x = 0: K maximum. x = ±A: v = 0, U maximum. x = A/√2: K = U. x = A/2: U = E/4, K = 3E/4.

35. Match List I with List II. List I: a. x = cos(πt) b. x = cos(2πt) c. x = cos(4πt) d. x = cos(πt/2) (x in m, t in s). List II: i. Period 0.5 s ii. Period 1 s iii. Period 2 s iv. Period 4 s.

    1. a-ii, b-iii, c-i, d-iv
    2. a-iii, b-ii, c-iv, d-i
    3. a-iv, b-iii, c-ii, d-i
    4. a-iii, b-ii, c-i, d-iv
    Answer

    D. a-iii, b-ii, c-i, d-iv

    T = 2π/ω: ω = π gives 2 s; 2π gives 1 s; 4π gives 0.5 s; π/2 gives 4 s.

36. Match List I with List II. List I: a. Two springs, each of constant k, in series b. Two springs, each of constant k, in parallel c. Springs k and 3k in series d. Springs k and 3k in parallel. List II: i. 2k ii. 3k/4 iii. 4k iv. k/2.

    1. a-ii, b-iii, c-iv, d-i
    2. a-i, b-iv, c-iii, d-ii
    3. a-iv, b-i, c-ii, d-iii
    4. a-iv, b-i, c-iii, d-ii
    Answer

    C. a-iv, b-i, c-ii, d-iii

    Series: k·k/2k = k/2; parallel: 2k; series with k and 3k: 3k²/4k = 3k/4; parallel: 4k.

37. Match List I with List II. List I: a. Free oscillation b. Damped oscillation c. Forced oscillation d. Resonance. List II: i. Amplitude decays with time ii. Oscillation at natural frequency without external periodic force iii. Oscillation under an external periodic force iv. Maximum amplitude when driving frequency equals natural frequency.

    1. a-ii, b-i, c-iv, d-iii
    2. a-i, b-ii, c-iii, d-iv
    3. a-ii, b-i, c-iii, d-iv
    4. a-iii, b-i, c-ii, d-iv
    Answer

    C. a-ii, b-i, c-iii, d-iv

    Free: natural frequency, no external force. Damped: amplitude falls. Forced: periodic external force. Resonance: ω_d = ω gives maximum amplitude.

38. Match List I with List II. List I: a. K against x b. U against x c. E against x d. a against x. List II: i. Parabola opening upward with minimum at x = 0 ii. Straight line through the origin with negative slope iii. Horizontal straight line iv. Parabola opening downward with maximum at x = 0.

    1. a-iv, b-i, c-ii, d-iii
    2. a-iv, b-i, c-iii, d-ii
    3. a-i, b-iv, c-iii, d-ii
    4. a-iii, b-iv, c-i, d-ii
    Answer

    B. a-iv, b-i, c-iii, d-ii

    K = ½k(A² − x²) is a downward parabola; U = ½kx² is an upward parabola; E is constant; a = −ω²x is a line with negative slope.

39. A block of mass 1 kg attached to a spring of force constant 100 N m⁻¹ oscillates with amplitude 0.1 m. What is the total mechanical energy of the oscillator?

    1. 0.5 J
    2. 5 J
    3. 1 J
    4. 0.05 J
    Answer

    A. 0.5 J

    E = ½kA² = ½ × 100 × 0.01 = 0.5 J. Check: v_max = Aω = 1 m s⁻¹, so ½mv_max² = 0.5 J.

40. A block of mass 0.4 kg on a spring of force constant 40 N m⁻¹ oscillates with amplitude 0.05 m. What is its maximum speed?

    1. 0.25 m s⁻¹
    2. 0.05 m s⁻¹
    3. 5 m s⁻¹
    4. 0.5 m s⁻¹
    Answer

    D. 0.5 m s⁻¹

    ω = √(40/0.4) = 10 rad s⁻¹; v_max = Aω = 0.5 m s⁻¹. Energy check: ½mv² = ½kA² gives v = A√(k/m) = 0.5.

41. The length of a simple pendulum is increased by 44%. By what percentage does its time period increase?

    1. 20%
    2. 12%
    3. 44%
    4. 22%
    Answer

    A. 20%

    T ∝ √l, so T′/T = √1.44 = 1.2, an increase of 20%.

42. A mass of 1 kg stretches a vertical spring by 0.1 m in equilibrium. Take g = 10 m s⁻². What is the time period of small vertical oscillations of this mass?

    1. 2π/5 s
    2. π/5 s
    3. π/10 s
    4. π s
    Answer

    B. π/5 s

    k = mg/Δl = 100 N m⁻¹; T = 2π√(1/100) = π/5 s. Check: T = 2π√(Δl/g) = 2π√(0.01) = π/5 s.

43. A particle in SHM has a speed of 8 cm s⁻¹ at x = 3 cm and 6 cm s⁻¹ at x = 4 cm. What is its time period?

    1. 4π s
    2. π/2 s
    3. 2π s
    4. π s
    Answer

    D. π s

    64 = ω²(A² − 9) and 36 = ω²(A² − 16). Subtracting: 28 = 7ω², so ω = 2 rad s⁻¹ (A = 5 cm). T = 2π/2 = π s.

44. The total energy of a spring-mass oscillator is 8 J. What is its potential energy when the displacement is half of the amplitude?

    1. 6 J
    2. 1 J
    3. 2 J
    4. 4 J
    Answer

    C. 2 J

    U = ½k(A/2)² = E/4 = 2 J; then K = 6 J.

45. A block of mass 0.25 kg is attached to a spring of force constant 100π² N m⁻¹. At what frequency of an external periodic force will the block show resonance?

    1. 10 Hz
    2. 10π Hz
    3. 20 Hz
    4. 5 Hz
    Answer

    A. 10 Hz

    ω = √(k/m) = √(400π²) = 20π rad s⁻¹, so f = ω/2π = 10 Hz. Resonance occurs when the driving frequency equals the natural frequency.