AIEEE 2004 | Simple Harmonic Motion Question 146 | Physics

A particle at the end of a spring executes $$S.H.M$$ with a period $${t_1}$$. While the corresponding period for another spring is $${t_2}$$. If the period of oscillation with the two springs in series is $$T$$ then

$${T^{ - 1}} = t_1^{ - 1} + t_2^{ - 1}$$

$${T^2} = t_1^2 + t_2^2$$

$$T = {t_1} + {t_2}$$

$${T^{ - 2}} = t_1^{ - 2} + t_2^{ - 2}$$

AIEEE 2004

MCQ (Single Correct Answer)

-1

A particle of mass $$m$$ is attached to a spring (of spring constant $$k$$) and has a natural angular frequency $${\omega _0}.$$ An external force $$F(t)$$ proportional to $$\cos \,\omega t\left( {\omega \ne {\omega _0}} \right)$$ is applied to the oscillator. The time displacement of the oscillator will be proportional to

$${1 \over {m\left( {\omega _0^2 + {\omega ^2}} \right)}}$$

$${1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$

$${m \over {\omega _0^2 - {\omega ^2}}}$$

$${m \over {\omega _0^2 + {\omega ^2}}}$$

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

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

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

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

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