First slide

Simple hormonic motion

Question

A simple harmonic oscillator is of mass 0.1kg. It is oscillating with a frequency of $\large \frac{5}{\pi }Hz$ . If its amplitude of vibration is 5cm find the force acting on the particle at its extreme position.

Easy

A

0.5N
B

2.5N
C

3.5N
D

4.5N

Solution

$\large M\, = \,0.1\,kg;\,f\, = \,\frac{5}{\pi }Hz;\,A\, = \,\frac{5}{{100}}m$
$\large \omega \, = \,2\pi f\, = \,2\pi \times \frac{5}{\pi }\, = \,10\,rad/\sec$
$\large {F_{extreme}}\, = \,m{a_{extreme}}\, = \,m\left( {{\omega ^2}A} \right)$
$\large \, = \,\left( {0.1} \right)\left( {100} \right)\frac{5}{{100}}\, = \,0.5\,Newton$

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