Simple Harmonic Motion(SHM)

If in an oscillatory motion the acceleration is directly proportional to displacement and is always directed towards the mean position then it is called SHM
If F be the force applied on the particle and x be the displacement then,
F α –x (-ve sign shows they are opposite in direction)
.-. F = -kx----------1 where k is force constants
If m be the mass of the particle then from Newtons second law of motion.
F = ma-------2
From 1 and 2
ma = -kx
a = -k/m*x
a = -ω^2x----4
This is the equation of simple harmonic motion

Motion of a helical(loded Spring)
Let us consider a negligible helical spring suspended through a rigid support. Now it will attach a mass in its free end then it will be elongated through same distance. let it be L. Then according to Hooke’s law, the restoring force is directly proportional to the elongation produces
F1 α L
F1 = -CL-------------1 (where –ve shows that the force is in the opposite direction of elongation.
Now let the mass be pulled down through a distance x and then realesed. The spring then starts to oscillate. Now elongation on spring will be l+x & restoring force is given as,
F2 = -C(l+x)--------------2
The resulting force is
F = F1 + F2 = C(L+x) + cl = -cx
.-. F = -Cx----------------3
Now
From Newtons second law of motion,
F=ma------------------4
From 3 and 4
ma = -Cx
a = -c/m * x------------5
equation 5 is similar to the equation of SHM. Hence loaded spring executes SHM.
Comparint the equation with SHM we can have
ω^2 = C/m
Now time period of oscillation T = 2 /√(c/m)

Mass attached in two spring
Let us consider a mass m be attached in between two strings s1 and s2 having force constant c1 and c2 respectively.
Let the mass and system be placed in a frictionless surface. Now suppose the mass be pulled toward the right direction through the distance x. Then clearly the spring S1 will be elongated by distance x and spring S2 will be compressed through distance x.
THerefore from Hook’s law restoring force acting on the mass will be
F = -C1x – C2x
F = -(C1+C2)x-------------1
This cause oscillation in mass from Newton’s second law of motion force acting on mass
F=ma-------------2
From 1 and 2
ma = -(C1+C2)x
a = -(C1+C2)/m *x--------3
This equation is similar to equation of SHM. Hence this executes SHM


Simple Pendulum
A point heavy mass suspended with mass less, weightless, extensible string through a rigid support. It is also called ideal pendulum.
Drawbacks of simple pendulum.
It is impossible to have heavy point mass.
It is impossible to have weightless string.
There will be relative motion between the bob and string at extreme position.
Length of pendulum, is difficult to measure exactly accurately.
Compound Pendulum
Compound pendulum is a rigid body of whatever shape capable to oscillate freely about a horizontal axis passing through it.
The point in which pendulum is suspended through rigid suppost is called point of suspension.
The distance between point of suspension and centre of gravity of the pendulum is called the length of pendulum.

Consider a compound pendulum having mass m and length l i.e SG = L. Let pendulum be displace through an angle ϴ. At this instant the weigh of pendulum mf act vertically downward direction. Then the restoring force that tends to bring pendulum torque acting on the body will be,
Ʈ = -mglsin ϴ------------1
Now,
If I be the moment of intertia and
α = a be angular accleration then the restoring torque will be given as
Ʈ = I * α = I * a--------------2
From equation 1 and 2
I * a = -mglsin ϴ
a = -mgl/I * ϴ --------------3
This equation 3 is also similar to equation of SHM. Hence compound also executes SHM.


Point of Suspension and point of oscillation are interchangeable. We have time period for a compound pendulum having length L as
T = 2 √((k^2/L + L)/g)
T = 2 π(√l1 + l)/g-------------1
such that k^2/L = L1
Now, let point of oscillation becomes point of suspension. Then length of pendulum
k^2/L = L1
Hence, time period in this case will be
T1 = 2 π(√k2/l1+l)/g
We have k^2 = LL1
T1 = 2π√LL1+L1/(l/2)
= 2 π√L1 + l)/g-------------2