Recent content by angelwentheng

then how to explain??:confused:

For a pendulum of length L and mass m, making an angle θ with the vertical, three forces act on the mass m. The vertical force due to gravity, mg. A horizontal restoring force (towards the equilibrium position) of mgsin(θ), and the tension, which is equal and opposite to the resultant of the...

external forces means f(t)...cz actually the original eqn will be mx'' + Bx' + kx = f(x) since there is no external force...therefore f(x)=o but why increasing the resistance?the question said directly proportional to velocity... erm...sorry i 'm nt physics students...so i might not...

External forces are zero, so the governing equation is homogeneous: mx"+Bx'+kx=0 x"=d²x/dt². x'=dx/dt x=displacement, positive to the right m=mass B=resistance proportional to the velocity k=stiffness (of a spring), which is resistance proportional to the displacement.

sorry I have no idea for this ques...can u help me?

A simple pendulum of length is oscillating through a small angle θ in a medium for which the resistance is proportional to the velocity. Obtain the differential equation of its motion and discuss the motion. sorry help me with these...I'm totally blank

oh...i am blur...sorry

Thanks~I'm wondering am I need to do the forced oscillation equation?which means using the spring equation f = kx

sorry for not noticing the sign

0.2(d^2 y)/(dt^2 )+ 1.2dy/dt + 2y = r(t) 0.2(d^2 y)/(dt^2 ) + 1.2dy/dt + 2y = 5cos4t 0.2m^2 + 1.2m + 2 = 0 m = (-b±√(b^2-4ac))/2a = (-(1.2)± √(〖(1.2)〗^2-4(0.2)(2)))/(2(0.2)) = (-1.2 ±0.4i)/0.4 = -3 ± i y_h(t) = e^(-3t) (Acosx+Bsinx) r (t) = 5cos4t y_p (t) = pcos4t +...

The differential eqn that governs the forced oscillation is shown below: 0.2(d^2 y)/(dt^2 )+ 1.2dy/dt + 2y = r(t) where r(t) is the external force Given that r(t) = 5cos4t with y(0) = 0.5 and y'(0) = 0. Find the equation of motion of the forced oscillation.. Please help me to solve by...