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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
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A simple pendulum is suspended inside a truck moving uniformly with a small acceleration of $‘2a’.$ Its mean position as an angle in radians with respect to the vertical is

$\begin{array}{1 1}(a)\;0\\(b)\;\frac{2a}{g} \\( c)\; 6.28 \\(d)\; None\;of\;the\;above \end{array}$

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There are two components of acceleration felt by the mass of a stationary pendulum inside the truck.
Vertical acceleration of $9.81 m/s^2$ due to gravity & inertial acceleration of 'a' opposite to direction of motion of the trunk.
The pendulum takes a mean position such that $T \sin \alpha= ma\;\& T \cos \alpha=mg$ where T is the tension in the string & is the angle string holding mass 'm' of the pendulum made to the vertical.
Thus $\tan \alpha=\large\frac{a}{g}$ or $\alpha$ in radians $=\large\frac{a}{g}$ for small angles.
Hence b is the correct answer
answered Dec 11, 2013 by meena.p

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