NDA (II) 2016 General Ability Test · Question 106 of 150

Correct — D, aₓ = – 3x. Simple harmonic motion is defined by two conditions on the acceleration together, and only this option meets both. The acceleration must be proportional to the first power of the displacement, and it must carry the opposite sign, so that it always points back towards the mean position. Matching aₓ = – 3x against the standard form a = – ω²x gives ω² = 3, so ω = √3 radians per second and the time period is T = 2π/√3 seconds — a real, finite period, which is the test that the motion genuinely oscillates.

• (a)aₓ = + 3x — The proportionality is right but the sign is fatal. A positive sign means the acceleration points away from the mean position, so the further the particle strays the harder it is pushed further out. Nothing brings it back and there is no oscillation at all — this describes a runaway, the behaviour of a ball balanced on top of a hill rather than sitting in a valley.
• (b)aₓ = + 3x² — Two faults. The relation is not linear, since x appears squared, and a squared term is positive for negative displacements as well as positive ones, so the acceleration points the same way on both sides of the centre. It can never be a restoring relation.
• (c)aₓ = – 3x² — The tempting near-miss, because the minus sign looks like the restoring sign. But squaring wipes out the sign of x, so this acceleration is directed the same way — negative — whether the particle is to the left or the right of the centre. On one side it restores and on the other it drives the particle further away, and it is not linear either, so it fails both tests.

A particle performs simple harmonic motion when its acceleration is directly proportional to its displacement from a fixed point and directed towards that point, that is a = – ω²x. Equivalently the restoring force obeys F = – kx, which is Hooke's law, and ω² = k/m. From this single relation follow the sinusoidal displacement, the time period T = 2π/ω, the frequency n = 1/T and the fact that the maximum speed is ωA and the maximum acceleration ω²A.

The whole item is a two-part checklist, and applying it in order settles the answer in seconds. First ask whether the power of x is one; that eliminates options (b) and (c) at a glance. Then ask whether the sign is negative; that eliminates option (a). Whatever survives is the harmonic one, and the coefficient it carries is ω². It is worth noticing that the number 3 is not the frequency but the square of the angular frequency, which is the source of a common slip on the follow-up numerical.

• Simple harmonic motion requires acceleration proportional to the first power of displacement AND directed opposite to it.
• Comparing with a = – ω²x, the coefficient of x is ω², so in aₓ = – 3x the angular frequency is √3 rad/s.
• The corresponding time period is T = 2π/ω = 2π/√3 seconds, independent of the amplitude.
• The restoring-force version of the same statement is Hooke's law, F = – kx, with ω² = k/m.

• Accepting a squared displacement term as harmonic because it carries a minus sign — the square destroys the direction information.
• Reading the coefficient of x as the angular frequency rather than as its square.

This is the GAT's favourite way to test whether a candidate knows the definition rather than the formulae, and it recurs with different coefficients, so treat the two-part checklist as the drill.

• practice — not a real PYQ

  A particle moves such that its acceleration is given by a = – 16x, where x is the displacement in metres. Its time period is

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

  Answer(b) π/2 s — comparing with a = – ω²x gives ω² = 16, so ω = 4 rad/s and T = 2π/4 = π/2 s.

• practice — not a real PYQ

  For a body to execute simple harmonic motion, the restoring force acting on it must be

  • (a)constant in magnitude and direction
  • (b)proportional to the displacement and directed towards the mean position
  • (c)proportional to the square of the displacement
  • (d)proportional to the velocity and opposite to it

  Answer(b) proportional to the displacement and directed towards the mean position — that is Hooke's law, F = – kx.