A-body-of-mass-20g-performs-simple-harmonic-motion-at-a-frequency-of-5Hz-at-a-distance-of-10cm-from-the-mean-position-its-velocity-is-200cm-s-calculate-i-Maximum-displacement-from-the-mean-posit

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Question Number 9635 by tawakalitu last updated on 22/Dec/16

$$\mathrm{A}\mathrm{body}\mathrm{of}\mathrm{mass}20\mathrm{g}\mathrm{performs}\mathrm{simple}\mathrm{harmonic}$$$$\mathrm{motion}\mathrm{at}\mathrm{a}\mathrm{frequency}\mathrm{of}5\mathrm{Hz},\mathrm{at}\mathrm{a}\mathrm{distance}$$$$\mathrm{of}10\mathrm{cm}\mathrm{from}\mathrm{the}\mathrm{mean}\mathrm{position}.\mathrm{its}\mathrm{velocity}$$$$\mathrm{is}200\mathrm{cm}/\mathrm{s}.\mathrm{calculate}$$$$\left(\mathrm{i}\right)\mathrm{Maximum}\mathrm{displacement}\mathrm{from}\mathrm{the}\mathrm{mean}$$$$\mathrm{position}$$$$\left(\mathrm{ii}\right)\mathrm{Maximum}\mathrm{velocity}$$$$\left(\mathrm{iii}\right)\mathrm{Maximum}\mathrm{potential}\mathrm{energy}.$$

Commented by tawakalitu last updated on 22/Dec/16

$$\mathrm{please}\mathrm{help}\mathrm{with}\mathrm{this}\mathrm{sires}.$$

Answered by mrW last updated on 22/Dec/16

$$\left(\mathrm{i}\right):10\mathrm{cm}$$$$\left(\mathrm{ii}\right):200+10\times 2\pi \times 5=514\mathrm{cm}/\mathrm{s}$$$$\left(\mathrm{iii}\right):\frac{1}{2}\times 0.1^2\times 0.02\times {\left(2\pi \times 5\right)}^2=0.1\mathrm{J}$$

Commented by tawakalitu last updated on 22/Dec/16

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{thanks}.$$

Answered by sandy_suhendra last updated on 22/Dec/16

$$\mathrm{m}=20\mathrm{gr}=0.02\mathrm{kg}$$$$\mathrm{f}=5\mathrm{Hz}$$$${\mathrm{y}}_1=10\mathrm{cm}\Rightarrow {\mathrm{v}}_1=200\mathrm{cm}/\mathrm{s}$$$$\left(\mathrm{i}\right)\mathrm{v}=\omega \sqrt{{\mathrm{A}}^2-{\mathrm{y}}^2}\Rightarrow \omega =2\pi \mathrm{f}$$$${\mathrm{v}}^2=4{\pi }^2{\mathrm{f}}^2({\mathrm{A}}^2-{\mathrm{y}}^2)$$$${200}^2=4{\pi }^2.5^2({\mathrm{A}}^2-{10}^2)$$$${\mathrm{A}}^2-100=\frac{400}{{\pi }^2}$$$${\mathrm{A}}^2=140.53$$$$\mathrm{A}=11.85\mathrm{cm}$$$$$$$$\left(\mathrm{ii}\right){\mathrm{v}}_{\max }=2\pi \mathrm{fA}=2\times 3.14\times 5\times 11.85=372.09\mathrm{cm}/\mathrm{s}=3.72\mathrm{m}/\mathrm{s}$$$$\left(\mathrm{iii}\right)\mathrm{PE}\max =\mathrm{KE}\max =\frac{1}{2}{\mathrm{mv}}_{\max }^2$$$$=\frac{1}{2}\times 0.02\times 3.{72}^2=0.14\mathrm{J}$$$$(\mathrm{KE}=\mathrm{kinetic}\mathrm{energy})$$

Commented by tawakalitu last updated on 22/Dec/16

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

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