Shm
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Muhammad Azhar [email protected] MS. Physics, NEDUET 03333406851 SHM & UNIFORM CIRCUAR MOTION Consider a particle performing uniform circular motion with angular velocity “ω”. At time {t} rsub {0} the radius OP makes an angle “ɸ” with x-axis. Later at time “t” the radius OP makes an angle “ωt+ɸ” with x-axis. The projection of this particle move back & forth around the centre. The position of projection of the particle is represented by OQ. At time “t” the position of projection is given by; x ( t) =Rcos( ωt +ɸ ) Differentiate with respect to “t” dx ( t ) dt = d dt [ Rcos ( ωt+ ɸ) ] v ( t) =−ωRsin ( ωt + ɸ) Now differentiate with respect to “t” dv ( t ) dt = d dt [−ω Rsin( ωt +ɸ) ] a ( t ) =−ω 2 Rcos ( ωt +ɸ ) ∴Rcos ( ωt +ɸ ) =x ( t ) a ( t ) =−ω 2 x ( t ) ∴ω=constant a ( t ) =−( constant ) x ( t) So a ( t ) ∝−x ( t) The relation shows that the projection of the particle performing uniform circular motion execute SHM. At t=0, and ɸ=0 Rcos ( ωt +ɸ ) =x m So a ( t ) =−x m ω 2 INSTANTANEOUS VELOCITY OF THE PROJECTION Consider a particle performing uniform circular motion with angular velocity “ω”. At time
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Transcript of Shm
Muhammad Azhar [email protected]. Physics, NEDUET 03333406851
SHM & UNIFORM CIRCUAR MOTION
Consider a particle performing uniform circular motion with angular velocity “ω”. At time {t} rsub {0} the radius OP makes an angle “ɸ” with x-axis. Later at time “t” the radius OP makes an angle “ωt+ɸ” with x-axis. The projection of this particle move back & forth around the centre. The position of projection of the particle is represented byOQ.
At time “t” the position of projection is given by;
x (t )=Rcos (ωt+ɸ)
Differentiate with respect to “t”
d x (t )dt
= ddt
[Rcos (ωt+ɸ )]
v ( t )=−ωRsin (ωt+ɸ )
Now differentiate with respect to “t”
d v (t )dt
= ddt
[−ωRsin (ωt+ɸ )]
a (t )=−ω2Rcos (ωt+ɸ )
∴Rcos (ωt+ɸ )=x (t )
a ( t )=−ω2 x (t )
∴ω=constant
a ( t )=−(constant ) x ( t )
So
a ( t )∝−x (t )
The relation shows that the projection of the particle performing uniform circular motion execute SHM.
At t=0, and ɸ=0
Rcos (ωt+ɸ )=xm
Soa ( t )=−xmω2
INSTANTANEOUS VELOCITY OF THE PROJECTION
Consider a particle performing uniform circular motion with angular velocity “ω”. At time {t} rsub {0} the radius OP makes an angle “ɸ” with x-axis. Later at time “t” the radius OP makes an angle “ωt+ɸ” with x-axis. The projection of this particle move back & forth around the centre. The position of projection of the particle is represented byOQ.
Muhammad Azhar [email protected]. Physics, NEDUET 03333406851
Here
vp→ Velocity of particle
vx→ Horizontal component of velocity of particle
v y→ Vertical component of velocity of particle
v→ Velocity projection of particle
The acceleration of projection is given by;
a=−ω2 x
d vdt
=−ω2 x
d vdx. d xdt
=−ω2 x
d vdx. v=−ω2 x
vdv=−ω2 xdx
Taking integral on both sides;
∫v0
v
vdv=−ω2∫x0
x
xdx
∫v0
v
vdv=ω2∫x
x0
xdx
v2
2│
0
v
¿ω2 x2
2│x
x0
v2=ω2(x02−x2)
v=ω√(x02−x2)
For maximum velocity;
x=0
vmax=ωx0
