Time period and frequency are the two quantities that define "how fast" an oscillation repeats — and NEET loves testing whether you confuse which parameters they depend on and which they do not.
Definitions (NCERT Class 11 Physics, Chapter 13, page 260). The time period T is the time for one complete oscillation. The frequency f is the number of complete oscillations per unit time. They are reciprocals: f = 1/T, with T in seconds and f in hertz (Hz). The angular frequency ω = 2πf = 2π/T has units rad/s.
The high-frequency trap: "period depends on amplitude." For ideal SHM — a mass on a Hooke's-law spring or a simple pendulum at small angle — the period is independent of amplitude. The spring period T = 2π√(m/k) contains only mass and spring constant. The pendulum period T = 2π√(L/g) contains only length and gravitational acceleration. Neither formula has amplitude A anywhere. Doubling the amplitude doubles the energy (E = ½kA²) but the oscillation takes exactly the same time per cycle. NEET exploits this by changing amplitude in the stem and offering distractors that scale T with A.
Second common confusion: "pendulum period depends on mass." It does not. In the derivation, gravitational mass (in the restoring force mg sin θ) cancels with inertial mass (in ma). So replacing a 50 g bob with a 200 g bob on the same string changes nothing about T.
Watch-out for the exam hall: when a question changes mass or amplitude and asks for the new period, the answer is almost always "period unchanged." Read the stem for what actually changed — length, g (elevator problems), or spring constant — because those do affect T.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The SI unit of frequency is:
Answer: B. Frequency f = 1/T is measured in hertz (Hz = s⁻¹), as defined in NCERT Class 11 Physics Chapter 13, page 261.
Why A is wrong: A is the unit of time period T, not frequency f.
Why C is wrong: C (rad/s) is the unit of angular frequency ω, not ordinary frequency f. ω = 2πf, so the units differ by the dimensionless factor 2π.
Why D is wrong: D (s⁻²) has dimensions of 1/time², which matches acceleration-like quantities, not frequency.
If the time period of an oscillator is 0.04 s, its frequency is:
Answer: B. f = 1/T = 1/0.04 = 25 Hz. Direct application of the reciprocal relation (NCERT Class 11 Physics Chapter 13, page 261).
Why A is wrong: A (0.025 Hz) results from taking 1/40 — misplacing the decimal so that 0.04 s is read as 40 s. The reciprocal of 0.04 is 25.
Why C is wrong: C (4 Hz) likely comes from dividing 1 by 0.25 instead of 0.04, a misread of the given period.
Why D is wrong: D (250 Hz) results from an extra factor of 10, possibly from misreading 0.04 as 0.004.
The angular frequency of an oscillator with time period T is:
Answer: D. ω = 2π/T by definition — one full cycle (2π radians) per period T (NCERT Class 11 Physics Chapter 13, page 264).
Why A is wrong: A (T/2π) inverts the formula entirely; it gives time per radian, not radians per second.
Why B is wrong: B (2πT) has dimensions of time, not rad/s. This inverts the relationship.
Why C is wrong: C (π/T) drops the factor of 2 — a full cycle is 2π radians, not π.
A mass-spring system has time period T. If the amplitude of oscillation is doubled, the new time period is:
Answer: A. T = 2π√(m/k) — amplitude does not appear. Period is unchanged (NCERT Class 11 Physics Chapter 13, page 267). This is a common trap: amplitude affects energy, not period.
Why B is wrong: B (T/2) incorrectly assumes an inverse proportionality between amplitude and period. No such relation exists in ideal SHM (trap: period-amplitude dependence claim).
Why C is wrong: C (2T) assumes direct proportionality T ∝ A. The spring period formula T = 2π√(m/k) has no amplitude term (trap: period-amplitude dependence claim).
Why D is wrong: D (4T) assumes T ∝ A². Energy scales as A², but period does not scale with amplitude at all.
A simple pendulum has time period 2 s. If the mass of the bob is tripled (keeping length and g unchanged), the new time period is:
Answer: D. T = 2π√(L/g) — mass of the bob does not appear. Period remains 2 s (NCERT Class 11 Physics Chapter 13, page 271). Gravitational and inertial mass cancel in the derivation.
Why A is wrong: A (2/3 s) assumes inverse proportionality T ∝ 1/m, which is incorrect — mass cancels completely (trap: pendulum mass-dependence claim).
Why B is wrong: B (2√3 s) assumes T ∝ √m, borrowing the form from the spring formula T = 2π√(m/k). The pendulum formula has L/g under the root, not m/k.
Why C is wrong: C (6 s) assumes direct proportionality T ∝ m. The pendulum period formula contains no mass term (trap: pendulum mass-dependence claim).
A spring of spring constant k is cut into two equal halves. A mass m is attached to one half. The time period of oscillation compared to the original spring is:
Answer: C. Cutting a spring in half doubles its spring constant (k' = 2k). Then T' = 2π√(m/k') = 2π√(m/2k) = T/√2. The period is reduced by factor 1/√2 (NCERT Class 11 Physics Chapter 13, page 263).
