- #1

Est120

- 54

- 3

- Homework Statement
- Find the frequency of

vibration of the system for small amplitude values (small angle aproximation)

- Relevant Equations
- sum of the torques around the axis of rotation

adding all the torques around the red circle position (taking clockside direction as positive ):

-M*g*L*sin(theta)-k*x*y=I *w (considering that the suspension bar is of negligible mass as the problem indicates )here "x" is the normal "x" of hooke's law (I don't know exactly what it is for a problem in two dimensions) "y" is the lever arm between kx and the red circle

for me:

x=s=h*theta and y=h (because I think that the spring force is represented by the red vector at "L" and therefore this is its lever arm)

which ends at:

M*g*L*sin(theta)+k*h^(2)*theta= -I* d^(2)/dt^(2) [theta]

then using the small angle aproximation and factoring:

d^(2)/dt^(2) [theta]=-(theta)*[(M*g*L+k*h^(2))/I] so is in the form d^(2)/dt^(2) [theta]=-W^(2)*theta

and W is = sqrt[(M*g*L+k*h^(2))/I]= sqrt[(M*g*L+k*h^(2))/M*L^(2)]

which is the correct answer but i got two questions

1.-

what is "x"? in one-dimensional hooke's law "x" it is simply the horizontal distance.

here is the arc length? or the distance (straight line) between the point of application of the force and the point of balance?

2.-

what is the lever arm "y" for this it would help me to know what is the direction of the spring force as the red vector or as the black

i don't speak a lot of english sorry