T = 2(pi)(sqrt(m/k))

Say you had to find the period of oscillation when k is doubled...what would you multiply or divide T by to mirror what you do on the LHS?

Would it be sqrt(1/2)T?

Can I recommend a read here

https://www.physicsforums.com/threads/simple-harmonic-motion-from-uniform-circular-motion.291504/

this should help you understand the problem a little more.

T is the time it takes to do 1 oscillation. just plug in your new k, can get your new time. I guess the question is very vague

I understand the question I'm just not sure how to duplicate what I'm doing on the RHS, on the LHS. This is purely for when you're given no values for T, m or k.

You don't have to do anything to the LHS, just plug in the larger 2k. and the new T is the periodicity

post the exact question, as what you have posted is quite vague

I'll use this question cos it seems a bit more clear:

The time period of a pendulum on Earth is 1.0 s. What would be the period of a
pendulum of the same length on a planet with half the density but twice the radius of
Earth?

A 0.5 s
B 1.0 s
C 1.4 s
D 2.0 s

So the gravitational field strength would be twice that of on Earth, so I would sub g = 2g into the equation T = 2*pi*sqrt(l/g)

But to find the new value of T I need to do to the LHS exactly what I did to the RHS right? So would I multiply T by sqrt(1/2)?

This is not a fixed solution. you need to use the earth information to find out the L of the pendulum.

1 = 2*pi*sqrt(l/9.81) Solve for l (part 1)

Once you find that, you can plug it into the next equation

T = 2*pi*sqrt(L_fp1/(9.81/2))

and solve for T

I would assume the answer is 1.4 as this is the only one that isn't really a perfect number.