A block weighing 10.0 Nis attached to the lower end of a vertical spring ($\mathbf{k}\mathbf{=}\mathbf{200}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$), the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of 2.00 Jas it passes through the point at which the spring is unstretched. (a) What is the period of the oscillation? (b) Use the law of conservation of energy to determine the maximum distance the block moves both above and below the point at which the spring is unstretched. (These are not necessarily the same.) (c) What is the amplitude of the oscillation? (d) What is the maximum kinetic energy of the block as it oscillates?

• The spring constant is,k=200 N/m.
• Weight of the block is,W=10 N.
• The kinetic energy of the block at equilibrium point is,E=2.0 J.

We can find the period of oscillation of the block using the formula for the period of oscillation for SHM. Then we can find the amplitude of a simple harmonic oscillator from the given total energy of the block using the law of conservation of energy. Next, we can find the maximum distance the block moves both above and below the point at which the spring is unstretched from it. Lastly, we can find the maximum K.E attained by the block from its given total energy.

Formulae:

The elastic P.E energy of the system,$\frac{1}{2}k{x}^2$ (i)

The period of oscillation for SHM, $T=2\mathrm{\pi }\sqrt{\frac{\mathit{m}}{\mathit{k}}}$ (ii)

The law of conservation of energy gives: E= constant(iii)

The mass of block is given by:

$$\begin{gathered} m=\frac{W}{g} \\ =\frac{10\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =1.02\mathrm{kg} \end{gathered}$$

The period of oscillation of block using equation (ii) is given as:

$$\begin{gathered} \mathrm{T}=2(3.142)\sqrt{\frac{1.02\mathrm{kg}}{200\mathrm{N}/\mathrm{m}}} \\ =0.448 \\ \approx 0.45\mathrm{s} \end{gathered}$$

Hence, the period of the oscillations is$0.45\mathrm{s}$.

The total mechanical energy of the block at unstretched position is given by the sum of kinetic and potential energy as:

$$\begin{gathered} E=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2 \\ =\frac{1}{2}(200\mathrm{N}/\mathrm{m}){(0.05\mathrm{m})}^2+2.0\mathrm{J} \\ =2.25\mathrm{J} \end{gathered}$$

The T.E energy at the topmost and bottom-most position of the block is only elastic P.E. Then, according to the conservation of energy, using equation (i) formula

$$\begin{gathered} \mathrm{E}=\mathrm{constant} \\ \mathrm{E}=\frac{1}{2}{{\mathrm{kx}}^2}_{\mathrm{m}} \\ 2.25\mathrm{J}=\frac{1}{2}{{\mathrm{kx}}^2}_{\mathrm{m}} \\ {\mathrm{x}}_{\mathrm{m}}=0.15\mathrm{m} \end{gathered}$$

By Hooke’s law, we get the displacement of the block as:

$$\begin{gathered} \mathrm{kx}=\mathrm{mg} \\ \mathrm{x}=\frac{\mathrm{mg}}{\mathrm{k}} \\ =\frac{10\mathrm{N}}{200\mathrm{N}/\mathrm{m}} \\ =0.05\mathrm{m} \end{gathered}$$

This is the distance between unstretched and equilibrium length of the spring

Hence, the maximum distance the block moves the point at which the spring is upstretched is,

$0.15\mathrm{m}-0.05\mathrm{m}=0.10\mathrm{m}$

And the maximum distance the block moves below the point at which the spring is upstretched is,

$0.15\mathrm{m}+0.05\mathrm{m}=0.20\mathrm{m}$

From part (b) we can write that the amplitude of oscillation of block is 0.15 m .

From part b we can write that the maximum kinetic energy of the block as it oscillates is 2.25 J .