Answer :
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 ]
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 ]
We have, [tex]\tau_{r}=-mglsin(\theta)[/tex], where
[tex]\tau_{r}[/tex] → restoring torque,
[tex]l=\sqrt{L^{2}-(\frac{d}{2})^{2}}[/tex],
[tex]\theta[/tex] → 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, [tex]\tau_{r}=I\alpha=-mglsin(\theta) [/tex],
[tex]\alpha=-\frac{mgl}{I}\theta[/tex], since angular displacement is small.
Here, [tex]I=ml^{2}[/tex].
Hence, [tex]\omega = \sqrt{\frac{g}{l}}[/tex].
Then [tex]T=2\pi\sqrt{\frac{l}{g}}=2\pi\sqrt{\frac{\sqrt{4L^{2}-d^{2}}}{2g}}[/tex].
[tex]\tau_{r}[/tex] → restoring torque,
[tex]l=\sqrt{L^{2}-(\frac{d}{2})^{2}}[/tex],
[tex]\theta[/tex] → 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, [tex]\tau_{r}=I\alpha=-mglsin(\theta) [/tex],
[tex]\alpha=-\frac{mgl}{I}\theta[/tex], since angular displacement is small.
Here, [tex]I=ml^{2}[/tex].
Hence, [tex]\omega = \sqrt{\frac{g}{l}}[/tex].
Then [tex]T=2\pi\sqrt{\frac{l}{g}}=2\pi\sqrt{\frac{\sqrt{4L^{2}-d^{2}}}{2g}}[/tex].