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Physics Question on Oscillations

The system's center of mass is suspended by a steel wire of torsional spring
Two small particles of mass 20 gm and 30 gm are connected with a rigid massless rod of length of 10 cm. The system's center of mass is suspended by a steel wire of torsional spring constant c=1.2×10−8c=1.2\times 10^{-8} N.m/rad. If the system is slightly rotated in its plane and the angular frequency of its oscillations in rad/sec is n×10−3n\times 10^{-3}, then write the value of n.

Answer

m1m2m1+m2=(20)(30)20+30=12gm=12×10−3kg\frac{m_1m_2}{m_1+m_2}=\frac{(20)(30)}{20+30}=12 gm =12\times10^{-3}kg
Icm=meqr2=(12×10−3)(0.1)2=12×10−5 kg.m2I_{cm}=m_{eq}r^2=(12\times10^{-3})(0.1)^2=12\times10^{-5}\,kg.m^2
T = 2πIcmC=ωn=CIcm=1.2×10−812×10−52\pi\sqrt{\frac{I_{cm}}{C}}=\omega_n=\sqrt{\frac{C}{I_{cm}}}=\sqrt{\frac{1.2\times 10^{-8}}{12\times 10^{-5}}}
ωn=10×10−3  rad/sec\omega_n=10\times10^{-3}\,\,rad/sec
Given that, the angular frequency of its oscillations in rad/sec is n×10−3rad/sec.n\times 10^{-3}rad/sec.
\Rightarrow\,\,$$n=10

So, the answer is 1010.