AIEEE 2006 | Simple Harmonic Motion Question 164 | Physics

Starting from the origin a body oscillates simple harmonically with a period of $$2$$ $$s.$$ After what time will its kinetic energy be $$75\% $$ of the total energy?

$${1 \over 6}s$$

$${1 \over 4}s$$

$${1 \over 3}s$$

$${1 \over 12}s$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

The function $${\sin ^2}\left( {\omega t} \right)$$ represents

a periodic, but not $$SHM$$ with a period $${\pi \over \omega }$$

a periodic, but not $$SHM$$ with a period $${{2\pi } \over \omega }$$

a $$SHM$$ with a period $${\pi \over \omega }$$

a $$SHM$$ with a period $${{2\pi } \over \omega }$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

Two simple harmonic motions are represented by the equations $${y_1} = 0.1\,\sin \left( {100\pi t + {\pi \over 3}} \right)$$ and $${y_2} = 0.1\,\cos \,\pi t.$$ The phase difference of the velocity of particle $$1$$ with respect to the velocity of particle $$2$$ is

$${\pi \over 3}$$

$${{ - \pi } \over 6}$$

$${\pi \over 6}$$

$${{ - \pi } \over 3}$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

If a simple harmonic motion is represented by $${{{d^2}x} \over {d{t^2}}} + \alpha x = 0.$$ its time period is

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

$${{2\pi } \over \alpha }$$

$$2\pi \sqrt \alpha $$

$$2\pi \alpha $$