Answer
0.002 s
Work Step by Step
For a physical pendulum $T=2\pi \sqrt {\frac{I}{mgL}}$ , where T - oscillation period, I - Rotational inertia of the system through the pivot, L - distance from the pivot to center of mass of the ball, g - gravitational acceleration ($9.8\space m/s^{2}$)
$T=2\pi \sqrt {\frac{I}{mgL}}-(1)$
According to the parallel axis theorem we can write,
$I=I_{cm}+mL^{2}$ (Please see the attached image)
$I=\frac{1}{2}mR^{2}+mL^{2}= m(\frac{2R^{2}}{5}+L^{2})-(2)$
(2)=>(1)
$T=2\pi\sqrt {\frac{m(\frac{1}{2}R^{2}+L^{2})}{mgL}}=2\pi\sqrt {\frac{(\frac{1}{2}R^{2}+L^{2})}{gL}}$
Let's plug known values for the initial condition.
$T_{1}=2\pi \sqrt {\frac{\frac{(3.175\times10^{-2}m)^{2}}{2}+[(30-3.175)\times10^{-2}m]^{2}}{9.8\space m/s^{2}\times(30-3.175)\times10^{-2}m}}$
$T_{1}=1.043\space s$
Let's plug known values for the second condition. (the nut is moved upward 1mm)
$T_{2}=2\pi \sqrt {\frac{\frac{(3.175\times10^{-2}m)^{2}}{2}+[(29.9-3.175)\times10^{-2}m]^{2}}{9.8\space m/s^{2}\times(29.9-3.175)\times10^{-2}m}}$
$T(2)=1.041\space s$
$T_{1}-T_{2}=(1.043-1.041)s=0.002\space s$
Period was changed by 0.002s
