d'x kx = m di dx 2 = ma, → F, sin ot-b dt (15.34) Again, the solution of this equation is rather lengthy and will not be presented. After the driving force on an initially stationary object begins to act, the ampli- tude of the oscillation will increase. The system of the oscillator and the surround- ing medium is a nonisolated system: work is done by the driving force, such that the vibrational energy of the system (kinetic energy of the object, elastic potential energy in the spring) and internal energy of the object and the medium increase. After a sufficiently long period of time, when the energy input per cycle from the driving force equals the amount of mechanical energy transformed to internal energy for cach cycle, a steady-state condition is reached in which the oscillations proceed with constant amplitude. In this situation, the solution of Equation 15.34 is x = A cos (wt + 6) (15.35) F/m bo (15.36) (² – w,)²

A small bird’s wings can undergo a maximum displacement amplitude of 4.60 cm (distance from the tip of the wing to the horizontal). If the maximum acceleration of the wings is 18.7 m/s2, and we assume the wings are undergoing simple harmonic motion when beating, what is the oscillation frequency of the wingtips?

Two particles oscillate in simple harmonic motion with amplitude A about the centre of a common straight line of length 2A. Each particle has a period of 1.5 s, and their phase constants differ by 4 rad. (a) How far apart are the particles (in terms of A) 0.5 s after the lagging particle leaves one end of the path? Enter the exact answer in terms of A. ab sin (a) Ω

Refer to attached.