A circular arc of mass m is connected with the help of two massless strings as shown in the figure in vertical plane. About point P, small oscillations are given in the plane of the arc. Time period of the oscillations of SHM will be :
Answers
Given:
A circular arc of mass m is connected with the help of two massless strings as shown in the figure in the vertical plane. About point P, small oscillations are given in the plane of the arc.
To find:
The time period of the oscillations of SHM will be :
Solution:
From given, we have,
An electric dipole of moment P is released in a uniform electric field E from the position of maximum torque
As the dipole experiences a torque gsinθ tending to bring itself back in the direction of the field, so on being released the dipole oscillates about an axis through its centre of mass and perpendicular to the field.
The equation of motion considering the moment of inertia I, we have,
d²θ/dt² =−gsinθ
For a smaller angular displacement/amplitude
sinθ ≈ θ
Thus d²θ/dt² = α = −(g/2I).θ=−ω²θ
where ω= (g/2I)
This is an S.H.M whose period of oscillation is T=π √2(I/g)
Therefore the period of oscillation is T =π √2(I/g)
Answer:
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