Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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)
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !