A seconds pendulum is suspended from the ceiling of a trolley moving horizontally with an acceleration of 4 m/s $^{2}$ . Its period of oscillation is:

1.90 s

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1.70 s

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2.30 s

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1.40 s

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Solution

The correct option is A 1.90 s
$g_{e} = \sqrt{g^{2} + a^{2}} = \sqrt{{(10)}^{2} + {(4)}^{2}}$
$= 10.77$ m/s $^{2}$
$T = 2 π \sqrt{\frac{l}{g_{e}}} = 2 π \sqrt{\frac{1}{10.77}}$
$= 1.90$ s

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A seconds pendulum is suspended from the ceiling of a trolley moving horizontally with an acceleration of 4 m/s2. Its period of oscillation is:

The correct option is A 1.90 sge=√g2+a2=√(10)2+(4)2 =10.77 m/s2T=2π√lge=2π√110.77 =1.90 s

A seconds pendulum is suspended from the ceiling of a trolley moving horizontally with an acceleration of 4 m/s $^{2}$ . Its period of oscillation is:

The correct option is A 1.90 s
$g_{e} = \sqrt{g^{2} + a^{2}} = \sqrt{{(10)}^{2} + {(4)}^{2}}$
$= 10.77$ m/s $^{2}$
$T = 2 π \sqrt{\frac{l}{g_{e}}} = 2 π \sqrt{\frac{1}{10.77}}$
$= 1.90$ s