At equilibrium, the springs are released, mass M is oscillating under the influence of three springs, k_(1),k_(2)andk_(3) as shown. Its frequency (upsilon) of oscillation is (1)/(2pi)sqrt((k_(eq))/(M)), where k_(eq) I such that

At equilibrium, the springs are released, mass M is oscillating under

1.	At equilibrium, the springs are released, mass M is oscillating under the influence of three springs, k_(1),k_(2)andk_(3) as shown. Its frequency (upsilon) of oscillation is (1)/(2pi)sqrt((k_(eq))/(M)), where k_(eq) I such that
Answer» <html><body><p>`k_(eq)=k_(1)+k_(2)+k_(3)` <br/>`k_(eq)=(k_(1)+k_(2)+k_(3))/(3)` <br/>`(1)/(k_(eq))=(1)/(k_(1))+(1)/(k_(2))+(1)/(k_(3))` <br/>`k_(eq)=(k_(1)+k_(2)-k_(3))` </p>Solution :When the mass M is <a href="https://interviewquestions.tuteehub.com/tag/displaced-440680" style="font-weight:bold;" target="_blank" title="Click to know more about DISPLACED">DISPLACED</a> <a href="https://interviewquestions.tuteehub.com/tag/slightly-3037487" style="font-weight:bold;" target="_blank" title="Click to know more about SLIGHTLY">SLIGHTLY</a> to the right, the resultant <a href="https://interviewquestions.tuteehub.com/tag/forces-16875" style="font-weight:bold;" target="_blank" title="Click to know more about FORCES">FORCES</a> acting on the mass M is as shown. <br/> `F_("net")=-(k_(1)+k_(2)+k_(3)).x,` <br/> Thus, `k_(eq)=k_(1)+k_(2)+k_(3)` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/MTG_NEET_GID_PHY_XI_C10_SLV_005_S01.png" width="80%"/></body></html>