NEET 2019 | Oscillations Question 16 | Physics

The displacement of a particle executing simple harmonic motion is given by

y = A₀ + A sin$$\omega $$t + B cos$$\omega $$t.

Then the amplitude of its oscillation is given by :

$$\sqrt {{A^2} + {B^2}} $$

A + B

A + $$\sqrt {{A^2} + {B^2}} $$

$$\sqrt {A_0^2 + {{\left( {A + B} \right)}^2}} $$

NEET 2018

MCQ (Single Correct Answer)

-1

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m s^–2 at a distance of 5 m from the mean position. The time period of oscillation is

2$$\pi $$ s

$$\pi $$ s

2 s

1 s

NEET 2017

MCQ (Single Correct Answer)

-1

A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is K'. Then they are connected in parallel and force constant is k''. Then k' : k'' is

1 : 9

1 : 11

1 : 14

1 : 6

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NEET Subjects

Biology

Zoology

Human Health and Diseases Body Fluids and Its Circulation Locomotion and Movement Neural Control and Coordination Reproduction in Organisms Reproductive Health Structural Organisation in Animals Digestion and Absorption Excretory Products and Their Elimination Chemical Coordination and Integration Human Reproduction Animal Kingdom Breathing and Exchange of Gases Evolution