Try this first
Answer: B. A periodic function takes the same value after one period: f(t) = f(t + T), and the period is the least interval for which this holds (NCERT Class 11 Physics Chapter 13, pages 262 and 264).
A is wrong: A is wrong because f(t) = −f(t + T) describes a function that returns with the opposite sign. A cosine does that after half a cycle, which is not its period.
C is wrong: C is wrong because f(t) = f(2t) compares the function at different instants scaled by 2 and says nothing about a fixed repeat interval T.
D is wrong: D is wrong because adding T to the value of the function is not repeating; a periodic function returns to the same value, not to a larger one.
Is sin ωt + cos 2ωt periodic? And is e^(−ωt)? The trap is deciding by eye: either "a sum of periodic terms has the period of its shortest term", or "anything that wiggles repeats".
A function is periodic if it returns to the same value after a fixed interval: f(t) = f(t + T). The period is the least such interval. The model case is f(t) = A cos ωt. Its argument increases by 2π over one cycle, so T = 2π/ω (NCERT Class 11 Physics Chapter 13, pages 261 and 262). A sine has the same period.
Combining terms:
Why the chapter cares: Fourier's result says any periodic function can be written as a superposition of sine and cosine functions of different time periods with suitable coefficients (page 262). The chapter also separates two ideas. Simple harmonic motion is not any periodic motion; its displacement is a sinusoidal function of time (page 262). And every oscillatory motion is periodic, but circular motion is periodic without being oscillatory (page 260).
Watch-out: to find the period of a sum, find each term's period and the smallest common multiple. Taking the shortest term's period fails: at that instant the other term has not returned to its starting value.
Given
f(t) = sin 2ωt + cos 3ωt, with ω = 5.0 rad s⁻¹.
Required
Show that f is periodic and find its period in seconds.
Concept
A sum of periodic terms repeats after the least interval that holds a whole number of every term's period. The period of each term follows from T = 2π/(angular frequency) (NCERT Class 11 Physics Chapter 13, pages 261 and 262).
Formula
T₁ = 2π/(2ω), T₂ = 2π/(3ω); T = n₁T₁ = n₂T₂ with the smallest whole numbers n₁, n₂.
Substitution
T₁ = 2π/(2 × 5.0) s, T₂ = 2π/(3 × 5.0) s. In terms of ω: T₁ = π/ω, T₂ = 2π/(3ω).
Calculation
T₁ = π/5.0 = 0.63 s and T₂ = 2π/15 = 0.42 s. We need n₁π/ω = n₂ 2π/(3ω), so 3n₁ = 2n₂. The smallest whole numbers are n₁ = 2 and n₂ = 3, giving T = 2π/ω = 2π/5.0 = 1.3 s. The numbers 2 and 3 and π are exact and do not count towards the significant figures.
Final answer
f(t) is periodic with period T = 2π/ω = 1.3 s (not T₁ = 0.63 s and not T₂ = 0.42 s).
Common trap
Taking the shortest term's period. At t = 0.42 s the cosine term has completed one cycle, but sin 2ωt has advanced by 2π × 0.42/0.63, which is 4π/3 and not a multiple of 2π, so f has not yet repeated.
Similar NEET-style question
Find the period of g(t) = sin 3ωt + cos 4ωt for ω = 2.0 rad s⁻¹. (Answer: the term periods are 2π/(3ω) and 2π/(4ω). Setting n₁ × 2π/(3ω) = n₂ × 2π/(4ω) gives 4n₁ = 3n₂, so n₁ = 3 and n₂ = 4, and T = 3 × 2π/(3ω) = 2π/ω = 2π/2.0 = 3.1 s.)
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Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
A function f(t) is periodic with period T. Which relation expresses this?
Answer: B. A periodic function takes the same value after one period: f(t) = f(t + T), and the period is the least interval for which this holds (NCERT Class 11 Physics Chapter 13, pages 262 and 264).
Why A is wrong: A is wrong because f(t) = −f(t + T) describes a function that returns with the opposite sign. A cosine does that after half a cycle, which is not its period.
Why C is wrong: C is wrong because f(t) = f(2t) compares the function at different instants scaled by 2 and says nothing about a fixed repeat interval T.
Why D is wrong: D is wrong because adding T to the value of the function is not repeating; a periodic function returns to the same value, not to a larger one.
Which statement about f(t) = A sin ωt + B cos ωt is correct?
Answer: D. A linear combination of sine and cosine of the same ωt is a periodic function with the same period T, and can be written as D sin(ωt + φ) (NCERT Class 11 Physics Chapter 13, page 262).
Why A is wrong: A is wrong because both terms have angular frequency ω, so the combination cannot repeat in half the time of a single cosine of the same ω.
Why B is wrong: B is wrong because periodicity does not need equal coefficients. A and B can have any values and the sum is still D sin(ωt + φ).
Why C is wrong: C is wrong because both terms already repeat after 2π/ω, so the sum repeats after that interval and not after twice it.
Which of the following functions of time is non-periodic? (ω is a positive constant.)
Answer: A. e^(−ωt) decreases monotonically with time and tends to zero as t → ∞, so it never repeats its value (NCERT Class 11 Physics Chapter 13, page 262).
