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`        A simple harminic oscillator has a period of 0.01sec.and an amplitude of 0.2m.the speed in m/s.at the center of oscillation is`
2 years ago

Arun
23311 Points
```							Dear student

Start by using this equation for the displacement x of the oscillator: x = A*sin(w*t) where A is the amplitude, w is the angular frequency, and t is time. speed = dx/dt = A*w*cos(w*t) From your data A = 0.2m, & w = 2*pi/T = 628rad/s (T is the period). The speed at the centre of the oscillation is the speed when t = 0, i.e. when cos(w*t) = 1 speed at centre = 0.2*628 = 125.7m/s = 40*pi m/s
Regards

```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions