 JEE Mains Previous Years Questions with Solutions

4.5
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1

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 $\delta = {{3\pi } \over 4}.$
2

AIEEE 2007

Two springs, of force constant ${k_1}$ and ${k_2}$ are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f.$ If both ${k_1}$ and ${k_2}$ are made four times their original values, the frequency of oscillation becomes A
$2f$
B
$f/2$
C
$f/4$
D
$4f$

Explanation

The two springs are in parallel.

$f = {1 \over {2\pi }}\sqrt {{{{K_1} + {K_2}} \over m}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$f' = {1 \over {2\pi }}\sqrt {{{4{K_1} + 4{K_2}} \over m}}$

$= {1 \over {2\pi }}\sqrt {{{4\left( {{K_1} + 4{K_2}} \right)} \over m}}$

$= 2\left( {{1 \over {2\pi }}\sqrt {{{{K_1} + {K_2}} \over m}} } \right)$

$= 2f\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ from eqn. $\left( i \right)$
3

AIEEE 2007

A particle of mass $m$ executes simple harmonic motion with amplitude a and frequency $v.$ The average kinetic energy during its motion from the position of equilibrium to the end is
A
$2{\pi ^2}\,m{a^2}{v^2}$
B
${\pi ^2}\,m{a^2}{v^2}$
C
${1 \over 4}\,m{a^2}{v^2}$
D
$4{\pi ^2}m{a^2}{v^2}$

Explanation

KEY CONCEPT : The instantaneous kinetic energy of a particle executing $S.H.M.$ is given by

$K = {1 \over 2}m{a^2}{\omega ^2}{\sin ^2}\omega t$

$\therefore$ average $K.E. = < K > = < {1 \over 2}m{\omega ^2}{a^2}{\sin ^2}\omega t >$

$= {1 \over 2}m\omega {}^2{a^2} < {\sin ^2}\omega t >$

$= {1 \over 2}m{\omega ^2}{a^2}\left( {{1 \over 2}} \right)$

$\left( \, \right.$ as $\left. { < {{\sin }^2}\theta > = {1 \over 2}} \right)$

$= {1 \over 4}m{\omega ^2}{a^2} = {1 \over 4}m{a^2}{\left( {2\pi v} \right)^2}$

$\left( \, \right.$ $\left. {\omega = 2\pi v} \right)$

or, $\,\,\,\,\, < K > = {\pi ^2}m{a^2}{v^2}$
4

AIEEE 2006

The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7$ $mm,$ is $4.4$ $m/s.$ The period of oscillation is
A
$0.01$ $s$
B
$10$ $s$
C
$0.1$ $s$
D
$100$ $s$

Explanation

Maximum velocity,

${v_{\max }} = a\omega ,\,\,\,\,\,{v_{\max }} = a \times {{2\pi } \over T}$

$\Rightarrow T = {{2\pi a} \over {{v_{\max }}}} = {{2 \times 3.14 \times 7 \times {{10}^{ - 3}}} \over {4.4}} \approx 0.01\,s$