 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2011

A mass $M,$ attached to a horizontal spring, executes $S.H.M.$ with amplitude ${A_1}.$ When the mass $M$ passes through its mean position then a smaller mass $m$ is placed over it and both of them move together with amplitude ${A_2}.$ The ratio of $\left( {{{{A_1}} \over {{A_2}}}} \right)$ is :
A
${{M + m} \over M}$
B
${\left( {{M \over {M + m}}} \right)^{{1 \over 2}}}$
C
${\left( {{{M + m} \over M}} \right)^{{1 \over 2}}}$
D
${M \over {M + m}}$

Explanation

The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.

$\therefore$ $M{v_1} = \left( {M + m} \right){v_2}$

$M{A_1}\sqrt {{k \over M}} = \left( {M + m} \right){A_2}\sqrt {{k \over {m + M}}}$

$\therefore$ $\left( {V = A\sqrt {{k \over M}} } \right)$

${A_1}\sqrt M = {A_2}\sqrt {M + m}$
$\therefore$ ${{{A_1}} \over {{A_2}}} = \sqrt {{{m + M} \over M}}$
2

AIEEE 2011

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean position is separated by distance ${X_0}\left( {{X_0} > A} \right)$. If the maximum separation between them is $\left( {{X_0} + A} \right),$ the phase difference between their motion is:
A
${\pi \over 3}$
B
${\pi \over 4}$
C
${\pi \over 6}$
D
${\pi \over 2}$

Explanation

For ${X_0} + A$ to be the maximum separation $y$ one body is at the mean position, the other should be at the extreme.
3

AIEEE 2009

If $x,$ $v$ and $a$ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $T,$ then, which of the following does not change with time?
A
$aT/x$
B
$aT + 2\pi v$
C
$aT/v$
D
${a^2}{T^2} + 4{\pi ^2}{v^2}$

Explanation

For an $SHM,$ the acceleration $a = - {\omega ^2}x$ where ${\omega ^2}$ is a constant. Therefore ${a \over x}$ is a constant. The time period $T$ is also constant. Therefore ${{aT} \over x}$ is a constant.
4

AIEEE 2007

The displacement of an object attached to a spring and executing simple harmonic motion is given by $x = 2 \times {10^{ - 2}}$ $cos$ $\pi t$ metre. The time at which the maximum speed first occurs is
A
$0.25$ $s$
B
$0.5$ $s$
C
$0.75$ $s$
D
$0.125$ $s$

Explanation

Here, $x = 2 \times {10^{ - 2}}\cos \,\pi \,t$

$\therefore$ $v = {{dx} \over {dt}} = 2 \times {10^{ - 2}}\,\pi \sin \pi t$

For the first time, the speed to be maximum,

$\sin \pi t = 1$ or, $\sin \pi t = \sin {\pi \over 2}$

$\Rightarrow \pi t = {\pi \over 2}\,\,\,$ or, $\,\,\,\,t = {1 \over 2} = 0.5\,\sec .$