First slide

Simple hormonic motion

Question

The potential energy of a particle oscillating on x-axis is given as U=20+(x–2)² where ‘U’ is in joules and ‘x’ is in metres. Its mean position is

Easy

A

0
B

1
C

2
D

4

Solution

$\large U\, = \,20 + {\left( {x - 2} \right)^2}\, \Rightarrow \,\frac{{dU}}{{dx}}\, = \,2\left( {x - 2} \right)$
at mean position, PE is minimum,
ie $\large \frac{{dU}}{{dx}}\, = \,0$
2 (x – 2) = 0 $\large \Rightarrow$ x – 2 = 0 $\large \Rightarrow$ x = 2

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