NDA (II) 2016 General Ability Test · Question 110 of 150

Correct — A, 2π/√3. Two standard results do all the work. The speed at displacement x is v = ω√(A² − x²) and the magnitude of the acceleration there is a = ω²x. With the amplitude A = 2 cm and the displacement x = 1 cm, the speed becomes ω√(4 − 1) = ω√3 and the acceleration becomes ω² × 1 = ω². The question says the two magnitudes are equal, so ω√3 = ω². Cancelling one factor of ω, which is legitimate because ω cannot be zero for an oscillation, leaves ω = √3 radians per second. The time period follows at once from T = 2π/ω = 2π/√3 seconds.

• (b)√3/2π — This is the reciprocal of the correct answer, and it is what you get by writing the period as ω/2π instead of 2π/ω. The quantity ω/2π is the frequency in hertz, not the period in seconds.
• (c)√3/π — This comes from mishandling the factor of two — for instance by writing T = π/ω, or by cancelling the 2 in 2π against the 2 cm of amplitude, which are unrelated quantities.
• (d)1/2π√3 — This is the frequency expressed as 1/T only if the period were 2π√3, so it is a doubly inverted form. Checking the units is the quickest defence: a period must come out in seconds, and 2π divided by an angular frequency in radians per second is the only combination here that does.

For a particle in linear simple harmonic motion of amplitude A and angular frequency ω, the displacement, speed and acceleration are linked by three relations that are worth memorising as a set: x = A sin(ωt + φ), v = ω√(A² − x²) and |a| = ω²x. The speed is greatest at the mean position and zero at the extremes; the acceleration behaves in exactly the opposite way. The time period T = 2π/ω does not depend on the amplitude at all, which is the property that makes harmonic oscillators useful as clocks.

Problems of this shape look intimidating because they seem to have too few numbers, but the amplitude and one displacement are always enough. Set the two expressions equal, and the amplitude and displacement fix ω on their own; the mass, the force constant and the units of length never enter. Note that the centimetres cancel between the two sides of ω√3 = ω², so there is no need to convert to metres — a conversion that costs time and, done on only one side, produces a wrong answer.

• Speed at displacement x in simple harmonic motion is v = ω√(A² − x²), maximum ωA at the mean position.
• Magnitude of acceleration at displacement x is ω²x, maximum ω²A at the extreme positions.
• Time period T = 2π/ω, and frequency n = 1/T = ω/2π.
• The period of a harmonic oscillator is independent of the amplitude, which is why pendulum clocks keep time as they run down.

• Inverting the period-frequency relation and answering ω/2π instead of 2π/ω.
• Converting centimetres to metres on only one side of the equation.
• Cancelling ω without noticing that you have implicitly rejected ω = 0, which is fine here but is worth stating.

The GAT usually pairs one conceptual simple harmonic motion item with one short numerical of exactly this type, so drill the substitution until it takes under a minute.

• practice — not a real PYQ

  A particle in simple harmonic motion has amplitude 5 cm and angular frequency 2 rad/s. Its speed when it is 3 cm from the mean position is

  • (a)4 cm/s
  • (b)6 cm/s
  • (c)8 cm/s
  • (d)10 cm/s

  Answer(c) 8 cm/s — v = ω√(A² − x²) = 2 × √(25 − 9) = 2 × 4 = 8 cm/s.

• practice — not a real PYQ

  In simple harmonic motion, the ratio of the maximum acceleration to the maximum speed equals

  • (a)the amplitude
  • (b)the angular frequency
  • (c)the time period
  • (d)the square of the amplitude

  Answer(b) the angular frequency — the maxima are ω²A and ωA, and their ratio is ω.