NEET 2019 | Oscillations Question 11 | Physics

NEET 2019

MCQ (Single Correct Answer)

-1

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

NEET 2017

MCQ (Single Correct Answer)

-1

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is

$${{\sqrt 5 } \over {2\pi }}$$

$${{4\pi } \over {\sqrt 5 }}$$

$${{2\pi } \over {\sqrt 3 }}$$

$${{\sqrt 5 } \over \pi }$$

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