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)

By above equation we get,

$\begin{array}{r}A=\sqrt{\frac{m{v}_x^2}{K}+x^2}\\ =\sqrt{\frac{2\mathrm{Kg}×(4\mathrm{m}/\mathrm{s}{)}^2}{315\mathrm{N}/\mathrm{m}}+(0.2\mathrm{m}{)}^2}\\ =0.38\mathrm{m}\end{array}$

b)

We know that,

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

Where is angular frequency and x is the displacement.

Angular frequency of the object in SHM is given as,

$\mathit{Ó\neg }\mathbf{=}\sqrt{\frac{\mathbf{K}}{\mathbf{m}}}$ ………………. (3)

From equation (1) and (2)

$a_x=-\frac{K}{m}x$ ………………. (4)

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

$a_{max}=\frac{K}{m}A$ ………………. (5)

role="math" localid="1668142489955" $\begin{array}{l}=\frac{315N/m}{2\mathrm{Kg}}×(0.38m)\\ =59.26m/s^2\end{array}$

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

F = -Kx ………………. (6)

Maximum force will be for x=-A,

$\begin{array}{l}{F}_{max}=KA\\ =315N/m×0.38m\\ =119.7N\end{array}$

Hence, the amplitude of the motion is 0.38 m, maximum acceleration is 59.26$m/s^2$ and maximum force during the SHM is 119.7 N