91Ó°ÊÓ

1. Mass of particle,$\mathrm{M}=10\mathrm{g}\mathrm{or}0.01\mathrm{kg}$
2. Amplitude of particle,${\mathrm{x}}_{\mathrm{maximum}}=2.0\mathrm{mm}\mathrm{or}0.002\mathrm{m}$
3. Maximum acceleration of the particle,$\mathrm{a}=8×{10}^3\mathrm{m}/{\mathrm{s}}^2$

Using the corresponding equations of SHM, we can find the required quantities.

$$\begin{gathered} \mathrm{Formulae}: \\ \mathrm{Acceleration}\mathrm{of}\mathrm{body}\mathrm{in}\mathrm{motion},\mathrm{a}={\mathrm{Ó\neg }}^2{\mathrm{x}}_{\mathrm{maximum}}.....\left(\mathrm{i}\right) \\ \mathrm{Angular}\mathrm{frequency}\mathrm{of}\mathrm{a}\mathrm{body}\mathrm{in}\mathrm{motion},\mathrm{Ó\neg }\mathit{=}\frac{\mathit{2}\mathit{Ï„}\mathit{Ï„}}{\mathit{T}}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(\mathrm{ii}\right) \\ \mathrm{The}\mathrm{velocity}\mathrm{of}\mathrm{a}\mathrm{body}\mathrm{in}\mathrm{motion},\mathrm{V}=\mathrm{aÓ\neg }........\left(\mathrm{iii}\right) \\ \mathrm{Total}\mathrm{mechanical}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{system},\mathrm{KE}=\frac{1}{2}{\mathrm{Mv}}^2.......\left(\mathrm{iv}\right) \\ \mathrm{Force}\mathrm{of}\mathrm{a}\mathrm{body}\mathrm{in}\mathrm{motion},\mathrm{F}=\mathrm{ma}.........\left(\mathrm{v}\right) \end{gathered}$$

Using equation (i), we get the angular frequency as:

$$\begin{gathered} 8×{10}^3={\mathrm{Ó\neg }}^2(0.002) \\ {\mathrm{Ó\neg }}^2=4000×{10}^3 \\ \mathrm{Ó\neg }=2000\mathrm{rad}/\sec \\ \mathrm{Now},\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{time}\mathrm{period}\mathrm{as}: \\ \mathrm{T}=\frac{2\mathrm{Ï„Ï„}}{\mathrm{Ó\neg }} \\ =\frac{2×3.14}{2000} \\ =3.14×{10}^{-3}\sec \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{period}\mathrm{is}3.14×{10}^{-3}\sec \end{gathered}$$

Using equation (iii), we get the velocity as:

$$\begin{gathered} v=2000(0.002) \\ =4.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (iv) and given values, the total kinetic energy of the system is given as:

$$\begin{gathered} KE\mathit{(}or\mathit{}{E}_{\mathit{t}\mathit{o}\mathit{t}\mathit{a}\mathit{l}})=\frac{1}{2}(0.01){\left(4\right)}^2 \\ =0.080\mathrm{J} \\ \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{total}\mathrm{mechanical}\mathrm{energy}\mathrm{is}0.080\mathrm{J}. \end{gathered}$$

Using equation (v) and the given values, we get the magnitude of force at maximum displacement as:

$$\begin{gathered} F=(0.01)(8×{10}^3) \\ =80\mathrm{N} \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{magnitude}\mathrm{of}\mathrm{force}\mathrm{is}80\mathrm{N} \end{gathered}$$

We know that when acceleration is maximum at an extreme point,amplitude is maximum. It is half at half maximum distance. So, using equation (v), we get the magnitude of force as:

$$\begin{gathered} F\mathit{=}(0.01)(4×{10}^3) \\ =40\mathrm{N} \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{force}\mathrm{at}\mathrm{half}\mathrm{it}\mathrm{maximum}\mathrm{displacement}\mathrm{is}40\mathrm{N} \end{gathered}$$