NDA (II) 2016 General Ability Test · Question 105 of 150

Correct — C, the acceleration is minimum when the speed is maximum. Everything in simple harmonic motion follows from a = −ω²x, so the size of the acceleration is fixed entirely by how far the particle is from the mean position. The speed, in contrast, is v = ω√(A² − x²), which is largest when x is zero. So both quantities peak at opposite ends of the swing: at the mean position the displacement is zero, the acceleration is therefore zero — its minimum — and the speed is at its maximum ωA. At the extreme positions the reverse holds, with the speed zero and the acceleration at its maximum ω²A.

• (a)It is always in the opposite direction to velocity — The acceleration always points back towards the mean position, but the velocity does not always point away from it. On the return journey — from an extreme position back towards the centre — the particle is moving towards the mean position and so is the acceleration, and the two are in the same direction, which is why the particle speeds up over that quarter of the cycle. The word 'always' is what breaks this option.
• (b)It is proportional to the frequency of oscillation — In a = −ω²x the angular frequency appears squared, not to the first power, so the acceleration scales with the square of the frequency. It also depends on the displacement, so it cannot be proportional to the frequency alone in any case.
• (d)It decreases as the potential energy increases — This gets the trend exactly backwards. The potential energy of a harmonic oscillator is ½mω²x², so it grows as the particle moves away from the centre — and the acceleration ω²x grows over the same journey. Both are largest at the extreme positions and both vanish at the mean position, so they rise and fall together.

Simple harmonic motion is the motion produced when the restoring force on a body is proportional to its displacement from a fixed mean position and directed back towards that position, F = −kx. Dividing by the mass gives a = −ω²x, where ω is the angular frequency and ω² = k/m. The displacement varies sinusoidally with time, the velocity is greatest at the centre and zero at the ends, and the acceleration is zero at the centre and greatest at the ends.

Almost every one-line question on this topic can be answered by writing a = −ω²x on the margin and reading the options against it. The minus sign kills any option claiming the acceleration points away from the mean position; the square on ω kills any option claiming a simple proportionality to frequency; and the presence of x tells you where the maxima and minima sit. The one genuine subtlety is in option (a) — 'towards the mean position' and 'opposite to the velocity' sound like the same statement but are not, because the velocity reverses twice a cycle while the direction of the acceleration is set purely by which side of the centre the particle is on.

• In simple harmonic motion the acceleration is a = −ω²x, always directed towards the mean position.
• Speed is maximum, equal to ωA, at the mean position, where the acceleration is zero.
• Speed is zero at the extreme positions, where the acceleration has its largest magnitude, ω²A.
• The total mechanical energy, ½mω²A², stays constant, with kinetic energy peaking at the centre and potential energy at the ends.

• Sliding from 'always directed towards the mean position' to 'always opposite to the velocity' — the two agree for only half of each cycle.
• Missing the square on the angular frequency in a = −ω²x.
• Assuming acceleration and potential energy move in opposite directions; in fact they rise and fall together.

Simple harmonic motion is a near-annual GAT topic, usually as one statement-checking item plus one short numerical, so keep a = −ω²x and v = ω√(A² − x²) fluent.

• practice — not a real PYQ

  For a particle in simple harmonic motion, the magnitude of the acceleration is maximum at

  • (a)the mean position
  • (b)the extreme positions
  • (c)half the amplitude
  • (d)every point equally

  Answer(b) the extreme positions — the acceleration is ω²x in magnitude, so it is largest where the displacement is largest.

• practice — not a real PYQ

  If the angular frequency of a simple harmonic oscillator is doubled while the amplitude is unchanged, the maximum acceleration becomes

  • (a)half
  • (b)twice
  • (c)four times
  • (d)unchanged

  Answer(c) four times — the maximum acceleration is ω²A, so doubling ω multiplies it by four.