1

Question

A uniform square plate of side a is hinged at one of its corners as shown. It is suspended such that it can rotate about horizontal axis. Find out its time period for small oscillations about its equilibrium position.

Open in App

Solution

The correct option is **B** 2π√2√2a3g

Moment of inertia of a square plate about one of its corners and perpendicular to its plane,

I=ICm+md2

I=ma26+m(a√2)2

Time period of the physical pendulum is

T=2π√Imgd

T=2π ⎷ma26+m(a√2)2mg.a√2

T=2π ⎷(a6+a2).√2g=2π√2√2a3g

Moment of inertia of a square plate about one of its corners and perpendicular to its plane,

I=ICm+md2

I=ma26+m(a√2)2

Time period of the physical pendulum is

T=2π√Imgd

T=2π ⎷ma26+m(a√2)2mg.a√2

T=2π ⎷(a6+a2).√2g=2π√2√2a3g

0

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program