1
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1 A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :
A
$${A \over 2}$$
B
$${A \over {2\sqrt 2 }}$$
C
$${A \over {\sqrt 2 }}$$
D
A
2
JEE Main 2018 (Online) 16th April Morning Slot
+4
-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)
A
$${1 \over {\sqrt 3 }}$$
B
$${1 \over 2}$$
C
$${2 \over 3}$$
D
$${3 \over {\sqrt 5 }}$$
3
JEE Main 2018 (Online) 16th April Morning Slot
+4
-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 :
A
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + c} \over {2b}}} \right)$$
B
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + b} \over {2c}}} \right)$$
C
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{2a + 3c} \over b}} \right)$$
D
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + 2b} \over {3c}}} \right)$$
4
JEE Main 2018 (Online) 15th April Evening Slot
+4
-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.
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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