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Question Number 12517 by tawa last updated on 24/Apr/17

$$\mathrm{A}\mathrm{spring}\mathrm{stretches}\mathrm{by}15\mathrm{cm}\mathrm{when}\mathrm{a}\mathrm{mass}\mathrm{of}300\mathrm{g}\mathrm{hangs}\mathrm{down}\mathrm{from}\mathrm{it}.$$$$\mathrm{if}\mathrm{the}\mathrm{spring}\mathrm{is}\mathrm{then}\mathrm{strethed}\mathrm{an}\mathrm{additional}10\mathrm{cm}\mathrm{and}\mathrm{realeased},\mathrm{calculate}$$$$\left(\mathrm{a}\right)\mathrm{the}\mathrm{spring}\mathrm{constant}$$$$\left(\mathrm{b}\right)\mathrm{Angular}\mathrm{velocity}$$$$\left(\mathrm{c}\right)\mathrm{The}\mathrm{amplitude}\mathrm{of}\mathrm{the}\mathrm{oscillation}$$$$\left(\mathrm{d}\right)\mathrm{The}\mathrm{maximum}\mathrm{velocity}$$$$\left(\mathrm{e}\right)\mathrm{The}\mathrm{maximum}\mathrm{acceleration}\mathrm{of}\mathrm{the}\mathrm{mass}$$$$\left(\mathrm{f}\right)\mathrm{The}\mathrm{period}\mathrm{T}\mathrm{and}\mathrm{frequency}\mathrm{f}$$

Answered by mrW1 last updated on 24/Apr/17

$$\left(a\right)k=F/\delta =mg/\delta =0.3\times 10/0.15=20N/m$$$$\left(b\right)w=\sqrt{\frac{k}{m}}=\sqrt{\frac{20}{0.3}}=8.1651/s$$$$\left(c\right)x_{max}=10cm$$$$\left(d\right)v_{max}=w\times x_{max}=8.165\times 0.1=0.816m/s$$$$\left(e\right)a_{max}=w\times v_{max}=8.165\times 0.816=6.662m/s^2$$$$\left(f\right)T=\frac{2\pi }{w}=\frac{2\pi }{8.165}=0.729sec$$$$f=\frac{1}{T}=1.371Hz$$

Commented by tawa last updated on 24/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{everytime}.\mathrm{i}\mathrm{really}\mathrm{appreciate}$$

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