Question

Assuming small oscillations, derive the differential equations of motion of the system shown in Fig. P3.10.

F(t) = Fo sin ω,t 3 2 a4 m3 1 2 a1 +a2 + аз + а,-1 FIG. P3.10

Answer 1

Answer:

Small Oscillation:

If a particle, originally in a position of equilibrium (we limit ourselves to the case of motions in one dimension), is displaced by a small amount, a force will tend to bring it back to its original position. We assume that this restoring force F is only a function of position, i.e., F = F( x) . It is then easily expanded in a Taylor series: F( x) = F0 + x dF dx ! " # $ % & 0 + 1 2! x 2 d2 F dx2 ! " # $ % & 0 + 1 3! x 3 d3 F dx3 ! " # $ % & 0 +… (2.1) where F0 is the value of F( x) at the origin (i.e., position of equilibrium), and dn F dxn ( )0 the nth derivative at the origin. Since the origin is defined to be the equilibrium point, we have F0 ! 0 . Furthermore, as we consider only small displacements from the origin, we will neglect all terms of second and higher powers of x. We can then rewrite equation (2.1) with F = !kx (2.2) where k ! "(dF dx)0 . (Strictly speaking, we should use ‘ ! ’ instead of ‘=’.) Because F is a restoring force (i.e., it brings the particle back toward its origin), its first derivative is negative and, therefore, k is positive. Systems that can be described with equation (2.2) obey Hooke’s Law (the restoring force is approximately linear).

The Simple Harmonic Oscillator If substitute Hooke’s Law (equation (2.2)) into the Newtonian equation of motion F = ma , we get ! x ! +!0 2 x = 0 (2.3) where we have defined a new quantity !0 2 " k m . (2.4) - 13 - The solution to equation (2.3) can be expressed with sinusoidal functions such as x(t) = Asin !0 ( t " # ) (2.5) or x(t) = Acos !0 ( t " #). (2.6) The constants A , ! (or ! ) are determined by the initial conditions of the problem. For example, if at t = 0 the particle is located at x(t = 0) ! x0 , and moving with a velocity x !(t = 0) ! x ! 0 , we find that (if we choose equation (2.5) as the solution) x0 = !Asin(" ) x ! 0 = A#0 cos(" ). (2.7) Equation (2.7) is a system of two equations with two unknown that is readily solved, and gives tan(! ) = "#0 x0 x ! 0 A = " x0 sin(! ) , (2.8) with x(t) given by equation (2.5). Incidentally, we can now appreciate that !0 is the angular frequency of the motion. It is related to frequency !0 by !0 = 2"#0 = k m #0 = 1 $ 0 = 1 2" k m (2.9) where ! 0 is the period. Note that these three quantities are independent of the amplitude A of the motion. Alternatively, we can express A as a function of the total energy E = T +U of the system. The kinetic energy T is given by T = 1 2 mx ! 2 = 1 2 m!0 2 A2 cos2 !0 ( t " # ) = 1 2 kA2 cos2 !0 ( t " # ), (2.10) - 14 - where we used equation (2.4) in the last step. The potential energy U is calculated from the work done in moving the particle a distance x. We start by calculating the incremental work dW necessary to move the particle by an amount dx against the force F (see equation (1.23) of chapter 1) dW = !F dx = kx dx. (2.11) We now calculate U = dW 0 x ! = kx 'dx ' 0 x ! = 1 2 kx 2 = 1 2 kA2 sin2 "0 ( t # $ ). (2.12) We finally get the total energy E E = T +U = 1 2 kA2 . (2.13) We get the general result that the total energy of the system is proportional to the square of the amplitude of the motion. It is also independent of time (we assumed that the system exhibits no losses), and of the phase constant ! .