HCI 08 Oscillations Solutions (Lecture Examples)

Oscillations 2022 Worked Examples

Example 8.1: Periodic Motion Which of the following cases are periodic oscillations? A and B are examples of periodic oscillations as we can identify a pattern that when repeated can reproduce theentire graph but C is not.

Example 8.2: Describing SHM in terms of time (a)(i)ݔ଴ = 0.12 m (ii)߱= ଶగ ் = ଶగ ଶ = 3.1rad s-1 (iii)ݒ଴ =߱ݔ଴ = 0.38 m s-1 (iv)ܽ଴ =߱ଶݔ଴ = 1.2 m s-2 (b)ݔ= 0.12sin (3.ݐ) c)(i) (c)(ii)

Example 8.3: Displacement-time Relationship for SHM (a) ݔ଴ = 3.5 m (b) ߱= 4.0 rad s-1 (c) ݔ= 3.5 sin 4.0(0.20) = 2.5 m (d) ݒ= 3.5 4.0 cos 4.0(0.50) = −5.8 m s-1

Example 8.4: Relationship between acceleration and displacement for SHM (a) ݔ଴ = 30 mm (b) From the graph, maximum acceleration ܽ଴ =߱ଶݔ଴ = 0.48 m s-2 hence ߱= ଴.ସ଼ ଴.଴ଷ = 4 rad s-1 (c) period ܶ= ଶగ ఠ = గ ଶ = 1.6 s

Example 8.5: Relationship between acceleration and displacement for SHM A B C D E x + + 0 - - v + - - - 0 a - - 0 + +

Example 8.6: Relationship between velocity and displacement for SHM (a) The object could be moving in either direction when it passes through the equilibrium position. (b) At zero velocity, the object is at the furthest point from the equilibrium position (maximum displacement) when its about to reverse direction, there are two such locations on either side of the equilibrium position. (c) Since maximum velocity is given by ݒ଴ =߱ݔ଴, taking readings from the graph, ߱= ଴.ଽ ଴.଴଴ଷ = 300rad s-1 Hence ݒ= ±߱ݔ ଴ ଶ −ݔଶ = ±300 0.003ଶ −ݔଶ

Example 8.7: Relationship between velocity and displacement for SHM (a) period, ܶ= ଵ ௙ = ଵ ଶ଴ = 0.050 s (b) At the equilibrium point, the object moves at maximum velocities, ݒ= ±߱ݔ଴ = ±݂ߨݔ ଴ = ±ߨ20 0.020 = ±2.51 m s-1 At position of maximum negative displacement, the object is instantly at rest.

Example 8.8: Phase Difference After A reached the maximum displacement (peak of graph), it takes a quarter of a cycle for B to also reach its maximum displacement. Hence A leads B by ߨ/2. It can also be stated that A lags B by 3ߨ/2.

Example 8.9: Relationship between Energies and Displacement (a) ܧ் = ଵ ଶ݉߱ଶݔ଴ ଶ = ଵ ଶ݉ ௞ ௠ݔ଴ ଶ = ଵ ଶ 20 0.03 ଶ = 0.009 J (b)(i) ܧ௄ = ଵ ଶ݉߱ଶݔ଴ ଶ −ݔଶ = ଵ ଶ 20 0.03ଶ − 0.02ଶ = 0.005 J (ii) ܧ௣ =ܧ் −ܧ௄ = ଵ ଶ݉߱ଶݔଶ = ଵ ଶ 20 0.02 ଶ = 0.004 J