JEE Mains Previous Years Questions with Solutions

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1

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.
2

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.
3

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, $\,\,\,\,t = {1 \over 2} = 0.5\,\sec .$
4

AIEEE 2007

A point mass oscillates along the $x$-axis according to the law $x = {x_0}\,\cos \left( {\omega t - \pi /4} \right).$ If the acceleration of the particle is written as $a = A\,\cos \left( {\omega t + \delta } \right),$ then
A
$A = {x_0}{\omega ^2},\,\,\delta = 3\pi /4$
B
$A = {x_0},\,\,\delta = - \pi /4$
C
$A = {x_0}{\omega ^2},\,\,\delta = \pi /4$
D
$A = {x_0}{\omega ^2},\,\,\delta = - \pi /4$

Explanation

Here,

$x = {x_0}\cos \left( {\omega t - \pi /4} \right)$

$\therefore$ Velocity, $v = {{dx} \over {dt}} = - {x_0}\omega \sin \left( {\omega t - {\pi \over 4}} \right)$

Acceleration,

$a = {{dv} \over {dt}} = - {x_0}{\omega ^2}\cos \left( {\omega t - {\pi \over 4}} \right)$

$= {x_0}{\omega ^2}\cos \left[ {\pi + \left( {\omega t - {\pi \over 4}} \right)} \right]$

$= {x_0}{\omega ^2}\cos \left( {\omega t + {{3\pi } \over 4}} \right)$ $\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

Acceleration, $a = A\cos \left( {\omega t + \delta } \right)$ $\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

Comparing the two equations, we get

$A = {x_0}{\omega ^2}$ and $\delta = {{3\pi } \over 4}.$