Question

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?

Short answer

1. Period of motion is$3.14\times {10}^{-3}\sec$
2. Maximum speed of particle is $4.0\mathrm{m}/\mathrm{s}$
3. Total mechanical energy of oscillator is 0.080 J
4. Magnitude of force of particle when it is at maximum displacement is $80\mathrm{N}$
5. Magnitude of force of particle when it is at half of maximum displacement is $40\mathrm{N}$

Step-by-step solution

The given data

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

Understanding the concept of Simple harmonic motion

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

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

(a) Calculation of period of motion

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

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

(b) Calculation of maximum speed

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

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

(c) Calculation of total mechanical energy

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

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

(d) Calculation of magnitude of force at maximum displacement

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

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

(e) Calculation of magnitude of force at half it maximum displacement

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:

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