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PHYS 218 sec. 517-520 Review Chap. 13 Periodic Motion What you have to know • • • • • Periodic motions (oscillations) Physical quantities describing periodic motions Equation of motion for simple harmonic motion Various examples of harmonic motion Pendulums • We skip the parts which have an asterisk (*) in the textbook. Physical quantities Restoring force Equilibrium point Amplitude (A): maximum displacement from equilibrium ® the maximum value of x Cycle: one complete round trip Period (T ): the time for one cycle, unit s (sec) Frequency ( f ): the number of cycles in unit time, unit Hz (= s- 1 ) Angular frequency (w): w = 2p f Basic relations f = 1 1 2p , T = , w = 2p f = T f T Simple harmonic motion (SHM) This is the simplest case when the restoring force is directly proportional to the displacement from the equilibrium In this case, the restoring force is in the form of Fx = - kx. d 2x Becasue F = ma = m 2 , dt d 2x k = x : This is the basic equation for SHM 2 dt m w does not depend on A Its solution is sinusoidal so sine or cosine function; x = A cos q = A cos (wt ) Therefore, the SHM can be described by d 2x = - w2 x 2 dt Then it follows that f = w 1 = 2p 2p k 1 m , T = = 2p m f k k w= m for SHM Ex 13.2 SHM A spring: 6 N force causes a displacement of 0.03 m Attach a 0.5 kg object to the end of the spring, pull it a distance of 0.02 m nad then release it Spring constant The first condition determines the spring constant F = kx Þ k = F x = 6.0 N = 200 N/m = 200 kg/s2 0.03 m w, f, T of this SHM Since m = 0.5 kg, w= f = k = m 200 kg/s 2 = 20 rad/s 0.5 kg w 20 rad/s = = 3.2 /s = 3.2 Hz, 2p 2p T= 1 1 = = 0.31 s f 3.2 Hz Displacement in SHM d 2x 2 x w = Equation of motion: 2 dt The general solution of this equation is x = A cos(wt + f ) A : amplitude, f : phase angle The two constants A and f can be determined by initial conditions such as the position and velocity at t = 0 At t = 0, x = x0 , where x0 = A cos f velocity in SHM dx d = ( A cos (wt + f )) = - Aw sin (wt + f ) dt dt acceleration in SHM vx = d 2 x dv ax = 2 = = - Aw 2 cos (wt + f ) = - w 2 x dt dt Therefore, the velocity oscillates between w A and - w A and the acceleration oscillates between w 2 A and - w 2 A. Displacement in SHM (2) At t = 0, x = x0 and v = v0 , where x0 = A cos f and v0 = - w A sin f 2 \ x = A cos f and v = (w A) sin f Þ 2 0 2 2 2 0 2 x02 v02 1 = cos f + sin f = 2 + 2 2 A w A Therefore, v02 A = x + 2 Þ w Also 2 2 0 A= v02 x + 2 w 2 0 æ v0 ÷ ö v0 - w A sin f ç ÷ = = - w tan f Þ f = arctan çç÷ x0 A cos f x è 0÷ ø 2 2 Energy in SHM In SHM, x = A cos (wt + f ), v = - Aw sin (wt + f ) Total energy 1 2 1 2 1 1 mv + kx = mA2 w2 sin 2 (wt + f )+ kA2 cos 2 (wt + f ) 2 2 2 2 1 k = kA2 {sin 2 (wt + f )+ cos 2 (wt + f )} ¬ using w2 = 2 m 1 It should be so because of energy constant = kA2 2 conservation E= Since x contains a sine function and v has a cosine function, when x = ± xmax , v = 0 and Since the energy relation gives when x = 0, v = ± vmax 1 2 1 2 1 2 kA = mv + kx , 2 2 2 we get v = ± k A2 - x 2 m Ex 13.4 Velocity, acceleration, and energy in SHM k = 200 N/m, m = 0.5 kg and the oscillating mass is released from rest at x = 0.02 m. Maximum and minimum velocity As we derived in the previous page, v= ± k A2 - x 2 . m Þ vmax = vmin = - k A= m Note that the minimum value is NOT 0! 200 N/m ´ (0.02 m) = 0.40 m/s 0.5 kg k A = - 0.40 m/s m Maximum acceleration k x, m kA (200 N/m)(0.02 m) = = = 8.0 m/s 2 m 0.50 kg Since a = - w2 x = amax = k x max m Ex 13.4 Cont’d The body has moved halfway to the center from its original position The position of the body is x = A / 2 = 0.01 m. Then, k v= A2 - x 2 = - 0.35 m/s m k a= x = - 4.0 m/s 2 m The body is moving from x=A/2 to x=0. Thus we choose the negative sign for the velocity. Ex 13.5 Energy and momentum in SHM m v1 M O This is a collision problem. Þ always consider momentum conservation! To use momentum conservation, we should first know the velocity of the body. The velocity can be obtained by energy consideration. Velocity at x=0 Before the collision, the total energy is E1 = At x = 0, U = 0, and E1 = 1 2 kA1 2 1 2 1 kA1 = Mv12 Þ 2 2 Collision By momentum conservation in x-direction, M M v1 + 0 = (M + m)v2 Þ v2 = v1 M+m v1 = k A1 M The putty m is moving in y-direction. Cont’d Ex 13.5 After the Collision Now the body (M + m) is moving with v2 at x = 0. Then the total energy becomes 2 æ M ö æ M ö÷ 1 1 E2 = (M + m)v22 = (M + m)çç v1 ÷ = çç E1 ÷ ÷ ÷ ÷ ç ç è ø è ø 2 2 M+m M+m The total energy can also be written in terms of the amplitude as E = therefore, ö1 2 1 2 æ M ÷ kA2 = çç kA1 Þ ÷ ÷ ç è ø 2 M+m 2 The period is T = 2p A2 = A1 mass spring constant Then the period after the collision is T = 2p M+m k M M+m 1 2 kA , 2 Cont’d Ex 13.5 If the Collision happens when M is at x=A The velocity of the block is then zero. There is no motion and the collision just causes the change of the mass from M to m. Since the amplitude does not change, the total energy of the system does not change. The period depends on the mass, so M+m k after the collision. T = 2p Vertical SHM oscillation equilibrium l l l Dl Dl - x F = kD l F = mg In equilibrium, k D l = mg F = kD l x F = mg Fnet = k (D l - x)- mg = - kx Therefore, d 2x k =x Þ dt m SHM Angular SHM q : angular displacement Equation of motion (from Newton's 2nd law) t = Ia If the restoring torque is proportional to the angular displacement t = - kq Therefore, we get d 2q k a = 2 = - q Þ SHM dt I The angular frequency is w = k I Simple pendulum Equation of motion along the x direction Fq = - mg sin q (cf x = Lq) q Therefore, L d 2x g = g sin q » g q = x Þ SHM 2 dt L The angular frequency is T x mg cos q mg sin q w= g L mg The Taylor expansion When q is small sin q ; q Physical pendulum Any real pendulum pivot You can treat this object as a point-like object which has the mass of the object at its center-of-mass position. q Then it becomes similar to the case of simple pendulum. d t = - (mg )(d sin q) = I a When q is small, sin q » q, d 2q mgd Þ = q 2 dt I angular frequency mg w= mgd I Þ SHM cf . For simple pendulum, I = md 2 then we have w = mgd = 2 md g d Damped oscillations Forced oscillations & resonance These topics will not be summarized here. But you should read the textbook for these topics.