Question A 500 g block is attached to a horizontal spring with a springconstant of 50 N/m. At t = 0.1 s, its positionis x = 0.2 m and its velocity is vx = 0.5 m/s.determine the amplitude, the pulsation and the phase constant ofthe system. Write the complete functionx(t) of this system and At what instant is the mass at x = -0.2 m with a velocity of vx =0.5 m/s for the first time since it was released?first time since it was released?

\begin{array}{l}x(t)=A \sin (\omega t+4) \\V(t)=\frac{d x(t)}{d t}=A \omega \cos (\omega t+\phi) \\a(t)=\frac{d v(t)}{d t}=-\omega^{2} A \cos (\omega t+\phi) \\a(t)=-\omega^{2}[A \cos (\omega t+4)] \\a=-\omega^{2} x\end{array}Here \omega=10 rodls\therefore a=-100 xAlgebraic equotion of a as a function of xAso\begin{array}{l}U^{2}=[A \omega \cos (\omega t+4)]^{2} \\V^{2}=A^{2} \omega^{2} \cos ^{2}(\omega t+4) \\U^{2}=\omega^{2} A^{2}\left[1-\sin ^{2}(\omega t+4)\right] \\U^{2}=\omega^{2}\left[A^{2}-A^{2} \sin ^{2}(\omega t+4)\right] \\V^{2}=\omega^{2}\left[A^{2}-x^{2}\right]\end{array}V=\omega \cdot \sqrt{A^{2}-x^{2}}Algebric equotion of V Here \omega=10 \operatorname{rad} / \mathrm{s} \& A=0.206 os a functim x. or A^{2}=0.0425\therefore V=10 \sqrt{0.0425-x^{2}}notural leyth(Equillibriom position)geverr Eqn of Oscillotim of sprirg-black Syster is given by;x(t)=A \sin (\omega t+\&)Here A= amplutude\omega= anjular frequen & 4= phase constantgiver: m=500 \mathrm{~g}=0.5 \mathrm{~g} \quad k=50 \mathrm{~N} / \mathrm{m}K=m \omega^{2} \Rightarrow \omega=\sqrt{k / m}=\sqrt{\frac{50}{0.5}}=10 \operatorname{rod} / \mathrm{s}veloaty V(t)=\frac{d x}{d t}V(t)=A \omega \cos (\omega t+\phi)\because A t t=0.1 \mathrm{sec} \quad x=0.2 \mathrm{~m} \quad \& \quad v=0.5 \mathrm{~m} / \mathrm{s}From(i)\begin{array}{cc}0.2=A \sin (0.1 \omega+4) & 0.5=A \times 10 \cos (10 \times 0.1+4) \\0.2=A \sin (0.1 \times 10+\phi) & 10 A \cos (q+1)=0.5 \\0.2=A \sin (4+1) & A \cos (\phi+1)=0.05 \\A^{2}=0.2^{2}+0.05^{2}=0.0425\end{array}Amplitude A=0.206 \mathrm{~m}=20.6 \mathrm{~cm}Answir.\begin{aligned}\sin (4+1) & =\frac{0.2}{A} \quad \cos (4+1)=\frac{0.05}{A} \\\sin (4+1) & =\frac{0.2}{0.206} \quad \cos (4+1)=\frac{0.05}{0.206} \\\therefore 4+1 & =61.33 \mathrm{rod} \\A & =75 \text { rodian } 0.33 \mathrm{rod} \\4 & =18.9^{\circ} \text { Answer }\end{aligned}Answerpulsotion or frequency f=\frac{\omega}{2 \pi}=\frac{10}{2 \pi}=\frac{1.6 \mathrm{~Hz}}{\text { Ansinan }} Ansiver.blten x=-0.2 \quad V=0.5 \mathrm{mls}\begin{array}{l}x=A \sin (\omega t+4) \\x=(0.206 m) \sin (10 t+0.33)\end{array}The complet functim-0.2=0.206 \sin (10 t+0.37)(Answer)\sin (10 t+0.33)=-\frac{0.2}{0.206}\begin{aligned}10 t+0.33 & =\sin ^{-1}\left(\frac{-0.2}{0.206}\right) \\& =2 \pi-1.33 \quad \text { As } \quad \forall>0\end{aligned}t=\frac{2 \pi-1.33-0.33}{10}=0.46 \mathrm{sec} ...