Physics - SHM Question with Solution | TestHub

For a uniform rod of length $L$, the center of mass (CM) is at $L/2$ from either end. The pivot $P$ is at a distance of $L/4$ from one end. The distance of the CM from the pivot is $d=L/2-L/4=L/4$. Given $L=1.2$ m, so $d=1.2/4=0.3$ m. The moment of inertia of the rod about its CM is $I_{CM}=\frac{1}{12}M{L}^2$. Using the parallel axis theorem, the moment of inertia about the pivot $P$ is $I_P=I_{CM}+M{d}^2$.$$I_P=\frac{1}{12}M{L}^2+M(L/4{)}^2=M{L}^2\left(\frac{1}{12}+\frac{1}{16}\right)=M{L}^2\left(\frac{4+3}{48}\right)=\frac{7}{48}M{L}^2$$ Substitute $M=2$ kg and $L=1.2$ m:$$I_P=\frac{7}{48}(2)(1.2{)}^2=\frac{7}{24}(1.44)=7\times 0.06=0.42{\text{ kg m}}^2$$ The time period of a compound pendulum is $T=2\pi \sqrt{\frac{I_P}{Mgd}}$.$$T=2\pi \sqrt{\frac{0.42}{(2)(10)(0.3)}}=2\pi \sqrt{\frac{0.42}{6}}=2\pi \sqrt{0.07}$$ We are given $T=2\pi \sqrt{\frac{X}{10g}}=2\pi \sqrt{\frac{X}{10(10)}}=2\pi \sqrt{\frac{X}{100}}$. Comparing the expressions for $T$: $\sqrt{0.07}=\sqrt{\frac{X}{100}}$.$$0.07=\frac{X}{100}\Rightarrow X=0.07\times 100=7$$