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प्रश्न
Assuming the expression for displacement of a particle starting from extreme position, explain graphically the variation of velocity and acceleration w.r.t. time.
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उत्तर
Displacement-time graph:
i.
At extreme position, α = π/2
Displacement, x = A cos ωt
Velocity time graph:
i.
At extreme position, α = π/2
Velocity of a particle is v = - Aω sin ωt
ii. Table:
Substituting ω = 2π/T in above equation,
| Time (t) | Phase (ωt) | Velocity (v) |
| 0 | 0 | 0 |
| T/4 | π/2 | -Aω |
| T/2 | π | 0 |
| 3T/4 | 3π/2 | Aω |
| T | 2π | 0 |
Graph:
Acceleration-time graph:
i. At extreme position,α = π/2
Acceleration of a particle is,
a =-Aω2 cos ωt
ii. Table:
Substituting ω = 2π/T in above equation
| Time (t) | Phase (ωt) | Velocity (v) |
| 0 | 0 | -Aω2 |
| T/4 | π/2 | 0 |
| T/2 | π | Aω2 |
| 3T/4 | 3π/2 | 0 |
| T | 2π | -Aω2 |
iii. Graph:
Conclusions:
i. Displacement, velocity and acceleration of S.H.M. are periodic functions of time.
ii. The displacement and acceleration curves are sine curves whereas velocity curve is
cosine curve (α = 0).
iii. The phase difference between displacement and acceleration is of π radian.
iv. The phase difference between displacement and velocity and velocity and acceleration
is of π/2 radian.
v. The displacement and acceleration is maximum at extreme position whereas velocity is
minimum at the same position.
vi. All curves repeat same path after phase of 2π radian.
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