Question

Consider two pendulums of length $L$ connected by a spring with spring constant $k$ placed at a distance $a$ from the pivot point of the pendulum.\ a) Using Lagrangian Mechanics, show that the equations of motion of the two pendulums are given by:\ $m_1L\ddot{\theta_1} + m_1gL\sin\theta_1 - a^2k(\theta_2 - \theta_1) = 0$\ $m_2L\ddot{\theta_2} + m_2gL\sin\theta_2 + a^2k(\theta_2 - \theta_1) = 0$\ b) Derive the same equations using Newton's 2$^{nd}$ Law.

Name: Consider two pendulums of length L connected by a spring with spring constant k placed at a distance a from the pivot point of the pendulum.a) Using Lagrangian Mechanics, show that the equations of motion of the two pendulums are given by:m1Lθ̈1̈ + m1gLsinθ1 - a^2k(θ2 - θ1) = 0m2Lθ̈2̈ + m2gLsinθ2 + a^2k(θ2 - θ1) = 0b) Derive the same equations using Newton's 2^nd Law.
Uploaded: 2021-12-26T10:00:15-08:00
Duration: 4 min
Channel: Pranay Shrivastava
Description: Consider two pendulums of length L connected by a spring with spring constant k placed at a distance a from the pivot point of the pendulum.a) Using Lagrangian Mechanics, show that the equations of motion of the two pendulums are given by:m1Lθ̈1̈ + m1gLsinθ1 - a^2k(θ2 - θ1) = 0m2Lθ̈2̈ + m2gLsinθ2 + a^2k(θ2 - θ1) = 0b) Derive the same equations using Newton's 2^nd Law.

          Consider two pendulums of length $L$ connected by a spring with spring constant $k$ placed at a distance $a$ from the pivot point of the pendulum.\
a) Using Lagrangian Mechanics, show that the equations of motion of the two pendulums are given by:\
$m_1L\ddot{\theta_1} + m_1gL\sin\theta_1 - a^2k(\theta_2 - \theta_1) = 0$\
$m_2L\ddot{\theta_2} + m_2gL\sin\theta_2 + a^2k(\theta_2 - \theta_1) = 0$\
b) Derive the same equations using Newton's 2$^{nd}$ Law.

Consider two pendulums of length L connected by a spring with spring constant k placed at a distance a from the pivot point of the pendulum.a) Using Lagrangian Mechanics, show that the equations of motion of the two pendulums are given by:m1Lθ̈1̈ + m1gLsinθ1 - a^2k(θ2 - θ1) = 0m2Lθ̈2̈ + m2gLsinθ2 + a^2k(θ2 - θ1) = 0b) Derive the same equations using Newton's 2^nd Law.

Added by Deborah P.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Pranay Shrivastava

12/26/2021

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a) Using Lagrangian Mechanics: Show more…

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Consider two pendulums of length L connected by a spring with spring constant k placed at a distance a from the pivot point of the pendulum. a) Using Lagrangian Mechanics, show that the equations of motion of the two pendulums are given by: mLÎ¸'' + mgLsinÎ¸ - ak(Î¸1 - Î¸2) = 0 mLÎ¸'' + mgLsinÎ¸ + ak(Î¸1 - Î¸2) = 0 b) Derive the same equations using Newton's 2nd Law.