[" ON.1)' Obtain follaning oquation from "],[" first brinciplas."],[qquad " (i) "omega=w_(0)+alpha" tin "=w_(0)t+(1)/(2)alpha t^(2)" ."]

Obtain the equation omega = omega_(0) alpha t from first principles.

Derive the three equation of rotational motion (i) omega = omega_(0) + at (ii) theta = omega_(0)t + 1/2alpha t^(2) (iii) omega^(2) = omega_(0)^(2) + 2alpha theta Under constant angular acceleration. Here symbols have usual meaning.

Drive the equations of rotatory motion, omega^2 - omega_0^2 = 2 alpha theta and theta= omega_0^t + 1/2 alpha t^2 , where every letter has its usual meaning.

A body is acted upon by a force vec(F) , given by vec(F)=-k[(cos omega t)hat(i)+(sin omega t) hat(i)] undergoes displacement, where the position vector vec(r) of the body is given by vec(r)=a[cos ( omega t+ alpha) hat(i)+ sin ( omega t+ alpha) hat(j)] . Find the work done by the force from time t=0 to time t=2pi//omega .

A body is acted upon by a force vec(F) , given by vec(F)=-k[(cos omega t)hat(i)+(sin omega t) hat(i)] undergoes displacement, where the position vector vec(r) of the body is given by vec(r)=a[cos ( omega t+ alpha) hat(i)+ sin ( omega t+ alpha) hat(j)] . Find the work done by the force from time t=0 to time t=2pi//omega .

If alpha, beta are the roots of x^(2) + px + q = 0, and omega is a cube root of unity, then value of (omega alpha + omega^(2) beta) (omega^(2) alpha + omega beta) is

The differentia equation obtained on eliminating A and B from y=A cos omega t+b sin omega t, is y^(+)y'=0 b.y^(-omega)-2y=0 c.y^(=)omega^(2)y d.y^(+)y=0

The equations of two waves acting in perpendicular direction are given as x = a cos (omega t + delta) " and " y = a cos (omega t + alpha) " where " delta = alpha + pi//2 the resultant wave represents

if 1,alpha_1, alpha_2, ……alpha_(3n) be the roots of equation x^(3n+1)-1=0 and omega be an imaginary cube root of unilty then ((omega^2-alpha_1)(omega^2-alpha).(omega^2-alpha(3n)))/((omega-alpha_1)(omega-alpha_2)……(omega-alpha_(3n)))= (A) omega (B) -omega (C) 1 (D) omega^2