Pendulum with horizontal spring

  • #1

Homework Statement



A 5kg sphere is connected to a thin massless but rigid rod of length L=1.3 m to form a simple pendulum. The rod is connected to a nearby vertical wall by a spring with a spring constant k= 75N/m, connected to it at a distance h=1.1 m below the pivot point of the pendulum. What is the angular frequency (in rad/s) of the system for small amplitude oscillations.



Homework Equations



[tex]\alpha[/tex] = [tex]\omega[/tex]2*[tex]\theta[/tex]
I=m*r2
Torque=f*d*sin[tex]\theta[/tex]

or small oscillations, sin[tex]\theta[/tex]=[tex]\theta[/tex] and cos[tex]\theta[/tex]=1



The Attempt at a Solution



sum the torque and set them equal to [tex]\alpha[/tex]I

gravitational torque= Lmg[tex]\theta[/tex]

if theta is the angle between the rod and its equilibrium position, the angle of the spring torque is (theta + [tex]\pi[/tex]/2), so the spring torque is
f*d*sin(theta + [tex]\pi[/tex]/2). The force is k*[tex]\Delta[/tex]x, or h*sin(theta + [tex]\pi[/tex]/2) and the distance is h, giving

torque from spring =k*h2*sin[tex]\theta[/tex]*sin(theta + [tex]\pi[/tex]/2)
or k*h2*[tex]\theta[/tex]*(theta + [tex]\pi[/tex]/2)

My final equation is

mL2[tex]\omega[/tex]2tex]\theta[/tex] = Lmg[tex]\theta[/tex] + k*h2*[tex]\theta[/tex]*(theta + [tex]\pi[/tex]/2)

The problem is that this gives me a [tex]\omega[/tex] that is dependant on [tex]\theta[/tex], which isn't possible. It should be constant. Can someone please tell me where I'm going wrong?

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
ehild
Homework Helper
15,510
1,897
sin(theta+pi/2)= cos(theta). For small angles, cos(theta) =1.

ehild
 

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