Question

A 100-coil spring has a spring constant of 400 N/m. It is cut into four shorter springs, each of which has 25 coils. One end of a 25-coil spring is attached to a wall. An object of mass 43.4 kg is attached to the other end of the spring, and the system is set into horizontal oscillation. What is the angular frequency of the motion?

Answer 1

The issue is whether a shorter spring has a different k from its parent. I don't know.
The angular velocity will be sqrt(k/m), but is k=400 N/m?

Answer 2

The spring constant should stay approximately the same for such a long spring.

Answer 3

But it is cut into quarters.

Answer 4

Sorry, after doing some calculation it's obvious that it should quadruple. One can prove it by considering the energy stored in the long spring vs the cut up springs.

Answer 5

It intuitively seemed that the short springs should be stiffer.

Answer 6

I look forward to your analysis.

Answer 7

Stretch out the long spring to distance x. It has potential energy
\[\frac{kx^2}{2}\]
considering the 4 spring parts, they each store
\[\frac{k'(\frac{x}{4})^2}{2}=\frac{k'x^2}{32}\]
The energy stored in the 4 parts must be equal to the energy stored in the compound spring
\[4\times\frac{k'x^2}{32}=\frac{kx^2}{2}\Leftrightarrow 4k=k'.\]
So the spring constant of the shorter strings are 4 times that of the long string.

Answer 8

I think you are right.

Let the short springs have F = k' x.

Put tension F on the assembly of four short springs:
x will become 4x, because
they will all feel the same tension.

x = F/k

meaning k = k'/4, softer.

Answer 9

I understand that the spring constant is now four times what it was than before, but that still doesn't help me find the angular frequency of the spring with the weight on it

Answer 10

Does that mean that I just calculate for x?

Answer 11

w = angular frequency = sqrt (k/m)

w = sqrt (1600/43.4) = 6.07 rad/sec

Sorry to have left you hanging, or oscillating.