91Ó°ÊÓ

a)

We know that,

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{K}{\mathit{x}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{K}{\mathit{A}}^{\mathbf{2}}$ …………… (1)

Here m is the mass of the object,is the velocity, K is the force constant of restoring force and A is the amplitude SHM.

$\begin{array}{r}E=\frac{1}{2}×0.15\mathrm{Kg}×(0.4\mathrm{m}/\mathrm{s}{)}^2+\frac{1}{2}×(300\mathrm{N}/\mathrm{m})×(0.012{)}^2\\ =0.034\mathrm{J}\end{array}$

From equation (1)

$\mathit{A}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{E}}{\mathbf{K}}}$ …………… (2)

role="math" localid="1668097274023" $\begin{array}{l}=\sqrt{\frac{2×0.034J}{300N/m}}\\ =1.5\mathrm{cm}\end{array}$

Maximum velocity is at the equilibrium position, hence total energy is kinetic energy, and potential energy is 0. Hence, at equilibrium position

$E=\frac{1}{2}m{v}_{{}^{max}}^2$ …………… (3)

From equation (3) is,

$\begin{array}{r}{\mathrm{v}}_{max}=\sqrt{\frac{2\mathrm{E}}{\mathrm{m}}}\\ =\sqrt{\frac{2×0.034\mathrm{J}}{0.15\mathrm{Kg}}}\\ =0.67\mathrm{m}/\mathrm{s}\end{array}$

Hence, total energy of SHM is 0.034 J, amplitude of SHM is 1.5 cm and maximum velocity during the motion is 0.67 m/s.