#angular frequency

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 161 N/m. The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2. The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2.

Well, lucky for you, the first point is for units. Spring constant k is measured in N/m. Hooray. Step 1 complete.

Now the thinking part...

Well, let’s see what we have: k1, the amplitude, and the max v.

Now, what kind of equations does that leave us with?

Well, A1 = 2A2 since the amplitude of spring 1 is twice as big as that of spring 2.

We know that maximum velocity is: A*ω

We also know that (angular frequency) ω = (m/k)^.5

Lastly, we know that the maximum velocities are the same in both cases.

Lots of information. But you’re smart.

Algebra time!! :)

Basically: A1*ω1 = 2A2*ω2

Which leads to… A1 * (m/161)^.5 N/m = 2A * (m/k)^.5

Divide by A (& square) and get… ((m/161)^.5)^2 = (2*(m/k)^.5)^2

Which is… m/161 = 4 * m/k

Then… mk/161 = 4m

mk = 4*161*m

mk = 644m

Divide by m because you can

[Click here for a PDF version of this post] Karl had a couple of fun filter problems that he chatted with me about. I didn’t remember how do find the bandwidth, nor the resonant frequency of such a circuit, and after he solved the problems, I tried them for myself. The circuits, as seen in fig. 1, were RC and LC respectively, and both ended up having the same transfer function. fig 1a. RC…

omega = angular frequency vector (rad/s)

T = period (s)

f = frequency (Hz)