### JEE Mains Previous Years Questions with Solutions

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### AIEEE 2002

In a simple harmonic oscillator, at the mean position
A
kinetic energy is minimum, potential energy is maximum
B
both kinetic and potential energies are maximum
C
kinetic energy is maximum, potential energy is minimum
D
both kinetic and potential energies are minimum.

## Explanation

$K.E = {1 \over 2}k\left( {{A^2} - {x^2}} \right);\,\,\,U = {1 \over 2}k{x^2}$
At the mean position $x=0$
$\therefore$ $K.E. = {1 \over 2}k{A^2} =$ Maximum and $U=0$
2

### AIEEE 2002

If a spring has time period $T,$ and is cut into $n$ equal parts, then the time period of each part will be
A
$T\sqrt n$
B
$T/\sqrt n$
C
$nT$
D
$T$

## Explanation

Let the spring constant of the original spring be $k.$

Then its time period $T = 2\pi \sqrt {{m \over k}}$ where $m$ is the mass of oscillating body.
When the spring is cut into $n$ equal parts, the spring constant of one part becomes $nk.$ Therefore the new time period,

$T' = 2\pi \sqrt {{m \over {nk}}} = {T \over {\sqrt n }}$