A$\mathbf{10}\mathbf{}\mathbf{g}$particle undergoes SHM with amplitude of $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mm}$, a maximum acceleration of magnitude$\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$, and an unknown phase constant$\varphi$.

(a) What is the period of the motion?

(b) What is the maximum speed of the particle?

(d) What is the total mechanical energy of the oscillator?

What is the magnitude of the force on the particle when the particle is at

(d) its maximum displacement and

(e) Half its maximum displacement?

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,$\mathrm{a}=8\times {10}^3\mathrm{m}/{\mathrm{s}}^2$

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

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

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

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

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:

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

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:

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