Tuesday, January 13, 2015

SHM Chapter 17 ke,pe,te vs t

a)    Variation with time of energy in simple harmonic motion

http://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_SHM17/SHM17_Simulation.xhtml




If the variation with time of displacement is as shown, then the energies should be drawn as shown.


recalling Energy formula
K E = 1 2 m v 2
P E = 1 2 k x 2 in terms of time t,
x = x0 sin(ωt)
differentiating with t gives
v = v0 cos (ωt)

K E = 1 2 m v 2 = 1 2 m ( x 0 ωc o s ( ωt ) ) 2 = 1 2 m (x 0 2 ω2 )c o s 2 ( ωt ) ) similarly
P E = 1 2 k x 2 = 1 2 k ( x 0 s i n ( ωt ) ) 2 = 1 2 ( m ω2 ) x 0 2 s i n 2 ( ωt ) )  
therefore total energy is a constant value in the absence of energy loss due to drag (resistance)
T E = K E + P E = 1 2 m ω2 x 0 2 ( s i n 2 ωt + c o s 2 ωt ) = 1 2 m ω2 x 0 2





this is how the x vs t looks together of the energy vs t graphs



the table shows some of the common values

general energy formula SHM energy formula when t = 0 when  t = T 4 when  t = T 2 when t = 3 T 4 when t = T
K E = 1 2 m v 2 K 1 2 m ω2 x 0 2 0 1 2 m ω2 x 0 2 0 1 2 m ω2 x 0 2
P E = 1 2 k x 2 P E = 1 2 ( m ω2 ) x 0 2 s i n 2 ( ωt ) ) 0
1 2 m ω2 x 0 2
0 1 2 m ω2 x 0 2 0
T E = K E + P E  T E = 1 2 m ω2 x 0 2 1 2 m ω2 x 0 2 1 2 m ω2 x 0 2
 
1 2 m ω2 x 0 2 1 2 m ω2 x 0 2 1 2 m ω2 x 0 2




Model:


http://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_SHM17/SHM17_Simulation.xhtml