### JEE Mains Previous Years Questions with Solutions

4.5
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1

### AIEEE 2005

Two simple harmonic motions are represented by the equations ${y_1} = 0.1\,\sin \left( {100\pi t + {\pi \over 3}} \right)$ and ${y_2} = 0.1\,\cos \,\pi t.$ The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is
A
${\pi \over 3}$
B
${{ - \pi } \over 6}$
C
${\pi \over 6}$
D
${{ - \pi } \over 3}$

## Explanation

${v_1} = {{d{y_1}} \over {dt}} = 0.1 \times 100\pi \cos \left( {100\pi t + {\pi \over 3}} \right)$

${v_2} = {{d{y_2}} \over {dt}} = - 0.1\pi sin\pi t = 0.1\pi cos\left( {\pi t + {\pi \over 2}} \right)$

$\therefore$ Phase diff. $= {\phi _1} - {\phi _2} = {\pi \over 3} - {\pi \over 2} = {{2\pi - 3\pi } \over 6} = {\pi \over 6}$
2

### AIEEE 2005

The function ${\sin ^2}\left( {\omega t} \right)$ represents
A
a periodic, but not $SHM$ with a period ${\pi \over \omega }$
B
a periodic, but not $SHM$ with a period ${{2\pi } \over \omega }$
C
a $SHM$ with a period ${\pi \over \omega }$
D
a $SHM$ with a period ${{2\pi } \over \omega }$

## Explanation

y = sin2$\omega$t

= ${{1 - \cos 2\omega t} \over 2}$

$= {1 \over 2} - {1 \over 2}\cos \,2\omega t$

$\therefore$ Angular speed = 2$\omega$

$\therefore$ Period (T) = ${{2\pi } \over {angular\,speed}}$ = ${{2\pi } \over {2\omega }}$ = ${\pi \over \omega }$

So it is a periodic function.

As y = sin2$\omega$t

${{dy} \over {dt}}$ = 2$\omega$sin$\omega$t cos$\omega$t = $\omega$ sin2$\omega$t

${{{d^2}y} \over {d{t^2}}}$ = $2{\omega ^2}$ cos2$\omega$t which is not proportional to -y.

Hence it is is not SHM.
3

### AIEEE 2004

In forced oscillation of a particle the amplitude is maximum for a frequency ${\omega _1}$ of the force while the energy is maximum for a frequency ${\omega _2}$ of the force; then
A
${\omega _1} < {\omega _2}$ when damping is small and ${\omega _1} > {\omega _2}$ when damping is large
B
${\omega _1} > {\omega _2}$
C
${\omega _1} = {\omega _2}$
D
${\omega _1} < {\omega _2}$

## Explanation

The maximum of amplitude and energy is obtained when the frequency is equal to the natural frequency (resonance condition)

$\therefore$ ${\omega _1} = {\omega _2}$
4

### AIEEE 2004

A particle of mass $m$ is attached to a spring (of spring constant $k$) and has a natural angular frequency ${\omega _0}.$ An external force $F(t)$ proportional to $\cos \,\omega t\left( {\omega \ne {\omega _0}} \right)$ is applied to the oscillator. The time displacement of the oscillator will be proportional to
A
${1 \over {m\left( {\omega _0^2 + {\omega ^2}} \right)}}$
B
${1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$
C
${m \over {\omega _0^2 - {\omega ^2}}}$
D
${m \over {\omega _0^2 + {\omega ^2}}}$

## Explanation

Given that, initial angular velocity = ${\omega _0}$

and at any instant time t, angular velocity = $\omega$

So when displacement is x then the resultant acceleration

f = $\left( {\omega _0^2 - {\omega ^2}} \right)x$

So the external force, F = $m\left( {\omega _0^2 - {\omega ^2}} \right)x$ ............(i)

But given that $F \propto \cos \omega t$

From (i) we get,

$m\left( {\omega _0^2 - {\omega ^2}} \right)x \propto \cos \omega t$ .........(ii)

From equation of SHM we know,

$x = A\sin \left( {\omega t + \phi } \right)$

When t = 0 then x = A

$\therefore$ A = $A\sin \left( \phi \right)$

$\Rightarrow A = {\pi \over 2}$

$\therefore$ $x = A\sin \left( {\omega t + {\pi \over 2}} \right) = A\cos \omega t$

Putting value of x in (ii), we get

$m\left( {\omega _0^2 - {\omega ^2}} \right)A\cos \omega t \propto \cos \omega t$

$\Rightarrow A \propto {1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$