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A simple pendulum (mass M and length L) is suspended from
Chapter 7, Problem 7.31(choose chapter or problem)
A simple pendulum (mass M and length L) is suspended from a cart (mass m) that can oscillate on the end of a spring of force constant k, as shown in Figure 7.15. (a) Write the Lagrangian in terms of the two generalized coordinates x and 0, where x is the extension of the spring from its equilibrium length. (Read the hint in 7.29.) Find the two Lagrange equations. (Warning: They're pretty ugly!) (b) Simplify the equations to the case that both x and 0 are small. (They're still pretty ugly, and note, in particular, that they are still coupled; that is, each equation involves both variables. Nonetheless, we shall see how to solve these equations in Chapter 11 see particularly 11.19.) Figure 7.15 7.31
Questions & Answers
QUESTION:
A simple pendulum (mass M and length L) is suspended from a cart (mass m) that can oscillate on the end of a spring of force constant k, as shown in Figure 7.15. (a) Write the Lagrangian in terms of the two generalized coordinates x and 0, where x is the extension of the spring from its equilibrium length. (Read the hint in 7.29.) Find the two Lagrange equations. (Warning: They're pretty ugly!) (b) Simplify the equations to the case that both x and 0 are small. (They're still pretty ugly, and note, in particular, that they are still coupled; that is, each equation involves both variables. Nonetheless, we shall see how to solve these equations in Chapter 11 see particularly 11.19.) Figure 7.15 7.31
ANSWER:Step 1 of 8
The generalized coordinates for the given system with two degree of freedom are 
Where
Let the origin is at the equilibrium position of the spring such that the distance of the cart from the origin is
As there is no motion of the cart takes place along the y-axis, thus, the net velocity of the cart is,
The x-component of position of bob in terms of the extension of the spring and length of the pendulum is,
The y-component of position of bob in terms of length of the pendulum is,
Here, the negative sign indicates that the bob is present below the x-axis (in the negative of y-axis).