## College Physics (4th Edition)

The amplitude is $3.0~cm$
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 $3.0~cm$