What Is the Oscillation Period of a Rod Attached to a Spring?

  • Thread starter bcjochim07
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In summary, the problem involves a 200g uniform rod attached to a horizontal spring, with a length of 20 cm and a spring constant of 3N/m. The rod's oscillation period can be calculated using the formula for angular frequency and the equation for period of a simple harmonic oscillator. The total restoring force for the pendulum-spring combination can be determined by using the formula F = ma = -kx.
  • #1
bcjochim07
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Homework Statement


A 200g uniform rod is pivoted at one end. The other end is attached to a horizontal spring. The spring is neither stretched nor compressed when the rod hangs straight down. What is the rod's oscillation period? You can assume that the rod's angle from the vertical is always small. The rod's length is 20 cm and the spring's constant is 3N/m.



Homework Equations





The Attempt at a Solution



Without the spring the rod's angular frequency would be sqrt[(.200)(9.80)(.10m)/(.0027kgm^2) Then I could go to period. Will someone tell me how I can get the period for the spring? The problem is I am used to doing problems with just a spring and a block, but I am not quite sure how to deal with the extended object. Would I just use omega= sqrt [k/m] with mass .200kg and k = 3.0 N/m? Then I could go to period, and maybe add the two period together?
 
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  • #2
For a simple harmonic oscillator the restoring force is F = ma = -kx. what is the total restoring force for the pendulum string combination? If that is in the form -kx you can use the equaton for the period of a simple harmonic oscillator.
 
  • #3




Hello, it seems like you are struggling with this problem. I can see that you are on the right track by using the equation for angular frequency, but it is important to also consider the effect of the spring on the rod's oscillation period. In this case, you can use the equation T = 2π√(I/mgd) where T is the period, I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of mass of the rod.

To calculate the moment of inertia for a rod pivoted at one end, you can use the equation I = (1/3)mL^2, where m is the mass of the rod and L is the length of the rod. In this case, the length of the rod is given as 20 cm, but it would be best to convert it to meters for consistency.

Once you have calculated the moment of inertia and the distance from the pivot point to the center of mass, you can plug these values into the equation for period. Remember to also include the effect of the spring by using the equation you mentioned, T = 2π√(m/k), where m is the mass and k is the spring constant.

I hope this helps you understand how to approach this problem. If you are still having trouble, don't hesitate to ask for more clarification. Good luck!
 

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