Answer
(a) $ 1.0 \, \mathrm{cm} $
(b) $ 0.86 \, \mathrm{cm} $
Work Step by Step
(a) At the lowest point with positive $y$-direction up, we can use Newton's second law together with the formula for radial acceleration and the relationship between tangential and angular speed to find the tension $T_1$:
$$ \Sigma F_y = T_1 - mg = ma_{\mathrm{rad}} = \frac{mv^2}{l} \Rightarrow T_1 = \frac{mv^2}{l} + mg = m(\omega^2l + g) = $$
$$ (12.0 \, \mathrm{kg}) \Bigg( \bigg( \frac{120 \times 2\pi \, \mathrm{rad}}{60 \, \mathrm{s}} \bigg)^2(0.70 \, \mathrm{m}) + 9.80 \, \mathrm{m/s^2} \Bigg) = 1444.1 \, \mathrm{N} $$
Now we can find the elongation $\Delta l$ using the equation for elastic modulus:
$$ Y_\mathrm{Al} = \frac{T_1l}{A \times \Delta l_1} \Rightarrow \Delta l_1 = \frac{T_1l}{Y_\mathrm{Al}A} = $$
$$ \frac{(1444.1 \, \mathrm{N})(0.70 \, \mathrm{m})}{(0.014 \times 10^{-4}\, \mathrm{m^2})(7.0 \times 10^{10} \, \mathrm{Pa})} = 0.010 \, \mathrm{m} = 1.0 \, \mathrm{cm}$$
(b) To find the elongation at the highest point, we do the same thing as in (a), except with positive $y$-direction down:
$$ \Sigma F_y = T_2 + mg = ma_{\mathrm{rad}} = \frac{mv^2}{l} \Rightarrow T_2 = m(\omega^2l - g) = 1208.9 \, \mathrm{N} $$
$$ \Delta l_2 = 0.0086 \, \mathrm{m} = 0.86 \, \mathrm{cm}$$
