Question

A system consists of a vertical spring with force constant k = 1,130 N/m, length L = 1.45 m, and object of mass m = 5.05 kg attached to the end (see figure). The object is placed at the level of the point of attachment with the spring unstretched, at position y_i = L, and then it is released so that it swings like a pendulum.

(a) Write Newton's second law symbolically for the system as the object passes through its lowest point. (Note that at the lowest point, r = L − y_f. Use the following as necessary: m, v, L, and y_f. Do not substitute numerical values; use variables only.)

(b) Write the conservation of energy equation symbolically, equating the total mechanical energies at the initial point and lowest point. (Use the following as necessary: m, L, k, y_f, and g. Do not substitute numerical values; use variables only.)

(c) Find the coordinate position of the lowest point.

(d) Will this pendulum's period be greater or less than the period of a simple pendulum with the same mass m and length L? Explain

Answer 1

The spring passes through vertical position the object is moving in a circular are of radius Also observe that y - Coordinate of the object at this point must be negative so the spring is stretched and an upward tension of magnitude greater than objects weight sigma F_y = -K y_f - m g = m v^2 /L - y_f Conservation of energy requires E = K E_1 + P E_g, f + PE = KE_f + PE_g, f + PE_s, t E = 0 + m g L + 0 = 1/2 mv^2 + m g + 1/2 k y^2_f m v^2 = 2 m g (L - y_f) - Ky^2_f from (a, m v^2 = -(L - y_f) (k y_f + m g) Substitute above into 2 m g (L - y_f) = -(L - y_f) (k y_f + m g) + k y^2_f After expanding and regrouping terms (2k) y^2_f + (3 mg - k L) y_f + (-3 m g L) = 0 Compare with quadratic equation a^2_y