Question

In the overhead view of Figure, a long uniform rod of mass 0.600 kgis free to rotate in a horizontal plane about a vertical axis through its center. A spring with force constant k=1850 N/mis connected horizontally between one end of the rod and a fixed wall. When the rod is in equilibrium, it is parallel to the wall. What is the period of the small oscillations that result when the rod is rotated slightly and released?

Short answer

The period of the small oscillations of the rod is 0.0653 s.

Step-by-step solution

The given data:

• The spring constant of the spring,$\mathrm{k}=1850\mathrm{N}/\mathrm{m}\underset{\mathrm{x}\to \infty }{\lim }$
• The mass of the rod,$\mathrm{m}=0.600\mathrm{kg}$

Understanding the concept of small oscillations:

When the rod attached to the spring is freely allowed to rotate about its rotational axis without any application of the force, and then the rod undergoes small oscillations with the effect of the torque being applied on its one end to balance the rod along the spring.

This torque creates a force that results in the motion of the body resulting in repeated oscillations of the body. Thus, using the equation of the period of the torsion pendulum and the torque, you can calculate the period of small oscillation set by rotating the rod.

Formulae:

The period of oscillations of a body in rotational motion,

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\mathrm{\kappa }}}$ ….. (i)

Where, I is the moment of inertia of the body, $\mathrm{\kappa }$is the torsion constant of the spring.

The moment of inertia of a rod about its central axis,

$\mathrm{I}=\frac{1}{12}{\mathrm{mL}}^2$ $\mathrm{I}=\frac{1}{12}{\mathrm{mL}}^2$ ….. (ii)

Where, m is the mass of the rod, L is the length of the rod.

The force applied on the spring system,

$\mathrm{F}=-\mathrm{kx}$ ….. (iii)

Where, k is the spring constant of the spring, x is the displacement of the spring.

The restoring torque acting on the body in spring oscillations,

$\mathrm{\zeta }=-\mathrm{\kappa \theta }$ ..... (iv)

Where, $\mathrm{\kappa }$is the torsion constant of the spring, $\mathrm{\theta }$is the angular displacement of the spring.

Calculate the period of the small oscillations that result when the rod is rotated slightly and released:

Let us consider the given figure.

From this figure, we infer that if the rod is rotated through some angle$\mathrm{\theta }$, the spring will contract with length x.

Using trigonometry, we get

$\mathrm{x}=\frac{\mathrm{L}}{2}\mathrm{sin\theta }$

For the small oscillations, you have

$\sin \theta =\theta$

So, the equation for compression length becomes

$\mathrm{x}=\frac{\mathrm{L}}{2}\mathrm{\theta }$

Now, according to Hook’s law, the force applied on the spring using the above condition in equation (iii) becomes:

$\mathrm{F}=-\mathrm{k}\times \frac{\mathrm{L}}{2}\mathrm{\theta }$

Thus, the torque applied by the rod on the spring is given by

$\mathrm{\zeta }=\mathrm{F}\times \mathrm{d}$

Here, F is the force applied and F is the distance from the spring to the pivot point.

As you consider the pivot point is the center of the rod, we get$\mathrm{d}=\frac{\mathrm{L}}{2}$and you have
$\mathrm{F}=-\mathrm{k}\times \frac{\mathrm{L}}{2}\mathrm{\theta }$

So, the torque acting on the spring is given by using these data as:

$$\begin{gathered} \mathrm{\zeta }=-\mathrm{k}\times \frac{\mathrm{L}}{2}\mathrm{\theta }\times \frac{\mathrm{L}}{2} \\ \mathrm{\zeta }=-\mathrm{k}\frac{{\mathrm{L}}^2}{4}\mathrm{\theta } \end{gathered}$$ ….. (I)

Comparing eq. (I) and (iv), you get the value of torsion constant of the spring as follows:

$\mathrm{\kappa }=\mathrm{k}\frac{{\mathrm{L}}^2}{4}$

Now, the equation for period of torsion pendulum is given using equation (iii) and the above torsion constant in equation (i) as follows:

$\begin{array}{rcl}\mathrm{T}& =& 2\mathrm{\pi }\sqrt{\frac{{\displaystyle \frac{1}{12}}{\mathrm{mL}}^2}{{\displaystyle \frac{{\mathrm{kL}}^2}{4}}}}\\ & =& 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{3\mathrm{k}}}\end{array}$

Using the given values in this equation, we get the value of the period of oscillations as follows:

$\begin{array}{rcl}\mathrm{T}& =& 2\mathrm{\pi }\sqrt{\frac{0.600\mathrm{kg}}{\left(3\times 1850\mathrm{N}/\mathrm{m}\right)}}\\ & =& 0.0653\mathrm{s}\end{array}$

Hence, the period of the small oscillations of the rod is 0.0653 s.