Now suppose the ball started from a point after than  = 0 say  then :
So here
= phase constant
# illustration :-
Two SHM's of amplitude  with initial hase constanta as  start with angular velocities w and 4w respectively. Find the time after which they are in phase.
Question:-
That they are in phase.
Solution :- "In phase" means having same phase. Which means they are at the same position instant & their velocities are also in same direction.
* Tip The problem can be solved easily using two techniques : (1) circle method for SHM
(2) Relative Concept.
Initial angle between two = 
So time taken = 
Dumb Question : Is only the functiony =  SHM?
Solution :- No, any function which satiesfies the condition that if y = y(t); then  , can represent the SHM.
Accelaration and velocity in SHM :
if  = 0 and 
The graphs if drawn are :-
[of  = 0] . All v, y, w have different Amplitude.
Differential Equation of accelation of particle SHM is
# Illustration : If equation of a motion is  Show that it is SHM and find maximum accelaration and maximum velocity.
Solution :-
Hence it is SHM.
Now
* Energy terms in SHM :
(a) Kinetic Energy :
(b) Potential Energy : 
Derivation : Work done by Force F during displacement from x to x + d is
Integrate both sides
Now  Potential Energy = - Work Done by force.
So, 
Let at 
So 
So total Energy in SHM :
 for a particular SHM with w
|
