A bob of mass m is tied with a light rope and is oscillating about the mean position M. If A and B are the extreme positions, then Horizontal acceleration of the bob at A is
Question
A bob of mass m is tied with a light rope and is oscillating about the mean position M. If A and B are the extreme positions, then Horizontal acceleration of the bob at A is
Solution
The horizontal acceleration of the bob at position A can be calculated using the formula for the acceleration of a simple harmonic oscillator, which is given by:
a = -ω²x
where:
- a is the acceleration,
- ω is the angular frequency of the oscillation, and
- x is the displacement from the mean position.
At the extreme position A, the displacement x is at its maximum value, which we can call A. Therefore, the acceleration of the bob at A is:
a = -ω²A
This acceleration is directed towards the mean position M, which is why there is a negative sign in the formula. The magnitude of the acceleration is ω²A.
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