# Damped SHMWatch

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#1
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?
0
3 weeks ago
#2
(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 .

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

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

1
#3
(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 .

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

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

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

thank you!
0
3 weeks ago
#4
(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.
1
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