1
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
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 $$\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$$
4
MCQ (Single Correct Answer)

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

Questions Asked from Simple Harmonic Motion

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