If a trolley accelerates horizontally with ac | vedclass

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(D) In the frame of reference of the trolley,a pseudo force $F_p = ma$ acts on the bob in the direction opposite to the acceleration of the trolley.Le

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$	heta=	an ^{-1}\left(\frac{a}{g}ight)$

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(D) In the frame of reference of the trolley,a pseudo force $F_p = ma$ acts on the bob in the direction opposite to the acceleration of the trolley.Let $m$ be the mass of the bob,$T$ be the tension in the string,and $	heta$ be the angle of deflection from the vertical.In equilibrium,the forces acting on the bob are:$1$. Tension $T$ along the string.$2$. Gravitational force $mg$ acting vertically downwards.$3$. Pseudo force $ma$ acting horizontally backwards.Resolving the forces along the direction of the string and perpendicular to it:Along the string: $T = mg \cos 	heta + ma \sin 	heta$Perpendicular to the string: $T \sin 	heta = ma \cos 	heta$ (This is not the standard way,let's use horizontal and vertical components).Alternatively,balancing forces in the horizontal and vertical directions:Horizontal: $T \sin 	heta = ma$Vertical: $T \cos 	heta = mg$Dividing the two equations:$\frac{T \sin 	heta}{T \cos 	heta} = \frac{ma}{mg}$$	an 	heta = \frac{a}{g}$$	heta = 	an ^{-1}\left(\frac{a}{g}ight)$

If a trolley accelerates horizontally with acceleration $a$,then the bob is displaced backward from its initial vertical position. The angular deflection of the bob in equilibrium is ..........

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