Two masses connected by a spring

Question

I have a question about a problem I saw on a website

A mass is attached to one end of a spring, and the other end of the spring is attached to an immovable wall. The system oscillates with period T. If the wall is removed and replaced with a second mass identical to the first, what is the new period of oscillation of the system?

My answer: $\frac{T}{\sqrt{2}}$

Website's answer (Incorrect thanks to trula): $T\sqrt{2}$

My thought process (pretty hand-wavy I know):

Because the spring constant is inversely proportional (I think) to length of the spring, the spring constant will be doubled if you halve the length of the spring, which is what is essentially happening with two identical massess. Because $T=2\pi\sqrt{\frac{m}{k}}$, if you double $k$, you divide $T$ by $\sqrt{2}$. I don't know why they got $T\sqrt{2}$, but if I am wrong, please correct my misunderstandings.

Thanks for reading if you made it this far, and if my question is unclear, please tell me what I can fix about it.

Answer 1

The approach you have used appears to be correct. The correct answer should be $\frac{T}{\sqrt 2}$. As you have noted, halving the length of spring with spring constant $k$ results in a doubling of the spring constant.

Since we know that $$T=2\pi\sqrt{\frac{m}{k}}$$ and if we let $k\rightarrow 2k$, then $$T=2\pi\sqrt{\frac{m}{k}} \rightarrow 2\pi\sqrt{\frac{m}{2k}}=\frac{2\pi}{\sqrt 2}\sqrt{\frac{m}{k}}=\frac{1}{\sqrt 2} 2\pi\sqrt{\frac{m}{k}}=\frac{1}{\sqrt 2}T$$

The answer posted - $T\sqrt 2$ - could very well be a typo as also pointed out by @trula in the comments above.

Answer 2

Equal masses, centre of mass does not move, stretch $x$ relative to centre of mass position for both masses, force exerted by spring $k\cdot 2x$, equation of motion for each of the masses of the form $-2kx = m\ddot x$, period is reduced by factor $1/\sqrt 2$.