Direction of a spring's force in more than one dimension

[Est120]

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

 

[Est120]

with that W value i just used the formula W=2*pi* f and f is the correct answer but that's not my doubt

 

[FactChecker]

I think that the problem wants you to assume a very small angle $\theta$. In that case, your questions are not significant.
In a more general case, you would want to determine the direction of the spring, the amount it would be stretched or compressed in that direction, and apply the spring force exactly in that direction.

 

[Est120]

I think that the problem wants you to assume a very small angle $\theta$. In that case, your questions are not significant.
In a more general case, you would want to determine the direction of the spring, the amount it would be stretched or compressed in that direction, and apply the spring force exactly in that direction.

but where does the spring force points?

 

[FactChecker]

but where does the spring force points?

In this problem, I don't think that they want you to do any geometry with anything more than a tiny angle, so you can assume that the spring and the spring force are horizontal.

In the more general case, you would have to figure out where the ends of the spring are and the force would be along the straight line between the ends.

 

[Lnewqban]

... 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?

Imagine the bar in its vertical position and no tension on the spring.
The direction of the spring is perpendicular to the bar.
The end of the bar describes a red semi-circle of radius h around the red point.
The left end of the spring (with no tension) describes a blue semi-circle around its right end or anchoring point.

In this case, the distance "x" that the spring stretches is very close to the horizontal distance between these two points:
1) The original point of spring-bar connection when the bar was in its vertical position and there was no tension on the spring (bar and spring are perpendicular to each other).
2) The point of spring-bar connection when the bar is angled at full amplitude and tension on the spring is maximum (bar and spring form an angle a little less than 90 degrees). Blue semi-circle crosses red one.

However, the effective force in your equation that produces the torque that helps the mass and bar to swing back must be perpendicular to the bar (red vector in the schematic).
That force deviates the angle $\Theta$ from horizontal.
For that reason, you input $kxcos(\Theta)$ in your equation.
Since for that small angle, the value of $cos(\Theta)$ is so close to 1, the difference is very small, as previously explained.

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

The lever arm for the red vector of magnitude $kxcos(\Theta)$ is always h for any position of the bar.

 

[hutchphd]

There is so much wrong here that you should start again. In particular the equations are dimensionally inconsistent (and largely illegible).

First draw a free body diagram !