JEE Main 2018 (Online) 16th April Morning Slot | Simple Harmonic Motion Question 77 | Physics

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)

-1

An oscillator of mass M is at rest in its equilibrium position in a potential
V = $${1 \over 2}$$ k(x $$-$$ X)². A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is : (M = 10, m = 5, u = 1, k = 1)

$${1 \over {\sqrt 3 }}$$

$${1 \over 2}$$

$${2 \over 3}$$

$${3 \over {\sqrt 5 }}$$

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)

-1

A particle executes simple harmonic motion and is located at x = a, b and c at times t₀, 2t₀ and 3t₀ respectively. The freqquency of the oscillation is :

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + c} \over {2b}}} \right)$$

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + b} \over {2c}}} \right)$$

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{2a + 3c} \over b}} \right)$$

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + 2b} \over {3c}}} \right)$$

JEE Main 2018 (Online) 15th April Evening Slot

MCQ (Single Correct Answer)

-1

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.

Parameters A $$ \ne $$ B, a = b; $$\delta $$ = 0;
Curve Parabola

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

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

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

JEE Main 2018 (Online) 15th April Morning Slot

MCQ (Single Correct Answer)

-1

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 :

$$\theta = {\tan ^{ - 1}}\pi \left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$$

$$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$$

$$\theta = {\tan ^{ - 1}}\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$$

$$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right)$$

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