In simple harmonic motion (SHM), the displacement x(t) of a particle from its equilibrium position is given by x(t) = A cos(ωt + φ), in which A is the amplitude of the displacement. The angular frequency ω is related to the period and frequency of the motion by ω = 2π/T = 2πν.
-- NCERT Class 11 Physics, Ch. 13, p. 272Displacement Function of Time
8 MCQs9-step worked example
Source: NCERT Oscillations and WavesOfficial key: NTA-verifiedLast updated: 8 Oct 2026
Try this first
In the chapter on oscillations, the NCERT uses the term "displacement" in a more general sense than the change of a position vector. Which description matches that usage?
- A.Only the straight-line distance of a body from its starting point
- B.Only the change in the position vector of a particle
- C.The change with time of any physical property under consideration, such as the angle of a pendulum from the vertical
- D.Only a quantity measured in metres
Tap to see the answer
Answer: C. The chapter extends displacement to any physical property that changes with time, for example the angle from the vertical for a simple pendulum or the voltage across a capacitor in an AC circuit (NCERT Class 11 Physics Chapter 13, page 261).
A is wrong: A is wrong because the chapter does not restrict displacement to distance from a start point; a block's displacement is measured from its equilibrium position, and the term covers non-position variables too (NCERT Class 11 Physics Chapter 13, page 261).
B is wrong: B is wrong because the chapter says explicitly that it uses displacement in a more general sense than the change of a position vector from the earlier kinematics chapter (NCERT Class 11 Physics Chapter 13, page 261).
D is wrong: D is wrong because the angle of a pendulum from the vertical and the voltage across a capacitor are both displacement variables, and neither is measured in metres (NCERT Class 11 Physics Chapter 13, page 261).
Displacement Function of Time, explained for NEET
A particle's displacement is x = 5 sin(πt + π/3) metres. What is its amplitude, and what is its period? Most of the marks here are lost before any physics starts: the unit is read wrongly, or ω is mistaken for the frequency.
In the oscillations chapter "displacement" is wider than "distance moved". It is the change with time of any physical property under consideration: the distance of a block from its equilibrium position, the angle a pendulum makes with the vertical, even the voltage across a capacitor in an AC circuit (NCERT Class 11 Physics Chapter 13, page 261). Like any displacement variable it takes both positive and negative values.
For periodic motion this displacement is a periodic function of time, and for SHM it is a sinusoidal one: x(t) = A cos(ωt + φ) (NCERT Class 11 Physics Chapter 13, page 262). Read the three constants off the expression:
- A, the amplitude, is the magnitude of the maximum displacement. It carries the unit of x, so a displacement written in metres has an amplitude in metres (page 263).
- ωt + φ is the phase, a time-dependent quantity; φ is its value at t = 0, the phase constant (page 263).
- ω, the angular frequency, comes from the period: ω = 2π/T, in rad s⁻¹ (page 264).
A sine form is not a different motion. Written as a cosine, a sine just carries a different phase constant. So for x = 5 sin(πt + π/3) m: A = 5 m, ω = π rad s⁻¹, T = 2π/π = 2 s.
Watch-out: read ω as the number multiplying t, never as the frequency. The frequency is ω/2π, and the period is its reciprocal. Check the unit of x before choosing between centimetres and metres.
How do you solve a Displacement Function of Time question? A worked example
- 1
Given
A particle's displacement is x = 8.0 sin(3.0π t + π/4) cm, with t in seconds.
- 2
Required
The amplitude, the period, the frequency, and the displacement at t = 0.
- 3
Concept
Displacement of a simple harmonic oscillator is a sinusoidal function of time. Compare the given expression with the standard form: the multiplier is the amplitude A, the coefficient of t is the angular frequency ω, and the constant added to ωt is the phase constant φ (NCERT Class 11 Physics Chapter 13, pages 262 to 264). A sine is a cosine with a shifted phase constant, so the same reading applies.
- 4
Formula
A = coefficient of the sine; ω = coefficient of t; T = 2π/ω; f = 1/T; x(0) = A sin φ for a sine form.
