A spring mass system (mass$$ ‘m’, spring constant ‘k’ and natural length $$l0$$ $$) rest in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of disc. If $$5g$$ mass of the body attached to the spring having spring constant $$20Nm$$−$$1$$. If the disc together with spring mass system, rotates about its axis with an angular velocity of $$5$$ rad.s−$$1$$ . kmω$$2$$ , then the percentage change in the length of the spring is

Question

A spring mass system (mass$$ ‘m’, spring constant ‘k’ and natural length $$l0$$ $$) rest in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of disc. If $$5g$$ mass of the body attached to the spring having spring constant  $$20Nm$$−$$1$$. If the disc together with spring mass system, rotates about its axis with an angular velocity of $$5$$ rad.s−$$1$$ . k>>mω$$2$$ , then the percentage change in the length of the spring is

Infinity Learn · Accepted Answer

kΔl←m→$$ml0+$$Δlω$$2$$ from free body diagram, forces acting on mass m are,(k=force$$ $$constant), restoring force of spring $$F=-$$$$kx$$$$=-k$$Δl, towards the centre here Δl is extension in length of spring and centrifugal force $$F=mw2l0+$$Δl, away from the centre, $$l0=unstretched$$ length of spring equating the above two equations kΔ$$l=mw2l0+$$Δl kΔ$$l-mw2$$Δ$$l=mw2l0$$ Δ$$l=mw2l0k-mw2$$ Δ$$ll0=mw2k-mw2$$≃$$mw2k$$ $$100($$Δ$$ll0)=(mw2k)100$$ Δ$$ll0$$%$$=(5x10-3x5220)100$$ Δ$$ll0$$%$$=0.625$$

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material