One-end-of-a-massless-spring-of-constant-100-N-m-and-natural-length-0-5-m-is-fixed-and-the-other-end-is-connected-to-a-particle-of-mass-0-5-kg-lying-on-a-frictionless-horizontal-table-The-spring-rema

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Question Number 21628 by Tinkutara last updated on 29/Sep/17

$$\mathrm{One}\mathrm{end}\mathrm{of}\mathrm{a}\mathrm{massless}\mathrm{spring}\mathrm{of}\mathrm{constant}$$$$100\mathrm{N}/\mathrm{m}\mathrm{and}\mathrm{natural}\mathrm{length}0.5\mathrm{m}\mathrm{is}$$$$\mathrm{fixed}\mathrm{and}\mathrm{the}\mathrm{other}\mathrm{end}\mathrm{is}\mathrm{connected}\mathrm{to}$$$$\mathrm{a}\mathrm{particle}\mathrm{of}\mathrm{mass}0.5\mathrm{kg}\mathrm{lying}\mathrm{on}\mathrm{a}$$$$\mathrm{frictionless}\mathrm{horizontal}\mathrm{table}.\mathrm{The}\mathrm{spring}$$$$\mathrm{remains}\mathrm{horizontal}.\mathrm{If}\mathrm{the}\mathrm{mass}\mathrm{is}\mathrm{made}$$$$\mathrm{to}\mathrm{rotate}\mathrm{at}\mathrm{an}\mathrm{angular}\mathrm{velocity}\mathrm{of}2$$$$\mathrm{rad}/s,\mathrm{find}\mathrm{the}\mathrm{elongation}\mathrm{of}\mathrm{the}\mathrm{spring}.$$

Answered by ajfour last updated on 30/Sep/17

$$m{\omega }^2(l+x)=kx$$$$x=\frac{m{\omega }^{2}l}{k-m{\omega }^2}=\frac{0.5\times 4\times 0.5}{100-0.5\times 4}$$$$=\frac{1}{98}m\approx 1.02cm.$$

Commented by Tinkutara last updated on 30/Sep/17

$$\mathrm{But}\mathrm{in}{1}^{\mathrm{st}}\mathrm{line}{\mathrm{shouldn}}^{\prime }\mathrm{t}\mathrm{there}\mathrm{be}k(l+x)$$$$\mathrm{because}\mathrm{now}\mathrm{radius}\mathrm{is}l+x?$$

Commented by ajfour last updated on 30/Sep/17

$$Nothinkagain,springforce$$$$isproportionaltochangeinlength.$$

Commented by Tinkutara last updated on 30/Sep/17

$$\mathrm{OK},\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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