2. Characteristics of Simple Harmonic Motion A body is in simple harmonic motion if its acceleration is proportional to its displacement from a fix point O, and is always directed towards that fixed point. a – x a = – 2 x Negative sign : Direction of a is opposite to x O -x o x o a a
3. Some Common Examples of Simple Harmonic Motion Vibrating Tuning fork A weight on a spring A boy on a swing 200 grams
4. Definition : SHM may be defined as the motion of a body which is subjected to a resultant force which is directly proportional to the displacement of the body from a fixed point and always directed towards that fixed point F – x F = – kx
5. O -x o x o Point of equilibrium : a = 0 F = 0 a x Measured from the point of equilibrium Amplitude : = x o =Maximum distance
7. Kinematics of Simple Harmonic Motion F = - kx ma = - kx a = k m - x a = - 2 x compare 2 = k m = angular frequency = 2 f = 2 T
8. F against x graph & a against x graph F x 0 x o -x o a x 0 x o -x o 2 x o - 2 x o F = -kx a = - 2 x
9. a = - 2 x d 2 x dt 2 = - 2 x x = x o sin ( t + ) General solution dx dt = x o cos ( t + ) v = d 2 x dt 2 = - 2 x o sin ( t + ) a = = - 2 x shown x = x o sin ( t + ) phase Phase constant
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11. graph : x - t , v - t , a - t x t 0 x = x o sin t x o -x o T 2T v T 0 - x o x o v = x o cos t t 2T a t 0 a = - 2 x o sin t 2 x o - 2 x o T 2T t=0, v= x o t=0, a=0
12. phase = t + /2 phase = t phase x = x o sin t v = x o cos t To compare the phase : They must all be expressed as same function (either as sine or cosine function) Phase difference : =( t+ ) - t 2 = 2 rad = x o sin ( t + /2) v leading x x t 0 x o -x o T 2T v T 0 - x o x o t 2T a t 0 2 x o - 2 x o T 2T
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14. compare the motion x t 0 x o -x o T 2T v T 0 - x o x o t 2T a t 0 2 x o - 2 x o T 2T
15. example : -x o x o 0 t=0 x x = x o sin ( t + ) General solution Substitute : t=0 , x = x o to find x o = x o sin ( (0) + ) sin = 1 = 2 Therefore , equation of x is: x = x o sin ( t + /2) t = x o cos t
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17. solution : x = 4.0 cos ( 2 t + ) 3 a) i) Displacement = ? t = 2.0 s ; x = 4.0 cos ( 2 (2) + ) 3 = 2.0 m ii) velocity = ? v = dx dt = -2 (4.0) sin ( 2 t + ) 3 v = -8.0 sin (4 + ) t = 2.0 s ; 3 = -21.8 ms -1
18. solution : iii) acceleration = ? a = dv dt = -16.0 2 cos ( 2 t + ) 3 t = 2.0 s ; a = -16.0 2 cos ( 2 (2) + ) 3 = -79.0 ms -2 iv) Phase = ? x = 4.0 cos ( 2 t + ) 3 Phase = 2 t + 3 t = 2.0 s ; Phase = 2 (2) + 3 = 13.6 rad
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20. solution : x = 4.0 cos ( 2 t + ) 3 b) i) frequency = ? x = x o cos ( t + ) General solution : compare = 2 2 f = 2 f = 1.0 Hz ii) Period = ? T = 1 f = 1.0 s
21. example : Figures (1) and (2) are the displacement-time graph and acceleration-time graph respectively of a body in simple harmonic motion. x,m 3 -3 t,s T 0 (1) a,ms -2 12 -12 t,s T 0 (2) What is the frequency of the motion? Write an expression to represent the variation of displacement x with time t.
22. solution : frequency = ? x o = 3 2 x o = 12 2 (3) = 12 substitute = 2 rad s -1 2 f = 2 f = 0.318 Hz x,m 3 -3 t,s T 0 (1) a,ms -2 12 -12 t,s T 0 (2)
23. solution : expression = ? x = x o sin ( t + ) General solution x o = 3 , = 2 ; x = 3 sin (2t + ) At t = 0, x = 3 3 = 3 sin (2(0) + ) sin = 1 = 2 rad Therefore, the expression is x = 3 sin (2t + ) 2 x = 3 cos 2t
24. variation of v with x a = dv = - 2 x dt dv x dx = - 2 x dx dt v dv = - 2 x dx v 2 2 = - 2 x 2 2 + C When x = x o , v = 0 0 = - 2 x o 2 2 + C C = 2 x o 2 2 Hence ; v 2 2 = - 2 x 2 2 2 x o 2 2 + v 2 = - 2 x 2 + 2 x o 2 v = x o 2 – x 2 v dv dx = - 2 x
25. Graph : v - x x o - x o x o -x o v x 0 -x o O x o v = x o 2 – x 2
29. energy in simple harmonic motion Total energy = E = K + U At point, x = 0 : K = ½ mv 2 = ½ m 2 (x o 2 – x 2 ) K = ½ m 2 (x o 2 – 0) = ½ m 2 x o 2 At point, x = 0 : U = 0 = maximum = ½ m 2 x o 2 + 0 = ½ m 2 x o 2 constant At point, x = x o : K = ½ m 2 (x o 2 –x o 2 ) = 0 = minimum -x o x o 0 x
30. energy in simple harmonic motion At any displacement,x E = K + U ½ m 2 x o 2 = ½ m 2 (x o 2 – x 2 ) + U = ½ m 2 x o 2 – ½ m 2 x 2 + U U = ½ m 2 x 2 -x o x o 0 x
31. Graph : E - x K max = ½ m 2 x o 2 U = ½ m 2 x 2 K min = 0 ; ; E total =½ m 2 x o 2 K=½ m 2 (x o 2 – x 2 ) U = ½ m 2 x 2 0 E x -x o x o -x o x o 0 x
32. Graph : F - x U = ½ m 2 x 2 F = – dU dx = – ½ (2) m 2 x = – m 2 x constant Therefore : F x -x o x o m x o -m x o -x o x o 0 x 0 F x
33. Graph : E - t x = x o sin ( t + ) General solution 0 = x o sin ( (0) + ) At t = 0 , x = 0 0 = sin = 0 x = x o sin t Therefore : The equation is -x o x o 0 x
34. Graph : E - t x = x o sin t v = dx dt = x o cos t K = ½ mv 2 = ½ m 2 x o 2 cos 2 t U = ½ m 2 x 2 = ½ m 2 x o 2 sin 2 t E total =½ m 2 x o 2 T 2T K U -x o x o 0 x 0 E t