91Ó°ÊÓ

a)

We know that velocity at any point in SHM is given by,

${\mathit{V}}_{\mathbf{x}}\mathbf{=}\mathbf{Â\pm }\sqrt{\frac{\mathbf{K}}{\mathbf{m}}}\sqrt{{\mathbf{A}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}$ ………………. (1)

Since, the blocks maximum speed occurs at x=0.

$V_{max}=\sqrt{\frac{K}{m}}A$ ………………. (2)

The amplitude of motion is,

$\begin{array}{r}A=\sqrt{\frac{m}{\mathrm{K}}}{v}_{max}\\ =\sqrt{\frac{2\mathrm{Kg}}{315\mathrm{N}/\mathrm{m}}}×12\mathrm{m}/\mathrm{s}\\ =0.96\mathrm{m}\end{array}$

b)

We know that,

${\mathit{a}}_{\mathbf{x}}\mathbf{=}\mathbf{-}{\mathit{Ó\neg }}^{\mathbf{2}}\mathit{x}$ ………………. (3)

Where is angular frequency and x is the displacement.

Angular frequency of the object in SHM is given as,

$\mathit{Ó\neg }\mathbf{=}\sqrt{\frac{\mathbf{K}}{\mathbf{m}}}$ ………………. (4)

From equation (1) and (2)

$a_x=-\frac{K}{m}x$ ………………. (5)

Block is moving in negative direction so, the maximum acceleration will be at x=-A.

$a_{max}=\frac{K}{m}A$ ………………. (6)

$\begin{array}{r}=\frac{315\mathrm{N}/\mathrm{m}}{2\mathrm{Kg}}×(0.96\mathrm{m}\\ =151.2\mathrm{m}/{\mathrm{s}}^2\end{array}$

The force on the object exerted by the spring is given by,

F = - Kx ………………. (7)

Maximum force will be for x=-A,

$\begin{array}{l}{F}_{max}=KA\\ =315N/m×0.96m\\ =302.4N\end{array}$

Hence, the amplitude of the motion is 0.96 m, maximum acceleration is 151.2$m/s^2$ and maximum force during the SHM is 302.4 N.