A bob of mass m is in equilibrium with the help of two in extensible strings connected to fixed support . The bob is slightly displace perpendicular to the plane of the figure and released.The time period of oscillation of bob is

(A) 2π \sqrt{\frac{L}{g}} (B) 2π \sqrt{\frac{d}{g}}

(C) 2π \sqrt{\frac{L+d}{g}} (D) 2π \sqrt{\frac{(\sqrt{4L^2-d^2})}{2g}}

Is the answer C? just tell me if I am correct, pls dont tell the answer
i don't know the answer it is a question from ftre and i have done with option D
which year?
and which class


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   We assume that the strings are both of same kind. The tensions in both of them is same, as their length is same and the bob hangs at the middle point (vertically) between the two points of suspension.

The vertical distance of the bob from ceiling = y = √[ L² - (d/2)² ]
                 y = 1/2*√(4L²-d²)

   Let us take the plane formed by the bob and two points of suspension of two strings.   As the bob is displaced perpendicular to the plane of the figure given, the tensions in the two strings always remain equal and balanced.  So the bob oscillates in the plane perpendicular to the plane of the figure given.

     The restoration force on the pendulum at any point is perpendicular to the tensions in the string as the bob moves along a circular arc.  The oscillations of the pendulum are similar to a simple pendulum with mass = m and length = y.

       restoration force = F = m g sin Ф = - k (y Ф)
                 k =  (mg/y) (sinФ/Ф)      = mg/y  for small Ф

       Here Ф = angle of displacement perpendicular to plane of the figure.
                 F = m a = m y α = m y d²Ф/dt² = - (mg/y)  y Ф
                               d²Ф/dt² = - (g/y) Ф

       SHM with  ω² = g/y,    ω = √(g/y)  => Time period T = 2π √(y/g)
   OR,       Time period = 2π √(m/k) = 2π √(y/g) =

                         T = 2 π √[ √(4L² - d²)/2g ]

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i hope my answer is correct.
yeah i have not written anything as when in the exam hall i was doing the question i also found some questions wrong
which exam ? is it in a JEE coaching class exam?
It is FIIT-JEE talent reward examination (FTRE)
There's a small calculation mistake. Option D is the correct one.
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We have, \tau_{r}=-mglsin(\theta), where
\tau_{r} → restoring torque,
\theta → Angle made by imaginary line connecting bob and ceiling with the vertical present in plane.
Don't worry about tension forces as they have no effect on restoring torque.
Now, \tau_{r}=I\alpha=-mglsin(\theta) ,
\alpha=-\frac{mgl}{I}\theta, since angular displacement is small.
Here, I=ml^{2}.
Hence, \omega = \sqrt{\frac{g}{l}}.
Then T=2\pi\sqrt{\frac{l}{g}}=2\pi\sqrt{\frac{\sqrt{4L^{2}-d^{2}}}{2g}}.
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