AIEEE 2004 | Simple Harmonic Motion Question 145 | Physics

The bob of a simple pendulum executes simple harmonic motion in water with a period $$t,$$ while the period of oscillation of the bob is $${t_0}$$ in air. Neglecting frictional force of water and given that the density of the bob is $$\left( {4/3} \right) \times 1000\,\,kg/{m^3}.$$ What relationship between $$t$$ and $${t_0}$$ is true

$$t = 2{t_0}$$

$$t = {t_0}/2$$

$$t = {t_0}$$

$$t = 4{t_0}$$

AIEEE 2004

MCQ (Single Correct Answer)

-1

The total energy of particle, executing simple harmonic motion is

independent of $$x$$

$$ \propto \,{x^2}$$

$$ \propto \,x$$

$$ \propto \,{x^{1/2}}$$

AIEEE 2004

MCQ (Single Correct Answer)

-1

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

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

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

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

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

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