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Question

(C) The car moves with an acceleration $a$. Given the position equation $x = \frac{g}{2} \sqrt{3} t^2$,we differentiate twice with respect to time $t$

Vedclass · Accepted Answer

$2\pi \sqrt{\frac{l}{2g}}$

Vedclass · Answer

(C) The car moves with an acceleration $a$. Given the position equation $x = \frac{g}{2} \sqrt{3} t^2$,we differentiate twice with respect to time $t$ to find the acceleration:$v = \frac{dx}{dt} = g \sqrt{3} t$$a = \frac{dv}{dt} = g \sqrt{3}$In the frame of the car,the pendulum experiences a pseudo-force in the direction opposite to the acceleration. The effective acceleration due to gravity $g_{eff}$ is the vector sum of the gravitational acceleration $g$ (downwards) and the pseudo-acceleration $a$ (backwards):$g_{eff} = \sqrt{g^2 + a^2} = \sqrt{g^2 + (g \sqrt{3})^2} = \sqrt{g^2 + 3g^2} = \sqrt{4g^2} = 2g$The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.Substituting $g_{eff} = 2g$,we get $T = 2\pi \sqrt{\frac{l}{2g}}$.Thus,the correct option is $C$.

$A$ simple pendulum is suspended in a car. The car starts moving on a horizontal road according to the equation $x = \frac{g}{2} \sqrt{3} t^2$. Find the time period of oscillation of the pendulum.

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