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Hrishant Goswami Grade: 10
`        The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2 cm, 1 ms-1 and 10 ms-2 at a certain instant. Find the amplitude and the time period of the motion. `
3 years ago

## Answers : (1)

Jitender Pal
365 Points
```										Sol. Given that, at a particular instant,
X = 2cm = 0.02m
V = 1 m/sec
A = 10 msec–2
We know that a = ω2x
⇒ ω = √(a/×) = √(10/0.02) = √500 = 10√5
T = 2π/ω = 2π/10√5 = 2 × 3.14/10 × 2.236 = 0.28 seconds.
Again, amplitude r is given by v = ω (√(r^2-X^2 ))
⇒ v2 = ω2 (r2 – x2)
1 = 500 (r2 – 0.0004)
⇒ r = 0.0489 = 0.049 m
∴ r = 4.9 cm.

```
3 years ago
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