1

JEE Main 2016 (Online) 9th April Morning Slot

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and I, respectively. At time t = 0 one particle has displacement A while the other one has displacement ${{ - A} \over 2}$ and they are moving towards each other. If they cross each other at time t, then t is :
A
${T \over 6}$
B
${5T \over 6}$
C
${T \over 3}$
D
${T \over 4}$

Explanation

Angular displacement ($\theta$1) of particle 1. from equilibrium,

${y_1}$ = A sin$\theta$1

$\Rightarrow$   A = Asin$\theta$1

$\Rightarrow$   sin$\theta$1 = 1 = sin ${\pi \over 2}$

$\therefore$   $\theta$1 = ${\pi \over 2}$

Similarly for particle 2 angular displacement $\theta$2 from equilibrium,

y2 = Asin$\theta$2

$\Rightarrow$   $-$ ${A \over 2}$ = Asin$\theta$2

$\Rightarrow$   sin$\theta$2 = $-$ ${1 \over 2}$ = sin$\left( { - {\pi \over 3}} \right)$

$\Rightarrow$   $\theta$2 = $-$ ${{\pi \over 3}}$

Relative angular displacement of the two particle,

$\theta$ = $\theta$1 $-$ $\theta$2

= ${{\pi \over 2}}$ $-$ $\left( { - {\pi \over 6}} \right)$

= ${{{2\pi } \over 3}}$

Relative angular velocity $=$ $\omega - \left( { - \omega } \right)$ = $2\omega$

If they cross each other at time t

then,   t = ${\theta \over {2\omega }}$ = ${{2\pi } \over {3 \times 2\omega }}$ = ${\pi \over {3 \times {{2\pi } \over T}}}$ = ${T \over 6}$
2

JEE Main 2016 (Online) 10th April Morning Slot

In an engine the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :
A
0.1 Hz
B
1.2 Hz
C
0.7 Hz
D
1.9 Hz

Explanation

Here,

Amplitude, A = 7 cm = 0.07 m

When washer is no longer stays in contact with the piston, then the normal force on the washer is = 0

$\therefore$   Maximum acceleration of the washer,

amax = $\omega$2A = g

$\Rightarrow$   $\omega$ = $\sqrt {{g \over A}}$ = $\sqrt {{{10} \over {0.07}}}$ = $\sqrt {{{1000} \over 7}}$

$\therefore$   Frequency of the piston,

f = ${\omega \over {2\pi }}$ = ${1 \over {2\pi }}\sqrt {{{1000} \over 7}}$ = 1.9 Hz
3

JEE Main 2017 (Online) 8th April Morning Slot

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is :
A
${1 \over 4}Hz$
B
${1 \over {2\sqrt 2 }}Hz$
C
${1 \over 2}Hz$
D
$2$ $Hz$

Explanation

For 1 kg block :

Here frequency of spring (f) = ${1 \over {2\pi }}\sqrt {{k \over m}}$

Given that, F = 1 Hz

$\therefore\,\,\,$ ${1 \over {2\pi }}\sqrt {{k \over 1}}$ = 1

$\Rightarrow $$\,\,\, k = 4\pi 2 N m-1 For 8 kg Block : Here two identical springs are attached in parallel. So, Keq = k + k = 2k \therefore\,\,\, Frequency of 8 kg block, F' = {1 \over {2\pi }}\sqrt {{{{k_{eq}}} \over {m'}}} = {1 \over {2\pi }}\sqrt {{{2k} \over 8}} = {1 \over {2\pi }}\sqrt {{{2 \times 4{\pi ^2}} \over 8}} = {1 \over {2\pi }} \times \pi = {1 \over 2} Hz 4 MCQ (Single Correct Answer) JEE Main 2017 (Online) 8th April Morning Slot The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s−1 . At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is {\pi \over 4}. A 500 m/s2 B 500 \sqrt 2 m/ s2 C 750 m/s2 D 750 \sqrt 2 m / s2 Explanation Mximum velocity, Vmax = a\omega Maximum acceleration, Amax = a\omega 2 Given that, {{a{\omega ^2}} \over {a\omega }} = 10 \Rightarrow$$\,\,\,$ $\omega$ = 10 s$-$1

Displacement, x = a sin ($\omega$t + ${\pi \over 4}$)

at t = 0, displacement x = 5

$\therefore\,\,\,$ 5 = a sin $\left( {{\pi \over 4}} \right)$

$\Rightarrow $$\,\,\, 5 = a \times {1 \over {\sqrt 2 }} \Rightarrow$$\,\,\,$ a = 5${\sqrt 2 }$

$\therefore\,\,\,$ Maximum acceleration,

Amax = a$\omega$2 = 5 ${\sqrt 2 }$ $\times$ (10)2

= 500 ${\sqrt 2 }$ m/s2