A particle is executing SHM along a straight line. Its velocities at distance x_1 and x_2 from the mean position are V_1 and V

Q: A particle is executing $SHM$ along a straight line. Its velocities at distance $x_1$ and $x_2$ from the mean position are $V_1$ and $V_2$ respectively. Its time period is

b In $SHM,$ velocities of a particle at distances $x_{1}$ and $x_{2}$ from mean position are given by ${V_{1}^{2}=\omega^{2}\left(a^{2}-x_{1}^{2}\right)}...(i)$ ${V_{2}^{2}=\omega^{2}\left(a^{2}-x_{2}^{2}\right)}...(ii)$ From equations $(i)$ and $(ii),$ we get $V_{1}^{2}-V_{2}^{2}=\omega^{2}\left(x_{2}^{2}-x_{1}^{2}\right)$ $\omega=\sqrt{\frac{V_{1}^{2}-V_{2}^{2}}{x_{2}^{2}-x_{1}^{2}}} $ $T=2 \pi \sqrt{\frac{x_{2}^{2}-x_{1}^{2}}{V_{1}^{2}-V_{2}^{2}}}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A particle is executing $SHM$ along a straight line. Its velocities at distance $x_1$ and $x_2$ from the mean position are $V_1$ and $V_2$ respectively. Its time period is

A$2\pi \sqrt {\frac{{{x_1}^2 + {x_2}^2}}{{{V_1}^2 + {V_2}^2}}}$
B$2\pi \sqrt {\frac{{{x_2}^2 - {x_1}^2}}{{{V_1}^2 - {V_2}^2}}} $
C$2\pi \sqrt {\frac{{{V_1}^2 + {V_2}^2}}{{{x_1}^2 + {x_2}^2}}}$
D$2\pi \sqrt {\frac{{{V_1}^2 - {V_2}^2}}{{{x_1}^2 - {x_2}^2}}} $

AIPMT 2015,JEE MAIN 2021, Diffcult

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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