Physics-Simple Harmonic Motion

Question

A block with mass M is attached to a vertical spring with a spring constant k. When the block is displaced from equilibrium and released, its period is T. A second identical spring k is added to the first spring in parallel. What is the period of oscillations when the block is suspended from two springs?

Answer 1

The relationship between the system mass, M_sys, the system spring constant, K_sys, and the period, T, is the following:

K_sys = 4π² M_sys / T²

If you'd like to know where this arises from, it is from that the natural frequency, ω_n, and has the following equations:

ω_n = √(K_sys / M_sys) = 2πf = 2π / T

After a little algebra and rearranging the equations for natural frequency, you arrive to the first equation I displayed.

So, in the first part, K_sys = k, M_sys = m, and T = T (redundant, I know, but you'll need this later on).

k = 4π² m / T²

In the second part, springs in parallel add normally together (k_eq = k₁ + k₂), so K_sys = 2k. And M_sys = m but this time T = x, where x is our unknown period. Setting up this systems equation looks like:

2k = 4π² m / x²

What to notice with this is that we have a replacement for k from the first system, so we can plug it in like so:

2 (4π² m / T²) = 4π² m / x²

Since 4π² m is on both sides, you can cancel them out like:

2/T² = 1/x²

Invert both sides:

T²/2 = x²

Square root both sides and you get that:

The period of the two identical springs in parallel x = T√(2)/2