Question

QUESTION : Refer the attached picture. ​

Answer 1

—

1

m(w2-2)

Initial angular velocity of particle wo and

at any instant t, angular velocity w

Therefore, for a displacement x, the

resultant acceleration

External force

F mf (w-w2) x..i)

Since, Fo coswt (given)

therefore

From eq. (ii) m (w2 -w2) x coswt. (iii)

Now, equation of simple harmonic motion

x A sin (wt + p) (iv)

at t 0; x A

.A A sin( 0 p )

> 0 =1/2

.. A sin wt =Acot wt (v)

Hence, from equations (ii) and (v), we finally

get

m(w-wA cos wt cos wt

1

mw-w2)

Initial angular velocity of particle wo and at

any instant t, angular velocity w Therefore,

for a displacement x, the resultant

acceleration

f

External force

F m(f (w -w2) x..(i)

Since, F coswt (given)

therefore

From eq. (ii) m (w2 -w2) x coswt (ii *...

Now, equation of simple harmonic motion

x = A sin (wtp) (iv)

at t 0; x A

. A A sin( 0 + 0)

A sin wt =Acot w

Hence, from equations (ii) and (v), we finally

get

mw2 -w) Acos wt a cos wt

1

Ax

mw?-w2)

HOPE THIS ANSWER WILL HELP YOU

Answer 2

Initial angular velocity of particle = ω0

and at any instant l, angular velocity = ω

Therefore, for a displacement x, the resultant acceleration