FurtherMaths2020
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based on the equation for damped SHM, shouldnt the period be: T = (2pi)/[(square root 2)(k)]? because we have 2k^2 in the place of w^2?
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ghostwalker
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(Original post by FurtherMaths2020)
based on the equation for damped SHM, shouldnt the period be: T = (2pi)/[(square root 2)(k)]? because we have 2k^2 in the place of w^2?
Looking at the equation for damped SHM:

If there was no damping, i.e. k=0, then yes, the period is \dfrac{2\pi}{\omega}.

However, with k>0, we have damping, and the effect of damping is to increase the period of oscillation.

Solving the damped SHM, you will find that the period is given by \dfrac{2\pi}{\sqrt{\omega^2-\frac{1}{4}k^2}}

If you now look at your example of the spring, and use the constants given there, the period is

\displaystyle\frac{2\pi}{\sqrt{(  2k^2)-\frac{1}{4}(2k)^2}}=\frac{2\pi}{  k}
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FurtherMaths2020
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(Original post by ghostwalker)
Looking at the equation for damped SHM:

If there was no damping, i.e. k=0, then yes, the period is \dfrac{2\pi}{\omega}.

However, with k>0, we have damping, and the effect of damping is to increase the period of oscillation.

Solving the damped SHM, you will find that the period is given by \dfrac{2\pi}{\sqrt{\omega^2-\frac{1}{4}k^2}}

If you now look at your example of the spring, and use the constants given there, the period is

\displaystyle\frac{2\pi}{\sqrt{(  2k^2)-\frac{1}{4}(2k)^2}}=\frac{2\pi}{  k}
\dfrac{2\pi}{\sqrt{\omega^2-\frac{1}{4}k^2}}

this equation or reasoning wasn't given nor explained in the textbook!

thank you!
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ghostwalker
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(Original post by FurtherMaths2020)
this equation or reasoning wasn't given nor explained in the textbook!
Not surprised; I had to dig out my volume of Intermediate Mechanics by Humphrey and Topping from 1971! ISBN 0582322375, though it's not going to conform to any current A-level syllabus.

thank you!
You're welcome.
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