CH2_oscillation

# CH2_oscillation - CHAPTER 2 OSCILLATIONS 2.1 Quick Review...

This preview shows pages 1–4. Sign up to view the full content.

1 CHAPTER 2 OSCILLATIONS 2.1 Quick Review of Simple Harmonic Motion (SHM) 2.1.1 Equation of Motion for SHM The equation of motion for the SHM is: or (2.1) where . The general solutions of this equation of motion are : (1) x(t) = A cos ( ω t + φ ); (2) x(t) = A sin ( t + ); (3) x(t) = A cos t + B sin t (4) x(t) = A exp( i t ) + B exp(- i t ) (2.2) All these four solutions are equivalent. 2.1.2 Energy of SHM 1) Kinetic energy (2.3) 2) Potential Energy (2.4)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 At the equilibrium position EP (i.e. x = 0), take PE=0, Therefore, total Energy E T = KE + PE (2.5) If there is no energy dissipation, E T is a constant. 2.1.3 Phasor Diagram Construct a rotating vector (or the position phasor) such that , where A is the amplitude of the SHM. The phasor is initially at position as shown in the figure and then rotating anti-clockwise with an angular velocity of ω . Similarly, we can also construct the velocity and the acceleration phasor. =Phasor of Position, =Phasor of Velocity and =Phasor of Acceleration Notice that: , ,
3 Example A 1kg mass is placed on a horizontal smooth table and linked a wall by a spring with force constant of 1Nm -1 . Let the equilibrium position of the particle be x=0. If the mass is placed at a position of x=-1m at t=0 and then released, (i) Prove that the subsequent motion is a SHM. (ii) Find the frequency and the phase of the SHM. Then find the functions x, v and a with respect to t. (iii) Repeat (ii) if initially the mass is placed at x=+0.5m and has an initial velocity of +0.1ms -1 . 2.2 Damped Oscillation O ther than the force that drives the particle in SHM, there exists another damping force which is against the motion of the particle. We write the damping force as , which is against the velocity. (2.6) or in a one dimensional case, ˙ ˙ x + 2 γ ˙ x + ω 2 x = 0 (2.7) where and = k m . Notice that γ is a measure of the effectiveness of the damping force. ω 0 is determined only by the oscillator and is called the natural frequency of the oscillator. The solution of equation (2.7) is . The characteristic equation is: λ 2 + 2 γλ + 2 = 0 Therefore, = ± 2 2 . The solution has three possibilities, namely,: (i) γ > ω , overdamped, λ 1 and λ 2 are real roots (ii) γ = ω , critically damped, real and equal roots (iii) γ < ω , underdamped, λ 1 and λ 2 are imaginary roots.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

CH2_oscillation - CHAPTER 2 OSCILLATIONS 2.1 Quick Review...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online