If the spring constant is doubled, what value does the period have for a mass on a spring?

Question

anonymous · Answer

@amistre64 Do you mind helping me out?

amistre64 · Answer

the formula is what:

F = -kx

how would you say that this relate to period?

amistre64 · Answer

or better yet:$$T=2\pi \sqrt{\frac{m}{k}}$$

anonymous · Answer

Science is a challenge for me. But thanks for the formula. all i do is substitute and solve?

amistre64 · Answer

yes, formulas are nice that way, just sub and solve :)

anonymous · Answer

Thank you :)

anonymous · Answer

Wait so if the spring constant is doubled does that mean the period will double also?

amistre64 · Answer

lets double it and find out

$$T=2\pi \sqrt{\frac{m}{2k}}$$
$$T=\frac{2}{\sqrt2}\pi \sqrt{\frac{m}{k}}$$
$$T=\sqrt2~~pi \sqrt{\frac{m}{k}}$$

amistre64 · Answer

thats definantly not a doubling effect

anonymous · Answer

No it isnt. Thank you for the explanation. So it will increase by $$\sqrt{2}$$

anonymous · Answer

Nevermind, i was wrong.

amistre64 · Answer

sqrt(2) is about 1.414, so im not sure if we could call this an increase in period "[S]prings with a greater spring constant (stiffer springs) have a smaller period; masses attached to these springs take less time to complete a cycle. Their greater spring constant means they exert stronger restoring forces upon the attached mass. This greater force reduces the length of time to complete one cycle of vibration." http://www.physicsclassroom.com/Class/waves/u10l0d.cfm the length of the period is decreased from 2pi to sqrt(2) pi

anonymous · Answer

That link is helpful, thank you for the help, really appreciate it :)

amistre64 · Answer

good luck