Question

Please answer all parts of the problem and e
A system consists of a vertical spring with force constant k = 1,060 N/m, length L = 1.25 m, and object of mass m = 5.70 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 system consists of a vertical spring with force constant k = 1,060 N/m, length L = 1.25 m, and object of mass m = 5.70 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 yL, and then it is released so that it swings like a pendulum L- 0 (a) Write Newton's second law symbolically for the system as the object passes through its lowest point. (Note that at the lowest point, rL- and y Do not substitute numerical values; use variables only.) Use the following as necessary: m, v, L (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, Ye 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? greater O less

Answer 1

a)

$\Sigma f_{y} = m.a_{y}$

b)

$mv^{2}=2mgh$

$hieght \rightarrow h=v^{2}/2g$

$\Delta L= L-y_{f}= mg/k$

=>L - y(f) =5.70 X 9.8/1060 =0.0526 x 10^2 meters

we know that L=1.5 m,

$y_{f}=$  5.26 - 1.5 = 3.76 m

c)

$period \rightarrow T=2\Pi /\omega$

$\omega=\sqrt{k/m}$

=>13.6 rad/sec

T= 2 X 3.14/13.6 = 0.46 sec.