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### 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 $$\,\,\, a = 5{\sqrt 2 } \therefore\,\,\, Maximum acceleration, Amax = a\omega 2 = 5 {\sqrt 2 } \times (10)2 = 500 {\sqrt 2 } m/s2 2 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 9th April Morning Slot A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm−1 and oscillates in a damping medium of damping constant 10−2 kg s−1 . The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to : A 2 s B 3.5 s C 5 s D 7 s 3 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Morning Slot A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are {\rho _1} and {\rho _2} \left( {{\rho _1} > {\rho _2}} \right), fill half the circle. The angle \theta between the radius vector passing through the common interface and the vertical is : A \theta = {\tan ^{ - 1}}\pi \left( {{{{\rho _1}} \over {{\rho _2}}}} \right) B \theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1}} \over {{\rho _2}}}} \right) C \theta = {\tan ^{ - 1}}\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right) D \theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right) ## Explanation As system is in equilibrium so the pressuse at A from both side of the liquid must be equal . (r cos \theta + r sin \theta ) \rho 2g = (r cos \theta - r sin \theta ) \rho 1g \Rightarrow$$\,\,\,$ ${{{\rho _1}} \over {{\rho _2}}} = {{\sin \theta + \cos \theta } \over {\cos \theta - \sin \theta }} = {{1 + \tan \theta } \over {1 - \tan \theta }}$

$\Rightarrow $$\,\,\, \rho 1 - \rho 1 tan\theta = \rho 2 + \rho 2 tan\theta \Rightarrow$$\,\,\,$ ($\rho$1 + $\rho$2) tan$\theta$ = $\rho$1 $-$ $\rho$2

$\Rightarrow $$\,\,\, tan\theta = {{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}} \Rightarrow$$\,\,\,$ $\theta$ = tan$-$1 $\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$
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### JEE Main 2018 (Online) 15th April Evening Slot

Two simple harmonic motions, as shown below, are at right angles. They are combined to form Lissajous figures.
x(t) = A sin (at + $\delta$)
y(t) = B sin (bt)

Identify the correct match below.
A
Parameters   A $\ne$ B, a = b; $\delta$ = 0;
Curve    Parabola
B
Parameters    A = B, a = b; $\delta$ = ${\raise0.5ex\hbox{\pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}$
Curve    Line
C
Parameters    A $\ne$ B, a = b; $\delta$ = ${\raise0.5ex\hbox{\pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}$
Curve    Ellipse
D
Parameters    A = B, a = 2b; $\delta$ = ${\raise0.5ex\hbox{\pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}$
Curve    Circle

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