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A simple pendulum 50 cm long is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration root(3) g m/s2 . The period of small oscillations of the pendulum about its equilibrium position is (g = pi^2 m/s2) :
ans: 1 sec.
How??

Deepthi , 8 Years ago
Grade 11
anser 1 Answers
Sarthak Gupta
Take vector summation of root3 g and g with angle 90° . g eff will be 2g. Then apply T= 2pi root(l/g eff)
Last Activity: 8 Years ago
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