A simple harmonic oscillator consists of a $\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{kg}$block attached to a spring ($\mathbf{k}\mathbf{=}\mathbf{200}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$). The block slides on a horizontal frictionless surface about the equilibrium point$\mathbf{x}\mathbf{=}\mathbf{0}$with a total mechanical energy ofrole="math" localid="1657274001354" $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{J}$. (a) What is the amplitude of the oscillation? (b) How many oscillations does the block complete inrole="math" localid="1657273942909" $\mathbf{10}\mathbf{}\mathbf{s}$? (c) What is the maximum kinetic energy attained by the block? (d) What is the speed of the block at$\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{m}$?

• The spring constant is,$k=200\mathrm{N}/\mathrm{m}$.
• Mass of the block is,$M=0.80\mathrm{kg}$.
• Total mechanical energy of the block at equilibrium point is, $E=40\mathrm{J}$.

A particle in simple harmonic motion has, at any time, the maximum kinetic energy
$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{max}}^{\mathbf{2}}$

A particle in simple harmonic motion has, at any time, potential energy

$\mathit{P}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$

If no friction is present, the mechanical energy$\mathit{E}\mathbf{=}\mathit{K}\mathbf{+}\mathit{U}$ remains constant even though Kand Uchange.

We can find the amplitude of a simple harmonic oscillator from the total energy which is equal to its P.E. Then we can find the number of oscillations completed by the block in 10 s from the period of oscillation. Next, we can find the maximum K.E attained by the block from its T.E. Lastly, we can find the speed of the block using the law of conservation of energy.

Formulae:

Translational kinetic energy of the system, $\frac{1}{2}M{v}^2$ (i)

Mechanical energy (or the potential energy) of the system,

$E=\frac{1}{2}k{x}_m^2$ (ii)

The period of oscillation for SHM,

$T=2\pi +\sqrt{\frac{m}{k}}$ (iii)

The angular speed of the system, $\omega =\frac{v}{R}$ (iv)

Using equation (ii), the amplitude of the block is given as:

$$\begin{gathered} x_m=\sqrt{\frac{2E}{k}} \\ =\sqrt{\frac{2(4\mathrm{J})}{200\mathrm{N}/\mathrm{m}}} \\ =0.20\mathrm{m} \end{gathered}$$

Therefore, the amplitude of a simple harmonic oscillator is $0.20\mathrm{m}$.

Using equation (iii), the period of oscillation is given as:

$$\begin{gathered} T=2(3.142)\sqrt{\frac{0.80\mathrm{kg}}{200\mathrm{N}/\mathrm{m}}} \\ =0.397 \\ \approx 0.4\mathrm{s} \end{gathered}$$

Now, the number of oscillations completed by block in 10 s is,

$$\begin{gathered} \begin{array}{l}N=\frac{Totaltime}{Timeperiod}\end{array} \\ =\frac{10\mathrm{s}}{0.4\mathrm{s}} \\ =25 \end{gathered}$$

Therefore, the number of oscillations completed by block in 10 s is $25$.

According to the law of conservation of energy,

Hence, the value of maximum kinetic energy attained by the block is 4.0 J

$$\begin{gathered} \mathrm{The}\mathrm{maximum}\mathrm{K}.\mathrm{E}\mathrm{attained}\mathrm{by}\mathrm{the}\mathrm{block}=\mathrm{Total}\mathrm{Energy} \\ =4\mathrm{J} \end{gathered}$$

According to the law of conservation of energy, the total energy of the system using equations (i) and (ii) is given as:

$\begin{array}{l}E=K+U\\ =\frac{1}{2}M{v}^2+\frac{1}{2}k{x}^2\end{array}$

Atrole="math" localid="1657273852067" $x=0.15\mathrm{m},$

role="math" localid="1657273743865" $$\begin{gathered} 4.0\mathrm{J}=\frac{1}{2}(0.80\mathrm{m}/\mathrm{s}){\mathrm{v}}^2+\frac{1}{2}(200\mathrm{N}/\mathrm{m}){(0.15\mathrm{m})}^2 \\ {\mathrm{v}}^2=\frac{4.0-2.25}{0.4} \\ \mathrm{v}=2.09\mathrm{m}/\mathrm{s} \\ \approx 2.1\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the block at role="math" localid="1657273794978" $x=0.15\mathrm{m}$is $2.1\mathrm{m}/\mathrm{s}$.