- 5
Substitution
A = 8.0 cm; ω = 3.0π rad s⁻¹; φ = π/4. T = 2π/(3.0π); f = 1/T; x(0) = 8.0 × sin(π/4).
- 6
Calculation
T = 2π/(3.0π) = 2/3.0 = 0.67 s. f = 1/0.67 s = 1.5 Hz. x(0) = 8.0 × 0.7071 = 5.7 cm. The 2, the π and the angle π/4 are exact and do not count towards the significant figures of the result.
- 7
Final answer
Amplitude 8.0 cm, period 0.67 s, frequency 1.5 Hz, and x = 5.7 cm at t = 0.
- 8
Common trap
Reading 3.0π as the frequency (it is ω in rad s⁻¹, which gives f = 1.5 Hz, not 3.0 Hz) and giving the amplitude in metres when x is given in centimetres.
- 9
Similar NEET-style question
A particle's displacement is x = 6.0 sin(5.0π t + π/6) cm. What are its period and its displacement at t = 0? (Answer: ω = 5.0π rad s⁻¹, so T = 2π/ω = 2π/(5.0π) = 0.40 s; x(0) = 6.0 × sin(π/6) = 6.0 × 0.50 = 3.0 cm.)
---
Can you answer these Displacement Function of Time MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the chapter on oscillations, the NCERT uses the term "displacement" in a more general sense than the change of a position vector. Which description matches that usage?
Show answer and why every option is right or wrong
Answer: C. The chapter extends displacement to any physical property that changes with time, for example the angle from the vertical for a simple pendulum or the voltage across a capacitor in an AC circuit (NCERT Class 11 Physics Chapter 13, page 261).
Why A is wrong: A is wrong because the chapter does not restrict displacement to distance from a start point; a block's displacement is measured from its equilibrium position, and the term covers non-position variables too (NCERT Class 11 Physics Chapter 13, page 261).
Why B is wrong: B is wrong because the chapter says explicitly that it uses displacement in a more general sense than the change of a position vector from the earlier kinematics chapter (NCERT Class 11 Physics Chapter 13, page 261).
Why D is wrong: D is wrong because the angle of a pendulum from the vertical and the voltage across a capacitor are both displacement variables, and neither is measured in metres (NCERT Class 11 Physics Chapter 13, page 261).
MCQ 2Easy RecallPractice
In the equation x(t) = A cos(ωt + φ) for simple harmonic motion, the time-dependent quantity (ωt + φ) is called
Show answer and why every option is right or wrong
Answer: A. The whole argument (ωt + φ) is the phase, and it fixes the state of motion (position and velocity) at time t (NCERT Class 11 Physics Chapter 13, page 263).
Why B is wrong: B is wrong because the phase constant (phase angle) is only φ, the value of the phase at t = 0, which is a fixed number and does not change with time (NCERT Class 11 Physics Chapter 13, page 263).
Why C is wrong: C is wrong because the amplitude A is the magnitude of the maximum displacement and is the multiplier of the cosine, not its argument (NCERT Class 11 Physics Chapter 13, page 263).
Why D is wrong: D is wrong because ω is the coefficient of t, a fixed constant related to the period by ω = 2π/T, not the full time-dependent argument (NCERT Class 11 Physics Chapter 13, page 264).
MCQ 3Direct ApplicationPractice
A particle's displacement is x = 3 cos(2πt) m, with t in seconds. What is its displacement at t = 0.25 s?
Show answer and why every option is right or wrong
Answer: D. At t = 0.25 s the argument is 2π × 0.25 = π/2, and cos(π/2) = 0, so x = 0. The particle is passing through its mean position (NCERT Class 11 Physics Chapter 13, page 262, for the form x = A cos(ωt + φ)).
Why A is wrong: A is wrong because x = 3 m is the value at t = 0, where cos 0 = 1. After a quarter of the period (T = 1 s here) the particle has moved to the mean position.
Why B is wrong: B is wrong because x = −3 m is the opposite extreme, reached after half a period (t = 0.5 s), not after a quarter.
