AIEEE 2006 | Simple Harmonic Motion Question 140 | 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

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

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

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

$$2\pi \sqrt \alpha $$

$$2\pi \alpha $$

AIEEE 2005

MCQ (Single Correct Answer)

-1

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

first decrease and then increase to the original value

first increase and then decrease to the original value

increase towards a saturation value

remain unchanged

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

$${\pi \over 3}$$

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

$${\pi \over 6}$$

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

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