Why A is wrong: A (unchanged) ignores that halving the spring changes its spring constant. A shorter spring is stiffer: k' = 2k, so T must change (trap: period-amplitude dependence thinking — confusing what does and doesn't affect T).
Why B is wrong: B (halved) assumes T ∝ 1/k, so doubling k halves T. Actually T ∝ 1/√k, giving T/√2, not T/2.
Why D is wrong: D (doubled) reverses the effect — a stiffer spring gives a shorter period, not a longer one.
The time period of a simple pendulum on Earth is 1 s. If the same pendulum is taken to a planet where g is four times that on Earth (length unchanged), the new time period is:
Answer: A. T = 2π√(L/g). If g → 4g, then T' = 2π√(L/4g) = T/2 = 0.5 s. Period scales as 1/√g (NCERT Class 11 Physics Chapter 13, page 268).
Why B is wrong: B (0.25 s) assumes T ∝ 1/g (linear inverse), giving T/4. The correct scaling is T ∝ 1/√g.
Why C is wrong: C (2 s) assumes T ∝ √g (period increases with g), which reverses the relationship. Stronger gravity means faster oscillation, shorter period.
Why D is wrong: D (4 s) assumes T ∝ g (direct proportionality), giving 4T. This contradicts the square-root dependence.
A body executes SHM with period T. The total energy of the oscillator is E. If the amplitude is doubled while keeping mass and spring constant the same, the new total energy and new time period are:
Answer: C. E = ½kA²; doubling A gives E' = ½k(2A)² = 4E. T = 2π√(m/k) is independent of amplitude, so T remains unchanged. Answer: 4E, T (NCERT Class 11 Physics Chapter 13, pages 263 and 7).
Why A is wrong: A (4E, 2T) gets energy correct but wrongly assumes T ∝ A. Period is independent of amplitude in ideal SHM (trap: period-amplitude dependence claim).
Why B is wrong: B (2E, 2T) gets energy wrong (E ∝ A², not A) and incorrectly assumes period depends on amplitude. T = 2π√(m/k) has no A term (trap: period-amplitude dependence claim).
Why D is wrong: D (2E, T) assumes E ∝ A (linear), but energy scales as A². Doubling amplitude quadruples energy.
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Pattern: Period, frequency and angular frequency from a count of oscillations — the definitional chain NEET tests directly.
Given
A body executing SHM completes 240 complete oscillations in 2.0 minutes. Its amplitude is 4.0 cm.
Required
(a) The time period T. (b) The frequency f. (c) The angular frequency ω. (d) The maximum speed of the body.
Concept
Period and frequency are defined directly by counting: the period is the time for ONE oscillation, so it is the total time divided by the number completed, and the frequency is its reciprocal — the number completed per second (NCERT Class 11 Physics Chapter 13, page 260).
Angular frequency is not a third independent quantity. It is the same rate expressed in radians, one full oscillation being 2π radians, so ω = 2πf. It is ω, not f, that appears in the equations of motion, which is why the conversion is worth doing immediately. The maximum speed then follows, since a particle in SHM passes the mean position at v = Aω.
Formula
• T = total time / number of oscillations• f = 1/T• ω = 2πf• v_max = Aω
Substitution
Convert the time first: 2.0 minutes = 120 s.• T = 120 / 240• f = 1 / T• ω = 2π × f• v_max = 0.040 × ω
Calculation
• T = 120 / 240 = 0.50 s• f = 1 / 0.50 = 2.0 Hz• ω = 2π × 2.0 = 4π = 12.6 rad/s• v_max = 0.040 × 12.6 = 0.503 m/s
Note on exact values: 240 oscillations is a count, the 60 s per minute is a definition, and 2π is a mathematical constant. None of them limits significant figures; the two-figure data do.
Final answer
(a) T = 0.50 s (b) f = 2.0 Hz (c) ω = 4π ≈ 12.6 rad/s (d) v_max ≈ 0.50 m/s
That v_max and T come out numerically close here is a coincidence of the chosen numbers, not a relationship — one is a speed in m/s and the other a time in s.
Common trap
The commonest slip is arithmetic rather than physical: using 2.0 as the time instead of 120 s, which makes T come out in minutes and every later quantity wrong by a factor of 60. Convert to SI before the first division.
The second is treating ω as interchangeable with f. They differ by 2π, and substituting f where ω belongs understates the maximum speed by that factor. If a formula came from a sine or cosine of time, the quantity inside it is ω.
Similar NEET-style question
A pendulum makes 30 oscillations in 45 s. Find its frequency and angular frequency. Approach: T = 45/30 = 1.5 s, so f = 1/1.5 = 0.67 Hz and ω = 2π × 0.67 = 4.2 rad/s.
Periodic 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. 260More in Oscillations and Waves: 6 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
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