Test Index

Oscillations Part 1

Section: Physics

© examsnet.com

Question : 91 of 100

Marks: +1, -0

A particle of mass

m

is attached to a spring (of spring constant

k

) and has a natural angular frequency

\omega_0

. An external force

F(t)

proportional to

\cos \omega t (\omega \neq \omega_0)

is applied to the oscillator. The time displacement of the oscillator will be proportional to

$\frac{1}{m (\omega_0^2+\omega^2)}$
$\frac{1}{m (\omega_0^2-\omega^2)}$
$\frac{m}{\omega_0^2-\omega^2}$
$\frac{m}{\omega_0^2+\omega^2}$

Solution:

Given that, initial angular velocity

=\omega_0

and at any instant time

t

, angular velocity

=\omega

So when displacement is

x

then the resultant acceleration

f= (\omega_0^2-\omega^2) x

So the external force,

F=m (\omega_0^2-\omega^2) x

....(1)

But given that

F \propto \cos \omega t

From (i) we get,

m (\omega_0^2-\omega^2) x \propto \cos \omega t

....(ii)

From equation of SHM we know,

x=A \sin (\omega t+\phi)

When

t=0

then

x=A

\therefore A=A \sin (\phi)

\Rightarrow A=\frac{\pi}{2}

\therefore x=A \sin \left(\omega t+\frac{\pi}{2}\right) =A \cos \omega t

Putting value of

x

in (ii), we get

m (\omega_0^2-\omega^2) A \cos \omega t \propto \cos \omega t

\Rightarrow A \propto \frac{1}{m (\omega_0^2-\omega^2)}

© examsnet.com

Go to Question:

Prev Question

Next Question