# Resonance Frequency and Centre of Gravity

Is there a relationship between these 2

As of now I am moving a piece blutac up and down a ruler to change my COG. I have connected the base of my ruler to a vibration generator and am measuring the highest amplitude to measure the resonance frequency.

The closer the COG is to to the base the higher my resonance frequency is.

• Do you have data from your experiment? Please can you post them into your question as you did with your earlier question? – sammy gerbil Feb 12 at 20:02
• yup I have added the data i collected! – ritzrori Feb 13 at 7:53
• Thanks. I guess you have calulated the position of the COG. Do you have the raw data - viz. mass of blu-tak and ruler and position of blu-tak? As BillN shows, resonance frequency depends on the moment of inertia as well as position of COG. – sammy gerbil Feb 13 at 17:38
• the ruler was 11.8 g and the blu tac os 6.7g – ritzrori Feb 13 at 18:11

If you are talking about a pendulum in a gravitational field, yes there is a relationship between the two. If you do a free-body diagram and analyze the situation, you will get a differential equation: $$\frac{\mathrm{d^2}\theta}{\mathrm{dt^2}}=\frac{-mgr}{\mathcal{I}}\sin\theta,$$ where

• List item $$\theta$$ is the angular position relative to vertical (direction of the gravitational field of magnitude $$g$$,
• $$t$$ is time,
• $$m$$ is the mass of the pendulum,
• $$r$$ is the distance of the center of mass from the pivot point, and
• $$\mathcal{I}$$ is the moment of inertial about the pivot point in the plane of oscillation.

Based on this equation the angular frequency of (small) oscillations will be $$\Omega = \sqrt{\frac{mgr}{\mathcal{I}}},$$ which will be in radians per second if all the other quantities are in SI units.

So, a change in the position of CoM due to a moveable mass could change the frequency as long as the moment of inertia changes at a different rate. For a moveable mass on a stick, that is the case.

• The OP is vibrating a ruler which is positioned vertically and clamped at the lower end in a variable frequency oscillator. – sammy gerbil Feb 12 at 21:17
• @sammygerbil It's an analogous situation with a torque-driven restoring action and the CM and MoI changing at different rates. It's still a type of pendulum and not a linear spring-mass system. Linear spring-mass doesn't have torque if you ignore the rotation of the spring. – Bill N Feb 13 at 15:28
• im assuming m in the formula is center of mass? – ritzrori Mar 15 at 22:57