Solve Torsional Pendulum Homework Statement

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A uniform meter stick oscillates with a period of 5 seconds when hung from a wire. After being shortened to 0.76 meters, the period of oscillation needs to be recalculated. The torsional constant is derived from the initial period and moment of inertia, leading to a final period of approximately 3.8 seconds. An alternative method suggests using the ratio of lengths to find the new period more efficiently. This approach confirms the calculated period is correct and simplifies the process significantly.
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Homework Statement



https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-15-SHM/torsion-pendulum/4.gif


A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation.

With what period does it now oscillate?

L = 1m
L(final) = 0.76m
T = 5s

Homework Equations



Moment of Inertia: I = (1/12)*M*L^2
period: T = 2pi*sqrt(I/K) Where K is the torsional constant

The Attempt at a Solution



I first found the torsional constant:

K = [(1/12)*M*(1^2)]/[(T/(2pi))^2] = 0.13459M

So no I have K = 0.13459M

T(final) = 2pi*[sqrt((1/12)*M*(0.76^2)/0.13159M)] = 3.8s

The answer looks correct but I'm not sure, any ideas? Thanks!
 
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instead of going for so much calculation...
u could have done it like this...

becauz..

T=2*pi*(I/K)^0.5

and u have I=(M*L^2)/12

so jus take the ratio

T1/T2=L1/L2 :smile:

where T2 is the time period when the length is 0.76
directly arrives at your answer ...
 
Oh yeah it does...cool.
 
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