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Question Number 33533 by NECx last updated on 18/Apr/18

A mass of 2kg is attached to a  spring with constant 8N/m.It is⊛  then displaced to the point x=2.  What time does it take for the block  to travel to the point x=1?  a)40s b)60s c)30s d)20s

$${A}\:{mass}\:{of}\:\mathrm{2}{kg}\:{is}\:{attached}\:{to}\:{a} \\ $$$${spring}\:{with}\:{constant}\:\mathrm{8}{N}/{m}.{It}\:{is}\circledast \\ $$$${then}\:{displaced}\:{to}\:{the}\:{point}\:{x}=\mathrm{2}. \\ $$$${What}\:{time}\:{does}\:{it}\:{take}\:{for}\:{the}\:{block} \\ $$$${to}\:{travel}\:{to}\:{the}\:{point}\:{x}=\mathrm{1}? \\ $$$$\left.{a}\left.\right)\left.\mathrm{4}\left.\mathrm{0}{s}\:{b}\right)\mathrm{60}{s}\:{c}\right)\mathrm{30}{s}\:{d}\right)\mathrm{20}{s} \\ $$

Answered by ajfour last updated on 18/Apr/18

x=2cos ωt = 2cos ((√(k/m)) t)  1=2cos ((√(8/2)) t)  ⇒  2t = (π/3)    ⇒   t=(π/6)s   .

$${x}=\mathrm{2cos}\:\omega{t}\:=\:\mathrm{2cos}\:\left(\sqrt{\frac{{k}}{{m}}}\:{t}\right) \\ $$$$\mathrm{1}=\mathrm{2cos}\:\left(\sqrt{\frac{\mathrm{8}}{\mathrm{2}}}\:{t}\right) \\ $$$$\Rightarrow\:\:\mathrm{2}{t}\:=\:\frac{\pi}{\mathrm{3}}\:\:\:\:\Rightarrow\:\:\:{t}=\frac{\pi}{\mathrm{6}}{s}\:\:\:. \\ $$

Commented by NECx last updated on 19/Apr/18

thank you so much sir Ajfour.Your  presence is really felt.

$${thank}\:{you}\:{so}\:{much}\:{sir}\:{Ajfour}.{Your} \\ $$$${presence}\:{is}\:{really}\:{felt}. \\ $$

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