A simple harmonic motion of a particle is represented as, y = 10 cos ωt 10. The acceleration of the particle at time t = π/2ω will be : (symbols here carry their usual meanings)

This item was dropped by UPSC, and the printed page shows why. The booklet gives the motion as y = 10 cos ωt 10 — a stray '10' sitting after cos ωt with no operator between them. It is not a scanning artefact. The English page (23-A) and the Hindi facing page (22-A) of booklet FWLB-F-GAI carry the identical defect, so the error went to press in both language versions and every candidate in the hall saw it. That leaves the expression open to more than one repair, and the repairs do not agree. Ignore the stray digit and the motion is y = 10 cos ωt; take it as an added constant and it is y = 10 cos ωt + 10; take it as a lost multiplication and it is y = 100 cos ωt. On all three of those readings the acceleration a = −ω²y is zero at t = π/2ω, because ωt = π/2 puts the particle exactly at the mean position of its oscillation. But read the digit as an absorbed coefficient of ωt — y = 10 cos 10ωt — and the acceleration becomes −1000ω² cos 5π = +1000ω², a value the paper does not offer at all. An item whose stem admits a reading with no answer among the options cannot be keyed, and UPSC voided it rather than penalise the candidates who read it that way.

• (a)10 ω — Fails on units before you calculate anything. With y in metres, 10ω has the dimensions of a velocity, not an acceleration, so it could not be the answer under any repair of the stem.
• (b)−10 ω² — This is the acceleration at an extreme of the motion, not at t = π/2ω. For y = 10 cos ωt the particle is at y = +10 when ωt = 0, and there a = −ω² × 10 = −10ω². At ωt = π/2 it has travelled a quarter of a cycle to the mean position, where the acceleration has fallen to zero.
• (c)0 — This is what the two most natural repairs of the printed line give, and a candidate who reached it was reasoning correctly. It is not marked right because the item was voided as a whole — with the operator missing there is no single stem to key, and the cos 10ωt reading leads somewhere else entirely.
• (d)10⁄ω — Dimensionally impossible as an acceleration — dividing a length by an angular frequency gives length × time. Like option (a), it is a filler built from the numbers in the stem.

Simple harmonic motion is defined by the relation a = −ω²y: the acceleration is proportional to the displacement from the mean position and directed back towards it. Everything about the standard version of this question follows from that one line. Where the displacement is zero the acceleration is zero and the speed is at its maximum; where the displacement is largest the acceleration is largest and the speed is zero.

For y = A cos ωt the particle starts at the extreme, y = A, at t = 0. A quarter of a period later ωt = π/2, cos ωt = 0, and it is passing through the mean position at full speed with no restoring force acting on it. That instant is exactly the t = π/2ω the question asks about, which is what makes the intended answer zero. The wider lesson is about how to handle a stem that will not parse. Two of the four options here can be discarded on dimensions alone — 10ω is a speed and 10/ω is a length multiplied by a time — which leaves a real contest only between −10ω² and 0, and those two differ by exactly a quarter of a cycle in when you evaluate the motion. That is a useful habit in any physics item: check the units of the options before doing any algebra. The printed defect itself is worth knowing about because the bank reproduces the paper as it was set. Where our earlier data quietly tidied the equation up, it was hiding the very thing that decided the item's fate.

• For simple harmonic motion, a = −ω²y; acceleration is zero at the mean position and maximum at the extremes.
• With y = A cos ωt the particle is at an extreme at t = 0 and at the mean position at t = π/2ω, a quarter of the period T = 2π/ω later.
• Speed and acceleration are out of step: speed peaks where acceleration vanishes, and vice versa.
• Adding a constant to y shifts the centre of oscillation but not the acceleration, since the second derivative of a constant is zero.
• Booklet FWLB-F-GAI prints 'y = 10 cos ωt 10' on both its English page 23-A and its Hindi page 22-A — a press defect, not a scan error.

• Evaluating the acceleration at t = 0 instead of at the instant the question names.
• Assuming the amplitude value must appear in the answer, when at the mean position the acceleration is zero regardless of amplitude.
• Spending time on options that dimensional analysis rules out in seconds.

Asked as a one-step substitution into a = −ω²y at a named instant — and, on this occasion, as a reminder that a mis-set stem can be worth more marks than a correct guess.

• practice — not a real PYQ

  A particle executes simple harmonic motion with y = 5 sin ωt. Its acceleration at t = 0 is

  • (a)5ω²
  • (b)−5ω²
  • (c)0
  • (d)5ω

  Answer(c) 0 — at t = 0 the sine is zero, so the particle is at the mean position where the displacement and hence the acceleration vanish.

• practice — not a real PYQ

  In simple harmonic motion, the magnitude of the acceleration of the particle is maximum when its

  • (a)speed is maximum
  • (b)displacement is zero
  • (c)displacement is maximum
  • (d)kinetic energy is maximum

  Answer(c) displacement is maximum — acceleration is proportional to displacement, so it peaks at the extremes where the speed is zero.