Why C is wrong: C is wrong because 1.5 m would need cos(2πt) = 0.5, i.e. an argument of π/3; the argument at 0.25 s is π/2.
MCQ 4Easy RecallPractice
The angular frequency ω of a simple harmonic motion of period T is given by
Show answer and why every option is right or wrong
Answer: B. Requiring x(t) = x(t + T) makes the argument change by 2π over one period, so ωT = 2π and ω = 2π/T, with unit rad s⁻¹ (NCERT Class 11 Physics Chapter 13, page 264).
Why A is wrong: A is wrong because it is upside down; a longer period means a smaller ω, so T must be in the denominator.
Why C is wrong: C is wrong because it makes ω grow with the period. A longer period means a slower oscillation and a smaller ω.
Why D is wrong: D is wrong because 1/T is the frequency in hertz. The angular frequency is 2π times the frequency (NCERT Class 11 Physics Chapter 13, page 264).
MCQ 5CalculationPractice
A particle executes SHM with x = 10 cos(4πt + π/3) cm, with t in seconds. What are its frequency and its displacement at t = 0?
Show answer and why every option is right or wrong
Answer: A. Here ω = 4π rad s⁻¹, so the frequency is ω/2π = 2.0 Hz. At t = 0 the displacement is x = 10 cos(π/3) = 10 × 0.5 = 5.0 cm (NCERT Class 11 Physics Chapter 13, pages 262 to 264).
Why B is wrong: B is wrong because 4.0 Hz reads the 4 in 4π as the frequency. That number sits inside ω = 4π rad s⁻¹, and the frequency is ω/2π = 2.0 Hz.
Why C is wrong: C is wrong because 8.7 cm is 10 sin(π/3), the value of a sine instead of the cosine in this expression. The displacement at t = 0 is 10 cos(π/3) = 5.0 cm.
Why D is wrong: D is wrong on both counts: 4.0 Hz confuses ω with the frequency, and 1.0 × 10¹ cm is the amplitude, which is reached only if the phase constant is zero.
MCQ 6Direct ApplicationPractice
Two simple harmonic motions x₁ = A cos(ωt + φ) and x₂ = B cos(ωt + φ) have the same ω and the same φ, but A ≠ B. Which statement is correct?
Show answer and why every option is right or wrong
Answer: C. The period depends only on ω (T = 2π/ω), and the phase constant is φ, so both are the same. The amplitudes A and B set only the extreme displacements (NCERT Class 11 Physics Chapter 13, page 263, Fig. 13.7(a)).
Why A is wrong: A is wrong because the period follows from ω alone (T = 2π/ω); a different amplitude does not change it.
Why B is wrong: B is wrong because the problem states the same φ for both motions, so the phase constants are identical.
Why D is wrong: D is wrong because the two motions are given with the same ω.
MCQ 7CalculationPractice
A particle's displacement is x = 3 sin(2πt) + 3 cos(2πt) cm. This is a simple harmonic motion of amplitude
Show answer and why every option is right or wrong
Answer: D. A sum A sin ωt + B cos ωt can be written as D sin(ωt + φ) with D = √(A² + B²) (NCERT Class 11 Physics Chapter 13, page 262). Here D = √(9 + 9) = 3√2 ≈ 4.2 cm, and the period is still 2π/ω = 1 s.
Why A is wrong: A is wrong because 6.0 cm adds the two amplitudes. The sine and cosine reach their peaks at different times, so their peaks do not add.
Why B is wrong: B is wrong because 3.0 cm keeps only one of the two terms, ignoring the other.
Why C is wrong: C is wrong because 18 cm is D², the sum of the squares, with the square root left out.
MCQ 8Direct ApplicationPractice
If x = 5 sin(πt + π/3) m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
Show answer and why every option is right or wrong
Answer: B. The amplitude is the coefficient of the sine, 5, in the unit of x, which is metres. Comparing with ω t gives ω = π rad s⁻¹, so T = 2π/ω = 2 s (NCERT Class 11 Physics Chapter 13, pages 262 to 264). This is an NTA question from 2024.
