# The relationship between time taken per oscillation and mass

In an experiment, a ruler is connected to the table and some weights are bounded to one end of the ruler. The ruler is then flicked and the time taken per oscillation is measured.

I have plotted a graph with the data I have collected, with the mass on the y-axis and time on x-axis. The graph produced appears to be a curve. I have tried altering the values on the x-axis; I have squared it, 1 over the square of it, square rooted it, and I found that the graph becomes linear when the values are squared. So the mass should be proportional to 1 over the square of the time taken.

I have tried finding out a mathematical relationship for this, but I am not sure if this is correct or not.

Well, if we make w=angular velocity, then w=θ/t, with θ being angular displacement and t being the time period. Since θ belongs in a circle, then it is safe to say that w=2π/t (?)

Also, if the force of an oscillation is proportional to -displacement (x), then it is true to say that F=-kx, with k being a constant.

Since F also = ma, then ma=-kx.

According the the simple harmonic wave equation for acceleration is a=-xw2sinwt. Since the formula for displacement(x) = x sinwt and a=-w2(x sinwt), then a=-xw2

So ma=-kx will become m(-xw2)=-kx, then using some algebra, m=k/w2. Since w=2π/t, then m=kT2/4π2.

Since k/4π2 is a constant, I can ignore that and say m is proportional to t2.

Is my reasoning true? I feel like I am wrong in quite a few spots.

Thanks for helping

Last edited:

## Answers and Replies

Gokul43201
Staff Emeritus
Science Advisor
Gold Member
Yes, this is essentially correct, but you take a few confusing steps because of a poor choice of notation. For instance, you use x to represent displacement as well as the amplitude (or maximum displacement). You should use different symbols for these - it will help keep things clear for you as well.

You write: x = x sin(wt), which looks nonsensical.

Better would be something like: x = A sin (wt), or x = x0 sin (wt).

Also, next time, a question like this is better suited for the Intro Physics section.

Thank you.

Yes thank you very much! It is so helpful!