NDA (I) 2017 General Ability Test · Question 120 of 149

Correct — B, ω = 2πf. Frequency counts complete cycles per second and is measured in hertz. Angular frequency measures the same motion in radians per second, and since one complete cycle corresponds to going once round a circle, that is 2π radians. So each cycle contributes 2π radians, and f cycles in a second contribute 2πf radians in a second, giving ω = 2πf. The same relation can be written through the period, since f = 1/T gives ω = 2π/T — the phase advances by 2π radians in one period, which is what a period means.

• (a)f = πω — This makes the frequency larger than the angular frequency by a factor of π, which is backwards on both counts. The angular measure is the larger of the two, and the factor is 2π and not π.
• (c)f = 2ω/π — The constant is wrong in size and in placement. Converting cycles to radians requires multiplying by 2π; there is no combination of the two in which 2/π appears.
• (d)f = 2πω — This is the correct relation written upside down. Starting from ω = 2πf and solving for f gives f = ω/2π, not 2πω. It is the commonest slip in the question, because the two symbols and the constant are all right and only the arrangement is wrong.

Angular frequency is the natural measure for anything that repeats, because simple harmonic motion is the shadow of uniform circular motion. Write the displacement as x = A sin(ωt + φ) and ω is the rate at which the phase angle grows, in radians per second. Frequency and period describe the same repetition in everyday units — cycles per second and seconds per cycle. The three are tied together by ω = 2πf = 2π/T, and every oscillation formula in the syllabus, from the pendulum's ω = √(g/L) to the spring's ω = √(k/m), is written in terms of ω for exactly this reason.

The question offers one correct arrangement and three scrambles of the same symbols, so the safest way through is a dimensional or numerical check rather than recall. Take a wheel turning once a second: f is 1 hertz, and in that second the angle sweeps through 2π radians, so ω is 2π radians per second — a bigger number than f. Any option that makes f the bigger quantity, or that produces a number other than 2π here, is wrong. That single test disposes of all three distractors in a few seconds.

• Angular frequency and frequency are related by ω = 2πf, with ω in radians per second and f in hertz.
• The period is the reciprocal of the frequency, T = 1/f, so ω = 2π/T.
• One complete cycle corresponds to a phase change of 2π radians, which is where the constant comes from.
• In simple harmonic motion x = A sin(ωt + φ), ω is the rate of change of phase and not a rate of rotation of any real object.

• Inverting the relation and writing f = 2πω.
• Using π instead of 2π; the factor comes from a full circle, not a half one.
• Quoting angular frequency in hertz — its unit is radian per second.

NDA asks for the relation between frequency, angular frequency and period, or gives one of the three and asks for another.

• practice — not a real PYQ

  A particle in simple harmonic motion has a time period of 0.5 s. Its angular frequency is

  • (a)π rad/s
  • (b)2π rad/s
  • (c)4π rad/s
  • (d)0.5π rad/s

  Answer(c) 4π rad/s — ω = 2π/T, and 2π divided by 0.5 is 4π.

• practice — not a real PYQ

  The SI unit of angular frequency is

  • (a)hertz
  • (b)radian
  • (c)radian per second
  • (d)second

  Answer(c) radian per second — frequency is counted in hertz, but angular frequency measures phase change and is quoted in radian per second.