$\mathbf{A}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}$ object on a horizontal frictionless surface is attached to a spring with$\mathit{k}\mathit{=}\mathit{1000}\mathit{}\mathbf{N}\mathbf{/}\mathbf{m}$. The object is displaced from equilibrium$\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$horizontally and given an initial velocity of$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$back toward the equilibrium position.

(a) What is the motion’s frequency?

(b) What is the initial potential energy of the block–spring system?

(c) What is the initial kinetic energy of the block–spring system?

(d) What is the motion’s amplitude?

1. Mass of spring,$m=5.00\mathrm{kg}$
2. Spring constant,$k=1000\mathrm{N}/\mathrm{m}$
3. Amplitude of the motion,$x_{\mathit{0}}=0.50\mathrm{m}$
4. Initial velocity of the motion,$v=10.0\mathrm{m}/\mathrm{s}$

Using the given information and the standard formula for frequency, potential energy, and kinetic energy of SHM, we can find the required quantities. Using the law of conservation of mechanical energy, we can find the amplitude.

$$\begin{gathered} \mathrm{Formulae}: \\ \mathrm{The}\mathrm{frequency}\mathrm{of}\mathrm{the}\mathrm{body}\mathrm{in}\mathrm{oscillation},f=\frac{1}{2\mathrm{\pi }}\frac{\sqrt{\mathrm{k}}}{\mathrm{m}}........\left(\mathrm{i}\right) \\ \mathrm{The}\mathrm{potential}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{body},\mathrm{U}=\frac{1}{2}{\mathrm{kx}}^2.........\left(\mathrm{ii}\right) \\ \mathrm{The}\mathrm{kinetic}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{body},K\mathit{=}\frac{1}{2}m{v}^{\mathit{2}}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{\left(}\mathrm{iii}\right) \\ \mathrm{The}\mathrm{total}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{system}\mathit{,}E\mathit{=}\mathrm{KE}+\mathrm{PE}......\left(\mathrm{iv}\right) \end{gathered}$$

Using equation (i) and the given values, the frequency of motion is given by:

$$\begin{gathered} f=\frac{1}{2(3.14)}\sqrt{\frac{1000}{5.0}} \\ =2.25\mathrm{Hz} \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{motion}’\mathrm{s}\mathrm{frequency}\mathrm{is}2.25\mathrm{Hz}. \end{gathered}$$

Using equation (ii) and the given values, we get the potential energy as:

$$\begin{gathered} U=\frac{1}{2}\times 1000\times {\left(0.50\right)}^2 \\ =125\mathrm{J} \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{initial}\mathrm{potential}\mathrm{energy}\mathrm{is}125\mathrm{J}. \end{gathered}$$

Using equation (iii) and the given values, we get the kinetic energy as:

$$\begin{gathered} \mathrm{K}=\frac{1}{2}\times 5.0\times {\left(10\right)}^2=250\mathrm{J} \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{initial}\mathrm{kinetic}\mathrm{energy}\mathrm{is}250\mathrm{J}. \end{gathered}$$

Using equation (iv) and the derived values, we get the total energy as:

$$\begin{gathered} E=125+250 \\ =375\mathrm{J} \\ \mathrm{Again},\mathrm{using}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{and}\mathrm{the}\mathrm{given}\mathrm{values},\mathrm{we}\mathrm{can}\mathrm{get}\mathrm{the}\mathrm{maximum}\mathrm{amplitude}\mathrm{of} \\ \mathrm{the}\mathrm{system}\mathrm{as}\mathrm{follows}: \\ 375=\frac{1}{2}\times 1000\times x_{\mathrm{m}}^2 \\ {x}_m^{\mathit{2}}=0.75 \\ {\mathrm{x}}_{\mathrm{m}}=0.866\mathrm{m} \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{amplitude}\mathrm{of}\mathrm{the}\mathrm{motion}\mathrm{is}0.866\mathrm{m}. \end{gathered}$$