Why B is wrong: B is wrong because sin ωt + cos ωt equals √2 sin(ωt + π/4), a sinusoid with period 2π/ω.
Why C is wrong: C is wrong because each term is periodic and the first term's period is a multiple of the other two, so the sum repeats with period 2π/ω.
Why D is wrong: D is wrong because a cosine with any constant phase φ is periodic with period 2π/ω.
The chapter quotes a result of the mathematician Fourier about periodic functions. It states that any periodic function can be expressed as
Answer: C. Fourier's result is that any periodic function is a superposition of sines and cosines of different time periods with suitable coefficients; this is why sinusoids are so important (NCERT Class 11 Physics Chapter 13, page 262).
Why A is wrong: A is wrong because one sine gives only a sinusoid. A general periodic function such as a sawtooth needs sines and cosines of several periods.
Why B is wrong: B is wrong because a polynomial grows without limit in t and cannot repeat, so it cannot represent a periodic function.
Why D is wrong: D is wrong because an exponentially decaying factor makes the function die away, which is the opposite of periodic.
What is the period of the function f(t) = cos 3ωt + sin 5ωt?
Answer: D. The two terms have periods 2π/(3ω) and 2π/(5ω). The least interval holding a whole number of both is 3 × 2π/(3ω) = 5 × 2π/(5ω) = 2π/ω (NCERT Class 11 Physics Chapter 13, page 262, the method of the example with three terms).
Why A is wrong: A is wrong because after 2π/(3ω) the first term repeats but sin 5ωt has advanced by 10π/3, which is not a multiple of 2π.
Why B is wrong: B is wrong because after 2π/(5ω) the second term repeats but cos 3ωt has advanced by 6π/5, which is not a multiple of 2π. The shortest term's period is not the period of the sum.
Why C is wrong: C is wrong because it adds the angular frequencies (3ω + 5ω). The period of a sum is not found from the sum of frequencies.
The function sin ωt + √3 cos ωt can be written in the form D sin(ωt + φ) as
Answer: A. With A = 1 and B = √3, D = √(A² + B²) = √(1 + 3) = 2 and φ = tan⁻¹(B/A) = tan⁻¹(√3) = π/3 (NCERT Class 11 Physics Chapter 13, page 262).
Why B is wrong: B is wrong because φ = π/6 comes from tan⁻¹(A/B), with the ratio inverted. The relation is tan φ = B/A.
Why C is wrong: C is wrong because (1 + √3) adds the two coefficients. A sine and a cosine peak at different instants, so D = √(A² + B²), not A + B.
Why D is wrong: D is wrong because 4 is D², the sum of the squares, with the square root left out.
The displacement of a particle is x = 3 sin ωt + 4 cos ωt. It can be written as D sin(ωt + φ). What are D and φ?
Answer: C. D = √(3² + 4²) = 5 and tan φ = B/A = 4/3, so φ = tan⁻¹(4/3), about 53° (NCERT Class 11 Physics Chapter 13, page 262). Check: 5 cos 53° ≈ 3 and 5 sin 53° ≈ 4.
Why A is wrong: A is wrong because 37° is tan⁻¹(3/4), with the ratio inverted. The relation is tan φ = B/A = 4/3.
Why B is wrong: B is wrong because 7 = 3 + 4 adds the coefficients. The amplitude of the combination is √(A² + B²) = 5.
Why D is wrong: D is wrong because 25 is D², the sum of the squares, with the square root left out.
Which of the following motions is periodic but not oscillatory?
Answer: B. Every oscillatory motion is periodic, but not every periodic motion is oscillatory. Circular motion is periodic, but it is not oscillatory because the body does not move to and fro about an equilibrium position (NCERT Class 11 Physics Chapter 13, page 260).
Why A is wrong: A is wrong because the block moves to and fro about its equilibrium position, so it is oscillatory as well as periodic.
Why C is wrong: C is wrong because the bob swings to and fro about the vertical, so it is oscillatory as well as periodic.
Why D is wrong: D is wrong because the NCERT uses exactly this case as its oscillation example: displaced a little, the ball performs oscillations in the bowl.
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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. 260Periodic motion repeats at regular intervals. Oscillatory motion is periodic motion about an equilibrium position. Time period T is the duration of one cycle; frequency ν = 1/T (Hz).
-- NCERT Class 11 Physics, Ch. 13, p. 260The 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. 261function f (t) is then periodic and its period, T, is given by ω π 2 = T (13.3b) Thus, the function f (t) is periodic with period T, f (t) = f (t+T )
-- NCERT Class 11 Physics, Ch. 13, p. 262The same result is obviously correct if we consider a sine function, f (t ) = A sin ωt. Further, a linear combination of sine and cosine functions like, f (t) = A sin ωt + B cos ωt (13.3c) is also a periodic function with the same period T.
-- NCERT Class 11 Physics, Ch. 13, p. 262The great importance of periodic sine and cosine functions is due to a remarkable result proved by the French mathematician, Jean Baptiste Joseph Fourier (1768–1830): Any periodic function can be expressed as a superposition of sine and cosine functions of different time periods with suitable coefficients.
-- NCERT Class 11 Physics, Ch. 13, p. 262Thus, 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. 262Since 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. 264No 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.
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