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Câu hỏi: Một vật dao động điều hòa khi vận tốc là $10 \pi \sqrt{3} \mathrm{~cm} / \mathrm{s}$ thì gia tốc là $2 \mathrm{~m} / \mathrm{s}$, còn khi vận tốc là $10 \pi \mathrm{cm} / \mathrm{s}$ thì gia tốc là $-2 \sqrt{3} \mathrm{~m} / \mathrm{s}$. Lấy $\pi^{2}=10$. Chiều dài quỹ đạo của vật là
A. 1 m
B. $2\pi \text{cm}$
C. $20 \mathrm{~cm}$.
D. $10 \mathrm{~cm}$.
& \dfrac{v_{1}^{2}}{v_{\max }^{2}}+\dfrac{a_{1}^{2}}{a_{\max }^{2}}=1 \\
& \dfrac{v_{2}^{2}}{v_{\max }^{2}}+\dfrac{a_{2}^{2}}{a_{\max }^{2}}=1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \dfrac{{{\left( 10\pi \sqrt{3} \right)}^{2}}}{v_{\max }^{2}}+\dfrac{{{2}^{2}}}{a_{\max }^{2}}=1 \\
& \dfrac{{{\left( 10\pi \right)}^{2}}}{v_{\max }^{2}}+\dfrac{{{\left( 2\sqrt{3} \right)}^{2}}}{a_{\max }^{2}}=1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{v}_{\max }}=20\pi \left( cm/s \right) \\
& {{a}_{\max }}=4m/{{s}^{2}}=400cm/{{s}^{2}} \\
\end{aligned} \right.$
$\omega =\dfrac{{{a}_{\max }}}{{{v}_{\max }}}=\dfrac{400}{20\pi }=\dfrac{20}{\pi }$ (rad/s)
$A=\dfrac{{{v}_{\max }}}{\omega }=\dfrac{20\pi }{20/\pi }\approx 10$ (cm)
$L=2A=2.10=20$ (cm).
A. 1 m
B. $2\pi \text{cm}$
C. $20 \mathrm{~cm}$.
D. $10 \mathrm{~cm}$.
$\left\{ \begin{aligned}& \dfrac{v_{1}^{2}}{v_{\max }^{2}}+\dfrac{a_{1}^{2}}{a_{\max }^{2}}=1 \\
& \dfrac{v_{2}^{2}}{v_{\max }^{2}}+\dfrac{a_{2}^{2}}{a_{\max }^{2}}=1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \dfrac{{{\left( 10\pi \sqrt{3} \right)}^{2}}}{v_{\max }^{2}}+\dfrac{{{2}^{2}}}{a_{\max }^{2}}=1 \\
& \dfrac{{{\left( 10\pi \right)}^{2}}}{v_{\max }^{2}}+\dfrac{{{\left( 2\sqrt{3} \right)}^{2}}}{a_{\max }^{2}}=1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{v}_{\max }}=20\pi \left( cm/s \right) \\
& {{a}_{\max }}=4m/{{s}^{2}}=400cm/{{s}^{2}} \\
\end{aligned} \right.$
$\omega =\dfrac{{{a}_{\max }}}{{{v}_{\max }}}=\dfrac{400}{20\pi }=\dfrac{20}{\pi }$ (rad/s)
$A=\dfrac{{{v}_{\max }}}{\omega }=\dfrac{20\pi }{20/\pi }\approx 10$ (cm)
$L=2A=2.10=20$ (cm).
Đáp án C.