what are the criterias for a motion to be SHM.... ANSWER in points...... explain it with example......... I mean if a displacement equation is given then price it to be SHM
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if force acting on particle directly proportional to displacement of its from mean position . but is directed in opposite to displacement then motion is known as SHM
according to above,
F directly proportional to (-y)
F= -ky {k is proportionality constant .
ma= -ky {Law of motion F=ma
a= -(k/m)y
let k/m =w^2 {w is known as angular vel.
now,
a= -w^2y
because a is acceleration so,
a=d^2y/dt^2
use this here ,
d^2y/dt^2 = -w^2y
multiply with both side (2dy/dt)
2dy/dt (d^2y/dt^2)= -w^2y (2ydy/dt )
d/dt{d^2y/dt^2}= -w^2dy^2/dt
d/dt {(dy/dt)^2} = - d (w^2y^2)/dt
now integrate
(dy/dt )^2=-w^2y^2+C
if y=A (maximum amplitude =A
dy/dt=0
then C=w^2A^2
now equation convert in ---------
dy/dt=wroot {A^2-y^2}
now use differential equation
y= Asin (wt+_@)
where @ is phase difference
according to above,
F directly proportional to (-y)
F= -ky {k is proportionality constant .
ma= -ky {Law of motion F=ma
a= -(k/m)y
let k/m =w^2 {w is known as angular vel.
now,
a= -w^2y
because a is acceleration so,
a=d^2y/dt^2
use this here ,
d^2y/dt^2 = -w^2y
multiply with both side (2dy/dt)
2dy/dt (d^2y/dt^2)= -w^2y (2ydy/dt )
d/dt{d^2y/dt^2}= -w^2dy^2/dt
d/dt {(dy/dt)^2} = - d (w^2y^2)/dt
now integrate
(dy/dt )^2=-w^2y^2+C
if y=A (maximum amplitude =A
dy/dt=0
then C=w^2A^2
now equation convert in ---------
dy/dt=wroot {A^2-y^2}
now use differential equation
y= Asin (wt+_@)
where @ is phase difference
abhi178:
shati this is so tough derivation please understand concept only okay
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