91Ó°ÊÓ

We know that,

${\mathit{a}}_{\mathbf{x}}\mathbf{=}{\mathit{Ó\neg }}^{\mathbf{2}}\mathit{x}$ ………………. (1)

Where is angular frequency and x is the displacement.

Angular frequency of the object in SHM is given as,

$\mathit{Ó\neg }\mathbf{=}\sqrt{\frac{\mathbf{K}}{\mathbf{m}}}$ ………………. (2)

From equation (1) and (2)

$a_x=-\frac{K}{m}x$ ………………. (3)

From equation (3)

a)

$$\begin{gathered} k=-\frac{\left(-12m/s^2\right)×0.3Kg}{0.24m} \\ =15N/m \end{gathered}$$

b)

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

${\mathit{V}}_{\mathbf{x}}\mathbf{=}\mathbf{Â\pm }\sqrt{\frac{\mathbf{K}}{\mathbf{m}}}\sqrt{{\mathbf{A}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}$ ………………. (4)

By above equation we get,

$\begin{array}{r}A=\sqrt{\frac{m{v}_x^2}{K}+x^2}\\ =\sqrt{\frac{0.3Kg×(4m/s{)}^2}{15N/m}+(0.24m{)}^2}\\ =0.61\mathrm{m}\end{array}$

c)

Velocity will be maximum when displacement x=0 that is at equilibrium position,

$\begin{array}{r}{\mathrm{V}}_{max}=Â\pm \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}\mathrm{A}\\ =Â\pm \sqrt{\frac{15\mathrm{N}/\mathrm{m}}{0.3\mathrm{Kg}}}×\left(0.61\mathrm{m}\right)\\ =Â\pm 4.3\mathrm{m}/\mathrm{s}\end{array}$

d) The maximum acceleration is given by

role="math" localid="1668141123909" $\begin{array}{r}{\mathit{a}}_{\mathbf{max}}\mathbf{=}{\mathit{Ó\neg }}^{\mathbf{2}}\mathbf{A}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ =\frac{\mathrm{K}}{\mathrm{m}}\mathrm{A}\\ =\frac{15\mathrm{N}/\mathrm{m}}{0.3\mathrm{Kg}}×\left(0.61\mathrm{m}\right)\\ =30.5\mathrm{m}/{\mathrm{s}}^2\end{array}$

Hence, The spring constant K = 15 N/m, the amplitude A=0.61 m, the maximum velocity $v_{max}=$4.3 m/s and maximum acceleration $a_{max}=30.5m/s^2$.