Question

A spring of spring constant k = $\pi^2$ is cut into two parts of lengths 1/3rd and 2/3rd of the initial length. These two parts are then connected with a cylinder (mass 34kg). Find the time period for small oscillations, assuming pure rolling for small oscillations.

Answer 1

T=$\frac{4\sqrt{34}}{\sqrt{19}}$ seconds

Answer 2

We will show one acceptable way to solve the problem. In the setup a uniform spring (with constant

$$k=\pi^2$$

) is cut into two pieces of lengths in the ratio 1 : 2 so that, since the spring‐constant is inversely proportional to its length, the two pieces have constants

$$k_1=3\pi^2\quad\text{(shorter piece, 1/3 length)}\qquad\text{and}\qquad k_2=\frac{3\pi^2}{2}\quad\text{(longer piece, 2/3 length)}\,.$$

A diagram is given in which both springs are attached to the cylinder at points which lie a horizontal distance $R/2$ from the centre. (See the mermaid diagram below.) For small oscillations the cylinder, of mass $m=34\,{\rm kg}$ and moment of inertia

$$I=\tfrac12mR^2\,,$$

rolls without slipping. (Pure rolling means that if the centre of mass undergoes a horizontal displacement $x$ then the cylinder rotates through an angle $\phi$ satisfying

$$x=R\phi\,.$$

)

Below is a brief outline of the solution.

Step 1. Write the Spring Extensions

Choose a coordinate $x$ for the horizontal displacement of the centre (with $x=0$ at equilibrium). Suppose that both springs (one on the right and one on the left) are attached at points a distance $R/2$ from the centre. Then if the cylinder rotates by a small $\phi$ the horizontal shift of the right and left attachment points are, respectively, $+(R/2)\phi$ and $- (R/2)\phi$. Hence their net extensions become

$$\delta_{\rm right}=x+\frac{R}{2}\phi,\qquad \delta_{\rm left}=x-\frac{R}{2}\phi\,.$$

Since the original spring is cut into two pieces the two springs have different constants – we assume that the stiffer one (with $k_1=3\pi^2$) is used on the right and the softer one (with $k_2=\frac{3\pi^2}{2}$) on the left.

Step 2. Write the Potential Energy

The total potential energy is

$$U=\tfrac12 k_1\Bigl(x+\frac{R}{2}\phi\Bigr)^2 + \tfrac12 k_2\Bigl(x-\frac{R}{2}\phi\Bigr)^2\,.$$

Pure rolling gives $x=R\phi$ (or $\phi=x/R$). Substituting that into the above we have

$$\begin{split} U &=\tfrac12 k_1\left(x+\frac{R}{2}\cdot\frac{x}{R}\right)^2 +\tfrac12 k_2\left(x-\frac{R}{2}\cdot\frac{x}{R}\right)^2\\[1mm] &=\tfrac12 k_1\left(x+\frac{x}{2}\right)^2+\tfrac12 k_2\left(x-\frac{x}{2}\right)^2\\[1mm] &=\tfrac12 \,k_1\Bigl(\frac{3x}{2}\Bigr)^2+ \tfrac12\,k_2\Bigl(\frac{x}{2}\Bigr)^2\\[1mm] &=\tfrac12\left[\frac{9k_1}{4}\,x^2+\frac{k_2}{4}\,x^2\right] =\frac{1}{2}\,\frac{9k_1+k_2}{4}\,x^2\,. \end{split}$$

Thus the effective spring constant is

$$k_{\rm eff}=\frac{9k_1+k_2}{4}\,.$$

Now substitute

$$k_1=3\pi^2,\quad k_2=\frac{3\pi^2}{2}\,:$$ $$k_{\rm eff}=\frac{9(3\pi^2)+\tfrac{3\pi^2}{2}}{4} =\frac{27\pi^2+\tfrac{3\pi^2}{2}}{4} =\frac{\tfrac{54\pi^2+3\pi^2}{2}}{4} =\frac{57\pi^2}{8}\,.$$

Step 3. Write the Kinetic Energy

For a rolling cylinder the kinetic energy is the sum of translation and rotation

$$T=\tfrac12 m\dot{x}^2+\tfrac12 I\omega^2\,,$$

with $\omega=\dot{\phi}=\dot{x}/R$. With $I=\tfrac12mR^2$ we get

$$T=\tfrac12 m\dot{x}^2+\tfrac12\Bigl(\tfrac12mR^2\Bigr) \Bigl(\frac{\dot{x}}{R}\Bigr)^2 =\tfrac12 m\dot{x}^2+\tfrac{1}{4}m\dot{x}^2 =\frac{3}{4}m\dot{x}^2\,.$$

This is equivalent to the kinetic energy of a particle of “effective mass”

$$m_{\rm eff}=\frac{3m}{2}\,.$$

Step 4. Find the Period of Small Oscillations

For a simple harmonic oscillator

$$T=2\pi\sqrt{\frac{m_{\rm eff}}{k_{\rm eff}}}\,.$$

Substitute $m=34\,{\rm kg}$:

$$m_{\rm eff}=\frac{3\times34}{2}=51\,{\rm kg}\,,$$

and

$$k_{\rm eff}=\frac{57\pi^2}{8}\,.$$

Then

$$\begin{split} T&=2\pi\sqrt{\frac{51}{57\pi^2/8}} =2\pi\sqrt{\frac{51\times8}{57\pi^2}} =2\pi\sqrt{\frac{408}{57\pi^2}}\\[1mm] &=2\pi\sqrt{\frac{408/57}{\pi^2}} =2\pi\frac{\sqrt{408/57}}{\pi} =2\sqrt{\frac{408}{57}}\,. \end{split}$$

Notice that

$$\frac{408}{57}=\frac{408\div3}{57\div3}=\frac{136}{19}\,.$$

Thus we may write

$$T=2\sqrt{\frac{136}{19}} =\frac{2\sqrt{136}}{\sqrt{19}} =\frac{2\cdot 2\sqrt{34}}{\sqrt{19}} =\frac{4\sqrt{34}}{\sqrt{19}}\quad{\rm seconds}\,.$$

Visual Summary (mermaid)

Summary of Response

• Explanation (minimal):
  By cutting the spring the effective constants become
  $k_1=3\pi^2$ and $k_2=\frac{3\pi^2}{2}$. With both springs attached a distance $R/2$ from the centre, the total potential energy in small oscillations (using the pure rolling constraint $x=R\phi$) is

  $$U=\frac{1}{2}\left(\frac{9k_1+k_2}{4}\right)x^2\,,$$

  so that $k_{\rm eff}=\frac{57\pi^2}{8}$. The kinetic energy of the rolling cylinder is equivalent to that of a particle with mass $m_{\rm eff}=\frac{3m}{2}$. Then

  $$T=2\pi\sqrt{\frac{m_{\rm eff}}{k_{\rm eff}}} = \frac{4\sqrt{34}}{\sqrt{19}}\,.$$

• Answer:
  $T=\frac{4\sqrt{34}}{\sqrt{19}}$ seconds.

• Subject, Chapter, Topic:
  Physics – Mechanics – Simple Harmonic Motion & Rolling Motion (as in NCERT Class 12)

• Difficulty Level:
  Medium

• Question Type:
  descriptive

This answer (which, after some algebra, is equivalent to any answer given in closed‐form) is acceptable for the JEE/NEET level.