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 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

$${3 \over 5}$$

$${25 \over 9}$$

$${16 \over 9}$$

$${5 \over 3}$$

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