BSC PHYSICS NOTES

1. SHM & UNIFORM CIRCUAR MOTION
Consider a particle performing uniform circular motion with angular
velocity “ω”. At time "𝑡0" the radius 𝑂𝑃̅̅̅̅ makes an angle “ɸ” with x-
axis. Later at time “t” the radius 𝑂𝑃̅̅̅̅ 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 𝑂𝑄̅̅̅̅.
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
[−ω𝑅𝑠𝑖𝑛(ωt + ɸ)]
𝑎( 𝑡) = −𝜔2
𝑅𝑐𝑜𝑠(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) = 𝑥 𝑚
So
𝑎( 𝑡) = −𝑥 𝑚 𝜔2
INSTANTANEOUS VELOCITY OF THE PROJECTION
Consider a particle performing uniform circular motion with angular
velocity “ω”. At time "𝑡0" the radius 𝑂𝑃̅̅̅̅ makes an angle “ɸ” with x-
axis. Later at time “t” the radius 𝑂𝑃̅̅̅̅ 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 𝑂𝑄̅̅̅̅.

2. Here
𝑣 𝑝→ Velocity of particle
𝑣 𝑥→ 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
𝑑𝑣
𝑑𝑡
= −𝜔2
𝑥
𝑑𝑣
𝑑𝑥
.
𝑑𝑥
𝑑𝑡
= −𝜔2
𝑥
𝑑𝑣
𝑑𝑥
. 𝑣 = −𝜔2
𝑥
𝑣𝑑𝑣 = −𝜔2
𝑥𝑑𝑥
Taking integral on both sides;
∫ 𝑣𝑑𝑣
𝑣
𝑣0
= −𝜔2
∫ 𝑥𝑑𝑥
𝑥
𝑥0
∫ 𝑣𝑑𝑣
𝑣
𝑣0
= 𝜔2
∫ 𝑥𝑑𝑥
𝑥0
𝑥
𝑣2
2
│0
𝑣
= 𝜔2
𝑥2
2
│ 𝑥
𝑥0
𝑣2
= 𝜔2
(𝑥0
2
− 𝑥2
)
𝑣 = 𝜔√(𝑥0
2 − 𝑥2)
For maximum velocity;
𝑥 = 0
𝑣 𝑚𝑎𝑥 = 𝜔𝑥0