Question

$\mathbf{A}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}$ 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?

Short answer

(a) Frequency of motion is$2.25\mathrm{Hz}$.

(b) Initial potential energy of block spring system is$125\mathrm{J}$.

(c) Initial kinetic energy is$250\mathrm{J}$ .

(d) The amplitude of the motion isrole="math" localid="1655092952658" $0.866\mathrm{m}$.

Step-by-step solution

The given data

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,$v=10.0\mathrm{m}/\mathrm{s}$

Understanding the concept of simple harmonic motion

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.

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

(a) Calculation of frequency of oscillation

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

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

(b) Calculation of initial potential energy

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

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

(c) Calculation of initial kinetic energy

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

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

(d) Calculation of motion’s amplitude

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

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