Physics - Calculus Question with Solution | TestHub

The original period is $T=2\pi \sqrt{\frac{L}{g}}$. The new length is $L^{\prime }=L(1+\alpha )$. The new period $T^{\prime }=2\pi \sqrt{\frac{L(1+\alpha )}{g}}=T\sqrt{1+\alpha }$. Using the binomial approximation $(1+x{)}^n\approx 1+nx$ for small $x$, we have $\sqrt{1+\alpha }=(1+\alpha {)}^{1/2}\approx 1+\frac{1}{2}\alpha$. So, $T^{\prime }\approx T(1+\frac{\alpha }{2})$. The fractional change in period is $\frac{\Delta T}{T}=\frac{T^{\prime }-T}{T}=\frac{T(1+\frac{\alpha }{2})-T}{T}=\frac{\alpha }{2}$.