AIEEE 2005 | Simple Harmonic Motion Question 142 | Physics

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

$${\pi \over 3}$$

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

$${\pi \over 6}$$

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

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 2004

MCQ (Single Correct Answer)

-1

Out of Syllabus

In forced oscillation of a particle the amplitude is maximum for a frequency $${\omega _1}$$ of the force while the energy is maximum for a frequency $${\omega _2}$$ of the force; then

$${\omega _1} < {\omega _2}$$ when damping is small and $${\omega _1} > {\omega _2}$$ when damping is large

$${\omega _1} > {\omega _2}$$

$${\omega _1} = {\omega _2}$$

$${\omega _1} < {\omega _2}$$

PREVIOUS NEXT

Questions Asked from Simple Harmonic Motion (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions