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Question-190257




Question Number 190257 by jlewis last updated on 30/Mar/23
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Commented by jlewis last updated on 30/Mar/23
Commented by mahdipoor last updated on 31/Mar/23
x=Asin(Bt+φ)   ⇒  x^(..) +wx=−AB^2 sin(Bt+φ)+w^2 Asin(Bt+φ)=0  ⇒B=w ⇒  x=Asin(wt+φ) ⇒ { ((x(0)=0 ⇒ φ=0 or π)),((x^. (0)=v ⇒v=Awcosφ)) :}  if:     x=Asin(wt+φ)  v>0 ⇒ φ=0    A=(v/w)  v<0 ⇒ φ=π    A=((−v)/w)  v=0 ⇒ A=0
$${x}={Asin}\left({Bt}+\phi\right)\:\:\:\Rightarrow \\ $$$$\overset{..} {{x}}+{wx}=−{AB}^{\mathrm{2}} {sin}\left({Bt}+\phi\right)+{w}^{\mathrm{2}} {Asin}\left({Bt}+\phi\right)=\mathrm{0} \\ $$$$\Rightarrow{B}={w}\:\Rightarrow \\ $$$${x}={Asin}\left({wt}+\phi\right)\:\Rightarrow\begin{cases}{{x}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\:\phi=\mathrm{0}\:{or}\:\pi}\\{\overset{.} {{x}}\left(\mathrm{0}\right)={v}\:\Rightarrow{v}={Awcos}\phi}\end{cases} \\ $$$${if}:\:\:\:\:\:{x}={Asin}\left({wt}+\phi\right) \\ $$$${v}>\mathrm{0}\:\Rightarrow\:\phi=\mathrm{0}\:\:\:\:{A}=\frac{{v}}{{w}} \\ $$$${v}<\mathrm{0}\:\Rightarrow\:\phi=\pi\:\:\:\:{A}=\frac{−{v}}{{w}} \\ $$$${v}=\mathrm{0}\:\Rightarrow\:{A}=\mathrm{0} \\ $$

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