AIEEE 2004 | Simple Harmonic Motion Question 147 | Physics

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

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

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

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

AIEEE 2003

MCQ (Single Correct Answer)

-1

A mass $$M$$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $$SHM$$ of time period $$T.$$ If the mass is increased by $$m.$$ the time period becomes $${{5T} \over 3}$$. Then the ratio of $${{m} \over M}$$ is

$${3 \over 5}$$

$${25 \over 9}$$

$${16 \over 9}$$

$${5 \over 3}$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

Two particles $$A$$ and $$B$$ of equal masses are suspended from two massless springs of spring of spring constant $${k_1}$$ and $${k_2}$$, respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of $$A$$ and $$B$$ is

$$\sqrt {{{{k_1}} \over {{k_2}}}} $$

$${{{{k_2}} \over {{k_1}}}}$$

$$\sqrt {{{{k_2}} \over {{k_1}}}} $$

$${{{{k_1}} \over {{k_2}}}}$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

The length of a simple pendulum executing simple harmonic motion is increased by $$21\% $$. The percentage increase in the time period of the pendulum of increased length is

$$11\% $$

$$21\% $$

$$42\% $$

$$10\% $$

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