Physics

1. 1.

   An oscillator executes simple harmonic motion (SHM) with angular frequency 5.0 rad s^-1. Calculate acceleration at displacement +0.080 m from equilibrium and explain the sign.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Square the angular frequency before multiplying by displacement.
   2. The restoring sign is part of the SHM condition and cannot be replaced by a positive magnitude alone.

   Marking points

   • Use a = -omega^2 x.
   • a = -2.0 m s^-2 (2 significant figures).
   • Acceleration is towards equilibrium, opposite the positive displacement.

   Examiner tip: Use angular frequency directly; do not multiply it by 2pi again.

2. 2.

   A mass 0.250 kg oscillates on a spring of stiffness 40.0 N m^-1. Find the period. Assume a massless spring and negligible damping.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Combine F = -kx with F = ma to identify omega^2 = k/m.
   2. A full cycle corresponds to 2pi radians, so divide 2pi by omega.

   Marking points

   • omega = sqrt(k/m).
   • T = 2pi sqrt(m/k).
   • T = 0.497 s (3 significant figures).

   Examiner tip: The ratio under the period square root is m/k, not k/m.

3. 3.

   A 0.200 kg oscillator executes simple harmonic motion (SHM) with amplitude 0.0500 m and omega = 10.0 rad s^-1. Find its speed and kinetic energy when its displacement from equilibrium is x = 0.0300 m. Neglect damping.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Energy conservation partitions the fixed total between displacement-dependent potential energy and kinetic energy.
   2. The square-root difference is sqrt(0.00250 - 0.000900) = 0.0400 m; multiply by omega, then use mv^2/2.

   Marking points

   • v^2 = omega^2(A^2 - x^2).
   • Speed is 0.400 m s^-1.
   • KE = 0.0160 J (3 significant figures).

   Examiner tip: Displacement alone fixes speed but not the direction of travel.

4. 4.

   An SHM displacement is x = 0.0400cos(4.00t) m, with t in seconds. Find the first positive time at which x = 0.0200 m and the signed velocity then, to 3 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The oscillator starts at maximum positive displacement and initially moves towards equilibrium.
   2. Choose the earliest positive phase pi/3, not 5pi/3; differentiation supplies the negative velocity sign.

   Marking points

   • cos(4t) = 1/2 first gives 4t = pi/3.
   • t = 0.262 s.
   • v = -0.160sin(4t) = -0.139 m s^-1.

   Examiner tip: Use radian mode because 4.00t is a phase in radians.

5. 5.

   A pendulum has length 0.800 m and small amplitude. After moving to a location with unknown g its measured time for 20 complete periods is 36.0 s. Estimate g and its percentage uncertainty if total timing uncertainty is +/-0.2 s and length uncertainty +/-0.005 m. Use a worst-case first-order uncertainty sum.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Dividing the total time and its uncertainty by twenty leaves their fractional uncertainty unchanged.
   2. The inverse square of T doubles its fractional contribution; sum positive magnitudes rather than subtract them.

   Marking points

   • T = 36.0/20 = 1.80 s; g = 4pi^2 L/T^2.
   • g = 9.75 m s^-2 (3 significant figures).
   • Fractional uncertainty = 0.005/0.800 + 2(0.2/36.0), or 1.7% to 2 significant figures.
   • The small-angle, point-bob model neglects damping and string mass.

   Examiner tip: Do not assign +/-0.2 s to each single period after dividing the total time.

6. 6.

   For weak damping the amplitude envelope is A(t) = A0 e^(-0.120t), with t in seconds and constant oscillator stiffness. Using a slowly varying envelope model in which cycle-averaged mechanical energy is approximately proportional to A^2, find the approximate time for that energy to fall to one quarter of its initial value. Explain why maximum forced-response amplitude need not occur exactly at the undamped natural frequency.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Within this cycle-averaged approximation, one-quarter energy corresponds to half the envelope amplitude, giving the same approximate time via ln(2)/0.120. This does not assert an exact instantaneous energy at every phase.
   2. A driven damped oscillator's displacement amplitude depends on both inertia-stiffness balance and energy dissipation, not just frequency matching.

   Marking points

   • In the weak-damping envelope model, cycle-averaged energy is approximately proportional to A^2, so E/E0 is approximately e^(-0.240t).
   • Approximate time t = ln(4)/0.240 = 5.78 s (3 significant figures).
   • Damping shifts the displacement resonance peak and broadens it; for weak damping it lies close to, but below, the undamped natural frequency.

   Examiner tip: For cycle-averaged energy in the weak-damping envelope model, double the amplitude decay constant, not the time.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.