10.1 The Ideal Spring and Simple Harmonic Motion

Transcription

1 10.1 The Ideal Spring and Simple Harmonic Motion TRANSPARENCY FIGURE restoring force F applied = (+)kx (10:1) Hooke s Law Restoring Force of an Ideal Spring The restoring force of an ideal spring is F = ( )kx (10:2) where k is the spring constant and x is the displacement of the spring from its unstrained length. The minus sign indicates that the restoring force always points in a direction opposite to the displacement of the spring. Ex. 1 puts in numbers for the magnitude of the force on a tire gauge spring.

2 Conceptual Ex. 2: Are Shorter Springs Sti er Springs? Fig Take a spring of 10 coils and constant k, and cut it in half to make 2 springs. Compress the 10 coil spring by 1 cm, or 0.1 cm per coil. Then compress a 5 coil spring by 1 cm, or 0.2 cm per coil. By (10:1=2) F = ( twice the force. )kx, the short springs must need Thus the shorter springs have a spring constant of 2k, i.e. inverse to the number of coils > Problems 8, 16 "SHM" - middle of page 269 (6th edition):

3 When the restoring force has the mathematical form given by F = ( )kx, the type of friction-free motion illustrated in Figure 10.5 is designated as simple harmonic motion. TRANSPARENCY FIGURE sinusoidal variation 10.2 Simple Harmonic Motion and the Reference Circle TRANSPARENCY FIGURE Displacement Since the angle varies with linearly time, =!t: x = A cos = A cos!t (10:3) If T is the period, (8:2)!! = 2 T (10:4) PLOT A cos 2t=T [Fig ]

4 frequency f is the number of cycles per second (hertz, Hz): f = 1 T (10:5) Thus! = 2 T = 2f (10:6) Velocity (v T = r!) [(8.9)] v = v x = v T cos ( + =2) = A! sin!t = v T sin!t (10:7) TRANSPARENCY FIGURE v max = v T = A! (10:8) Ex. 3: Conceptual Ex. 4: Moving lights - NOT SHM Acceleration a c = r! 2

5 a = a x = a c cos ( + ) = A! 2 cos!t (10:9) N.B.This means that the components of displacement an dacceleration are exactly opposite: a x =! 2 x (10:) TRANSPARENCY FIGURE a max = a c = A! 2 (10:10) Ex. 5: Loudspeaker Frequency of Vibration Use N2: F x = ma x with (10:2) F x = ( (10:3) for x and (10:9) for a: )kx, and F = ( )kx = ka cos!t = m A! 2 cos!t Cancel A cos!t: k = m! 2 or

6 ! = k m 1 2 (10:11) Ex. 6: Vibrating Platform Bjarni Trygvason, astronaut Energy and Simple Harmonic Motion Taking the magnitude of the average force as F = 1 2 kx0 + kx f and displacement magnitude as x 0 x f leads to: W elastic = 1 2 kx kx2 f (10:12) compare with (6:4) that leads to "mgh". De nition of Elastic Potential Energy The elastic potential energy PE elastic is the energy that a spring has by virtue of being stretched or compressed. For an ideal spring that has a spring constant k and is

7 stretched or compressed by an amount x relative to its unstrained length, the elastic potential energy is: PE elastic = 1 2 kx2 (10:13) TRANSPARENCY FIGURE CONCEPTS AT A GLANCE The Conservation of Mechanical Energy E = 1 2 mv I!2 + mgh kx2 (10:14) Ex. 7: object on a horizontal spring Conceptual Ex. 8: Changing mass of a SHO [->Q.6, P.32] Ex The Pendulum

8 TRANSPARENCY FIGURE For small angles:! = 2f = 1 g 2 L (10:16) Ex. 10: Keeping Time [Damped Harmonic and Driven motions OMIT section 10.5, look at Fig ] [(Elasticity and) Simple Harmonic Motionl OMIT sections 10.7, 10.8.]

9 1 ALTERNATIVE DERIVATION OF THE EQUATION FOR THE SIM- PLE PENDULUM The derivation in Cutnell & Johnson makes use of torque AND moment of inertia. You may be comfortable with their approach. If not we can avoid the moment of inertia as follows: Take the position of the bob to be at an angle to the vertical. Examine the forces tangential (direction- ) and perpendicular (direction r ) to its (circular) path of motion: F r = T mg cos = 0 (no motion as xed length) F = mg sin (acceleration along arc is ( )g sin )

10 We now observe that for small angles (less than about 10 degrees) the arc path is essentially at or horizontal ( x ) so that F x ' F. The horizontal displacement from the equilibrium position ( = 0, x = 0) is x = L sin [the actual displacement along the arc is between L sin and L tan, but these are essentially the same for small angles.] A critical step in any derivation for the simple or small angle pendulum is a simpli cation of sin and/or tan for small angles: the result is in each case. [see the triangle insert] Thus F x ' F (= mg sin ) ' mg and x = L sin ' L, so that a x =! 2 x, where a x = F x m ' g sin ' g; becomes g =! 2 L, and nally upon cancelling on each side: g =! 2 L or! 2 = g=l (10.16). [! = g L 1 2 ]

11 1.1 Caption for the triangle insert Take a thin triangle for small. If the hypoteneuse has unit length, then the adjacent side is cos and the opposite side is sin. Now because of the small angle the adjacent side is almost the same as the hypoteneuse, i.e. unit length, i.e. cos ' 1, and the opposite side is essentially an arc of a circle of unit radius, and therefore by the de nition of radian measure,, i.e. sin ' 1 With these approximations tan = cos sin '. We present below a table of selcted values to illustrate the range of angles over which the small angle approximation is useful - it boils down to the accuracy required.

12 (deg) (rad) sin tan =360 = 1: : : =360 = 3: : : =360 = 5: : : =360 = 6: : : =360 = 8: : : =360 = : : : =360 = : : : =360 = : : : =360 = : : : =360 = : : 5 : 577