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

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  position  (ii) Maximum velocity  (iii) Maximum potential energy.

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

Commented by tawakalitu last updated on 22/Dec/16

please help with this sires.

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

Answered by mrW last updated on 22/Dec/16

(i):  10cm  (ii):  200+10×2π×5=514 cm/s  (iii): (1/2)×0.1^2 ×0.02×(2π×5)^2 =0.1 J

$$\left(\mathrm{i}\right):\:\:\mathrm{10cm} \\ $$$$\left(\mathrm{ii}\right):\:\:\mathrm{200}+\mathrm{10}×\mathrm{2}\pi×\mathrm{5}=\mathrm{514}\:\mathrm{cm}/\mathrm{s} \\ $$$$\left(\mathrm{iii}\right):\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{0}.\mathrm{1}^{\mathrm{2}} ×\mathrm{0}.\mathrm{02}×\left(\mathrm{2}\pi×\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0}.\mathrm{1}\:\mathrm{J} \\ $$

Commented by tawakalitu last updated on 22/Dec/16

God bless you sir. thanks.

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

Answered by sandy_suhendra last updated on 22/Dec/16

m=20 gr=0.02 kg  f=5 Hz  y_1 =10 cm ⇒ v_1 =200 cm/s  (i) v=ω(√(A^2 −y^2  ))    ⇒  ω=2πf      v^2 =4π^2 f^2 (A^2 −y^2 )   200^2 =4π^2 .5^2 (A^2 −10^2 )  A^2 −100=((400)/π^2 )  A^2 =140.53  A=11.85 cm    (ii) v_(max) =2πfA=2×3.14×5×11.85=372.09 cm/s=3.72 m/s  (iii) PE max=KE max=(1/2)mv_(max) ^2                   =(1/2)×0.02×3.72^2 =0.14 J  (KE=kinetic energy)

$$\mathrm{m}=\mathrm{20}\:\mathrm{gr}=\mathrm{0}.\mathrm{02}\:\mathrm{kg} \\ $$$$\mathrm{f}=\mathrm{5}\:\mathrm{Hz} \\ $$$$\mathrm{y}_{\mathrm{1}} =\mathrm{10}\:\mathrm{cm}\:\Rightarrow\:\mathrm{v}_{\mathrm{1}} =\mathrm{200}\:\mathrm{cm}/\mathrm{s} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{v}=\omega\sqrt{\mathrm{A}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:}\:\:\:\:\Rightarrow\:\:\omega=\mathrm{2}\pi\mathrm{f} \\ $$$$\:\:\:\:\mathrm{v}^{\mathrm{2}} =\mathrm{4}\pi^{\mathrm{2}} \mathrm{f}^{\mathrm{2}} \left(\mathrm{A}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right) \\ $$$$\:\mathrm{200}^{\mathrm{2}} =\mathrm{4}\pi^{\mathrm{2}} .\mathrm{5}^{\mathrm{2}} \left(\mathrm{A}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \right) \\ $$$$\mathrm{A}^{\mathrm{2}} −\mathrm{100}=\frac{\mathrm{400}}{\pi^{\mathrm{2}} } \\ $$$$\mathrm{A}^{\mathrm{2}} =\mathrm{140}.\mathrm{53} \\ $$$$\mathrm{A}=\mathrm{11}.\mathrm{85}\:\mathrm{cm} \\ $$$$ \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{v}_{\mathrm{max}} =\mathrm{2}\pi\mathrm{fA}=\mathrm{2}×\mathrm{3}.\mathrm{14}×\mathrm{5}×\mathrm{11}.\mathrm{85}=\mathrm{372}.\mathrm{09}\:\mathrm{cm}/\mathrm{s}=\mathrm{3}.\mathrm{72}\:\mathrm{m}/\mathrm{s} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{PE}\:\mathrm{max}=\mathrm{KE}\:\mathrm{max}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{mv}_{\mathrm{max}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{0}.\mathrm{02}×\mathrm{3}.\mathrm{72}^{\mathrm{2}} =\mathrm{0}.\mathrm{14}\:\mathrm{J} \\ $$$$\left(\mathrm{KE}=\mathrm{kinetic}\:\mathrm{energy}\right) \\ $$

Commented by tawakalitu last updated on 22/Dec/16

Thanks sir. i really appreciate.   God bless you.

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

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