A simple harmonic oscillator is of mass $$0.100\ kg$$. It is oscillating with a frequency of $$\dfrac{5}{\pi}$$ Hz. If its amplitude of vibration is $$5\ cm$$, the force acting on the particle at its extreme position is
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Solution
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The expression for force and the its relation with amplitude is given as:
$$ F = m {\omega}^{2} A$$
We know that the angular velocity is given as:
$$\omega = 2\pi f = 10 rad / sec$$
Substitute the value in the expression for force for the oscillator:
$$F = 0.1 \times 100\times \dfrac{5}{100}$$
$$ = 0.5 N $$
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