Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A spring mass system (mass m, spring constant k and natural length l ) rests in equilibrium on a horizontal disc.The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about its axis with an angular velocity $\omega$ , $(k> > m\omega ^{2})$ the relative change in the length of the spring is best given by the option :
Option: 1 $\frac{m\omega ^{2}}{3k}$
Option: 2 $\frac{m\omega ^{2}}{k}$
Option: 3 $\sqrt{\frac{2}{3}}\left ( \frac{m\omega ^{2}}{k} \right )$
Option: 4 $\frac{2m\omega ^{2}}{k}$

Answers (1)

best_answer

As natural lentgh=l₀

Let elongation=x

Mass m is moving with angular velocity $\omega$ in a radius r

where $r=l_{0}+x$

Due to elongation x spring force is given by $F_{s}=Kx$

And $F_{C}=m\omega ^{2}r=m\omega ^{2}(l_{0}+x)$

as $F_{C}=F_{s}$

$Kx=m\omega ^{2}(l_{0}+x)$

$\Rightarrow x=\frac{m\omega ^{2}l_{0}}{K-m\omega ^{2}}$

Using $k> > m\omega ^{2}$

So, $\frac{x}{l_{0}}$ is equal to $\frac{m\omega ^{2}}{k}$

Hence the correct option is (2).

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 Registration Live: NTA session 1 application last date today; official website, syllabus

anam-khan

JEE Main 2025 session 1 registration ends today at jeemain.nta.nic.in; application correction from November 26

anam-khan

JEE Main 2025 registration ends tomorrow; changes introduced in exam this year

Latest Question