Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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. Time period of the oscillations of the SHM will be:

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. Time period of the oscillations of the SHM will be:
 

Grade:12th pass

1 Answers

Phanindra
13 Points
12 days ago

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)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free