Solve:
#d^2x//dt^2 + g sin theta t =0#
if
# theta=g//l# , and g and l are constants?

Question

Isabella Ackerman · Accepted Answer

#x(t) = - g sin(g/1) t^2/2 + At + B# You want to know how to solve: #(d^2x)/(dt^2) = - g sin(g/1) # #implies (d x)/(dt) = - g sin(g/1) t + A# #implies x(t) = - g sin(g/1) t^2/2 + At + B#

Ezra Adams · Answer

# x = l^2/g \ sin theta t + At + B #

We have: # (d^2x)/(dt^2) + g sintheta t = 0# where #theta=g/l#, a constant which we could type as: # (d^2x)/(dt^2) = - g sintheta t # It is possible to "separate the variables" and obtain: # (dx)/(dt) = int \ - g sintheta t \ dt # Additionally, integrating provides us with: # (dx)/(dt) = g/theta cos theta t + A # And when we keep doing so, we obtain: # x = int \ g/theta cos theta t + A \ dt # So that: # x = g/theta^2 sin theta t + At + B # # \ \ = g/(g/l)^2 sin theta t + At + B # # \ \ = g l^2/g^2 \ sin theta t + At + B # # \ \ = l^2/g \ sin theta t + At + B #

Luke Alvarez · Answer

undefined The solution to the differential equation $ \frac{{d^2x}}{{dt^2}} + g \sin(\frac{g}{l} t) = 0 $, where $ \theta = \frac{g}{l} $ and $ g $ and $ l $ are constants, involves solving the second-order linear homogeneous ordinary differential equation. This equation doesn't have a general analytical solution in terms of elementary functions. However, it can be solved numerically or approximated under specific conditions or using numerical methods.

Solve: #d^2x//dt^2 + g sin theta t =0# if # theta=g//l# , and g and l are constants?