A block is in simple harmonic motion (SHM ) on the end of the spring with position given  by x = 5cos  $(wt+\frac{\pi }{4})$  in cm. If the total mechanical energy used is 100J to achieve maximum  displacement, then the potential energy at time, t = 0 is _____ (J) 

Total mechanical energy for a block in SHM is E  $E=\frac{1}{2}$ mw2 A2 Given E = 100J, 
A = 5cm = 5 × 10 -2 m.
So . $100=\frac{1}{2}m{w}^2\times {(5\times {10}^{-2})}^2$
 $\Rightarrow \frac{1}{2}m{w}^2=\frac{100}{25\times {10}^{-4}}$
  $\Rightarrow \frac{1}{2}m{w}^2=4\times {10}^4$
At t = 0, displacement x is given by x = 5 cos  $wt+\frac{\pi }{4}$
               X (t = o) = 5 cos  $\left(\frac{\pi }{4}\right)$ = $\frac{5}{\sqrt{2}}$ cm
Now, potential energy at  $x=\frac{5}{\sqrt{2}}\times {10}^{-2}m$
Is  $U=\frac{1}{2}m{w}^2{x}^2$
 $$\begin{gathered} U=4\times {10}^4{(\frac{5}{\sqrt{2}}\times {10}^{-2})}^2 \\ U=\frac{4\times 25}{2}=50J \end{gathered}$$