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Extreme
position
Mean
position
O
Extreme
positionE M E
Extreme
position
Mean
position
O
Extreme
position
AGAR ISE CARTESIAN SE
UNDERSTAND KARE TO EK
EXTREME POINT + AND –VE.
- +
P called left
extreme or –ve
extreme.
R called right
exteme or + ve
extreme.
R
Isme bhi 2 Type
displacement hoga
particle agar left side ka
distanct cover karega to
–ve displacement and
RHD + displacement.
One osclllation = mean positin se start and
cover right extreme then left exteme then
again come mean position.
Koi bhi spring jab
movement karta hai to
oske displacement,
velocitiy and
acceleration me time ke
based continuous
changes ata hai jise
graph se show kiya hai.
Jise meths ke zarie
show karte hai.
SHM me 2 type hai linear
and angular.
Linear means straight
motion and angular mean
making angle motion.
Now remaining concept is
same because again
distance and force are in
opposite directin so same
forumula
F = -kx
F = ma
But Acc. to
newton second
law Force
formula is
F=ma.But Acceleration is a = dv/dt
Velocity is v = dx/dt
Now put v ka value
in acceleration
forumula dv me.
Left extreme
position
Right extreme
positionMean
position
PX
a = ?
V = ?
x = ?
Straight line me
object motion karte
ek point P per rukta
hai. Jiska distance x
& time t.
With help of x we
can fine
acceleration,
velocity and
distance.
We already found
a = d2x/dt2
- Sign indicate
acceleration and
displacement are in
opposite direction.
We know angular
velocity is const.
then…
S.H.M. As a Projection of U.C.M. on any Diameter
A O B
C
D
Let AB and CD be
the diameter of the
circle and ‘O’ be
the Centre of the
circle.
S.H.M. As a Projection of U.C.M. on any Diameter
AO
B
C
D
At time t=0,
the particle P0
is at point D
P0
From ‘P0’ draw
a perpendicular
on diameter AB
M0
S.H.M. As a Projection of U.C.M. on any Diameter
AO
B
C
DP0
M0
Now if particle
performing
UCM
P1
M1
M2P2
S.H.M. As a Projection of U.C.M. on any Diameter
A O B
C
D
Consider, a particle
performing U.C.M
with an angular
velocity ω along a
circle of radius ‘a’.
a
ω
S.H.M. As a Projection of U.C.M. on any Diameter
Suppose the
particle start from
position Po.
α
‘α’ is the initial
phase angle or
epoch.
Let, in time ‘t’ the
particle reaches
point P
θ = ?
S.H.M. As a Projection of U.C.M. on any Diameter
Magnitude matlab angle
, distance, a, v ka
measure asani se
karsakte hai. Aur saath
saath direction pata
karsakte hai motion
kidhar horaha hai.
OB wala particle zero se start
hua hai. Isliye angle 0. and DB
wala object distance cover
karke 90 degree angle bana
raha hai.
YEH wo phase hai jaha per
particle apni initial stage per
hai abhi movement start nahi
ki isliye angle 0 means mean
postion.
Simple pedulam me ek
heavy mass jise bob kehte
hai. String jisse bob tigh
kiya hai.
Pendulam ki length bob se
rigid ka distance hai.