Answer
The amplitude is $3.0~cm$
Work Step by Step
We can find the amplitude:
$E = \frac{1}{2}mv_m^2$
$E = \frac{1}{2}m(A~\omega)^2$
$A^2 = \frac{2E}{m\omega^2}$
$A = \sqrt{\frac{2E}{m~\omega^2}}$
$A = \sqrt{\frac{2E}{m~(\sqrt{\frac{g}{L}})^2}}$
$A = \sqrt{\frac{2EL}{mg}}$
$A = \sqrt{\frac{(2)(0.015~J)(0.75~m)}{(2.5~kg)~(9.80~m/s^2)}}$
$A = 0.030~m$
$A = 3.0~cm$
The amplitude is $3.0~cm$
