2. s
θ
PP’
K.E = 0
V = 0
P.E = max
= mgh
P.E = 0
V = max
K.E = max
=1/2 mv2
h
3. Useful formulae
Period of a simple pendulum
T = 2π √l/g
l = lenght of simple pendulum
g = acceleration due to gravity
T = period
Angular velocity ω = θ/t
Θ = angle turned by body
t = time takken
Linear velocity v = rω
4. Acceleration of SHM
Angular acceleration is the rate of change of
angular velocity with time
α = a/r
a = αr
Frequency – number of complete revolutions
per min
f = 1/T
5. Energy of SHM
shm
θ
PP’
K.E = 0
V = 0
P.E = max
= mgh
P.E = 0
V = max
K.E = max
=1/2 mv2
h
6. In an elastic spring, F =kx; F being the force
used to stretch the spring by a distance, and
k = force constant
Work done = ½ force x displacement x
Kx/2 x X = ½ kx
Therefore energy in the spring = ½ kx2
The maximum P.E = ½ kx2
7. K.E at any instant = ½ mv2
At position of maximum displacement v = 0
therefore K.E = 0
At that point the total energy = K.E + P.E = 0
+ ½ kx2
Total energy = ½ kx2
8. At equilibrum when K.E is max, P.E = 0
Total energy will be = K.E + P.E = ½ mv2 + 0
= ½ mv2
But at any other positions besides position of
maximum displacement and position of
equilibrum
total energy = KE + PE = ½ mv2 + ½ kx2
9. At position P and P’ which is when there is
maximum displacement TE = PE + 0 = mgh
θ
PP’
K.E = 0
V = 0
P.E = max
= mgh
P.E = 0
V = max
K.E = max
=1/2 mv2
h
