1.

A 5 kg collar is attached to a spring of spring constant 500 N m^(-1). It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate (a) the period of oscillation, (b) the maximum speed and (c) maximum acceleration of the collar.

Answer» <html><body><p></p>Solution :(a) The period of <a href="https://interviewquestions.tuteehub.com/tag/oscillation-1139998" style="font-weight:bold;" target="_blank" title="Click to know more about OSCILLATION">OSCILLATION</a> as given by Eq. (14.21) is, <br/> `T=2pisqrt(m/k)=2pisqrt((5.0kg)/(500Nm^(-1)))` <br/> `=(2pi//10)s` <br/> `=0.63s` <br/> (b) The <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of the collar executing SHM is given by, <br/> `v(t)=Aomega` <br/> `=0.1xxsqrt(k/m)` <br/> `=0.1xxsqrt((500Nm^(-1))/(<a href="https://interviewquestions.tuteehub.com/tag/5kg-326499" style="font-weight:bold;" target="_blank" title="Click to know more about 5KG">5KG</a>))` <br/> `=<a href="https://interviewquestions.tuteehub.com/tag/1ms-1803776" style="font-weight:bold;" target="_blank" title="Click to know more about 1MS">1MS</a>^(-1)` <br/> and it occurs at x = 0 <br/> (c) The acceleration of the collar at thedisplacement x (t ) from the equilibrium is given by, <br/> `a(t)=-omega^(2)xt` <br/> `=-k/mx(t)` <br/> Therefore, the maximum acceleration is, <br/> `a_("max")=omega^(2)A` <br/> `(500Nm^(-1))/(5kg)xx0.1m` <br/> `=<a href="https://interviewquestions.tuteehub.com/tag/10ms-267176" style="font-weight:bold;" target="_blank" title="Click to know more about 10MS">10MS</a>^(-2)` <br/> and it occurs at the extremities.</body></html>


Discussion

No Comment Found

Related InterviewSolutions