Why A is wrong: A is wrong because the amplitude is 5 m, not 5 cm: x is given in metres. The period 2 s is right, since ω = π rad s⁻¹ and T = 2π/ω = 2 s.
Why C is wrong: C is wrong on both counts. The amplitude is 5 m (x is in metres), and with ω = π the period is T = 2π/ω = 2 s, not 1 s.
Why D is wrong: D is wrong because the amplitude 5 m is right, but T = 1 s would need ω = 2π. Here ω = π rad s⁻¹, so T = 2π/π = 2 s.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
What to remember before solving Displacement Function of Time questions
8 NCERT lines
Key Fact
Phase relation in SHM
Velocity leads displacement by π/2: v = -A ω sin(ωt+φ). Acceleration leads displacement by π: a = -A ω² cos(ωt+φ) = -ω² x. KE max at x=0; PE max at x=±A.
-- NCERT Class 11 Physics, Ch. 13, p. 267Definition
Simple harmonic motion (SHM)
Oscillation in which the restoring force is directly proportional to displacement from equilibrium and directed back toward equilibrium: F = -k x. Equation: a = -ω² x. Solution: x(t) = A cos(ω t + φ).
-- NCERT Class 11 Physics, Ch. 13, p. 260In this chapter, we use the term displacement in a more general sense. It refers to change with time of any physical property under consideration. For example, in case of rectilinear motion of a steel ball on a surface, the distance from the starting point as a function of time is its position displacement. The choice of origin is a matter of convenience. Consider a block attached to a spring, the other end of the spring is fixed to a rigid wall [see Fig.13.2(a)]. Generally, it is convenient to measure displacement of the body from its equilibrium position. For an oscillating simple pendulum, the angle from the vertical as a function of time may be regarded as a displacement variable [see Fig.13.2(b)]. The term displacement is not always to be referred Fig. 13.2(a) A block attached to a spring, the other end of which is fixed to a rigid wall. The block moves on a frictionless surface. The motion of the block can be described in terms of its distance or displacement x from the equilibrium position. Fig.13.2(b) An oscillating simple pendulum; its motion can be described in terms of angular displacement θ from the vertical. in the context of position only. There can be many other kinds of displacement variables. The voltage across a capacitor, changing with time in an AC circuit, is also a displacement variable.
-- NCERT Class 11 Physics, Ch. 13, p. 261The displacement can be represented by a mathematical function of time. In case of periodic motion, this function is periodic in time. One of the simplest periodic functions is given by f (t) = A cos ωt (13.3a)
-- NCERT Class 11 Physics, Ch. 13, p. 261Thus, simple harmonic motion (SHM) is not any periodic motion but one in which displacement is a sinusoidal function of time.
-- NCERT Class 11 Physics, Ch. 13, p. 262This time-dependent quantity, (ωt + φ) is called the phase of the motion. The value of plase at t = 0 is φ and is called the phase constant (or phase angle).
-- NCERT Class 11 Physics, Ch. 13, p. 263Key Fact
Periodicity x(t) = x(t + T) gives w = 2 pi / T, the angular frequency of SHM, in rad per second
Since the motion has a period T, x (t) is equal to x (t + T). That is, A cos ωt = A cos ω (t + T ) (13.6) Now the cosine function is periodic with period 2π, i.e., it first repeats itself when the argument changes by 2π. Therefore, ω(t + T ) = ωt + 2π that is ω = 2π/ T (13.7) ω is called the angular frequency of SHM. Its S.I. unit is radians per second.
-- NCERT Class 11 Physics, Ch. 13, p. 264Displacement Function of Time: NEET previous year questions (PYQs) with answers
14 questions in Oscillations and Waves
No question in our NEET 2020–2025 set targets this topic directly.
More in Oscillations and Waves: 6 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Sources
NCERT refs: Class 11 Physics Chapter 13, p.261 | Class 11 Physics Chapter 13, p.262
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Every past-paper question from this chapter:
Practice the whole chapter →