JEE Main 2019 (Online) 9th January Evening Slot | Simple Harmonic Motion Question 121 | Physics

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to :

0.77

0.57

0.37

0.17

JEE Main 2019 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)

-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 :

$${A \over 2}$$

$${A \over {2\sqrt 2 }}$$

$${A \over {\sqrt 2 }}$$

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) 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)

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

$${1 \over 2}$$

$${2 \over 3}$$

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

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