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`        A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm gong towards positive x-direction. Write the equation for the displacement of the particle at t = 4 s. `
3 years ago

Kevin Nash
332 Points
```										Sol. Given, r = 10cm.
At t = 0, x = 5 cm.
T = 6 sec.
So, w = 2π/T = 2π/6 = π/3 Sec-1
At, t = 0, x = 5 cm.
So, 5 = 10 sin (w × 0 + ∅) = 10 sin ∅ [y = r sin wt]
Sin ∅ = 1/2 ⇒ ∅ = π/6
∴ Equation of displacement x = (10cm) sin (π/3)
(ii) At t = 4 second
X = 10 sin [π/3 × 4 × π/6] = 10 [8π+π/6]
= 10 sin (3π/2) = 10 sin (π = π/2) = - 10 sin (π/2) = - 10
Acceleration a = – w2x = – (π2/9) × (- 10) = 10.9 = 0.11 cm/sec.

```
3 years ago
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