Question

In figure$,k=100N/m,M=1kgandF=10N.$ Write the potential energy of the spring when the block is at the left extreme during simple harmonic motion moving with $2m/s$.

Answer 1

$F=k{x}_0$
$\therefore$ $x_0=\frac{F}{k}=\frac{10}{100}=0.1$ m

$E_P=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}_0^2$
$=\frac{1}{2}\times 1\times {\left(2\right)}^2+\frac{1}{2}\times 100\times {\left(0.1\right)}^2=2.5$ J

From P to Q
Work done by applied force=change in mechanic energy.
$\therefore$ $FA=E_Q-E_P$
$\therefore$ $\left(10\right)A=\frac{1}{2}k{(A+x_0)}^2-2.5$
$=\frac{1}{2}\times 100{(A+0.1)}^2-2.5$
Solving this equation, we get
$A=0.2$ m

$U_Q=\frac{1}{2}k{(A+x_0)}^2$
$=\frac{1}{2}\times 100{(0.2+0.1)}^2$
$=4.5$ J