Springy rod

Moment of inertia of such a rod is equal to $\frac{ML^{2}}{3}$ This means that this situation is equal to a situation in which a point size mass equal to $\frac{M}{3}$ is pivoted at a distance of L from the floor. IF WE IGNORE GRAVITATIONAL PULL, then using $M = \frac{2}{3} kg$ and $K=450N/m$ we can find the value of time period.

Answer will be 241.7 ms

The length of the rod is not needed to solve the problem in the approximation that the amplitude of oscillations is very small (as noted by the words "very slightly"). To begin the problem, note that the moment of inertia of the rod about the hinge is $I=\frac{1}{3}ML^{2}=\frac{2}{3}L^{2}$ . Next, using the approximation that the amplitude of oscillations is small, the extension of the spring is $x=L\theta$ (this approximation assumes the stretching distance is close to straight). Now, we use torque to equate two quantities:

$\tau=I\alpha=\overrightarrow{r} \times \overrightarrow{F}$

We care about the magnitude of the force, so we simplify the cross product into $FLsin\theta=(kL)L\sin \theta=kL^{2}\sin \theta$ (magnitude of $\overrightarrow{r}$ is $L$ ). Since the value of $\theta$ is small, we can make the approximation $\sin \theta \approx \theta$ . Since the force is a restoring force (it opposes the direction of the displacement from the equilibrium position), we must also place a negative sign on the right side on the equation. Finally, replacing $\omega$ with $\frac{d^{2}\theta}{dt^{2}}$ and $I$ with $\frac{2}{3}L^{2}$ :

$\frac{2}{3} L^{2} \frac{d^{2}\theta}{dt^{2}}=-kL^{2}\theta$

$\Rightarrow \frac{d^{2}\theta}{dt^{2}}=-\frac{3}{2}k \theta$

Notice that $L$ cancels out, and that the expression resembles one that follows Hooke's law. Thus, the solution to the differential equation yields the angular frequency to be $\omega=\sqrt{\frac{3k}{2}}$ . Since $\omega=2\pi f=\frac{2\pi}{T}$ , the period is $T=2\pi \sqrt{\frac{2}{3k}}=2\pi \sqrt{\frac{2}{3(450)}} \approx 0.242 s = \boxed{242 ms}$ .

Very slightly doesent state dt yu dnt have to take gravity into consideration.......its fine if yu say sinx=x wen x<<1degree......yu have to find out de length of de rod.....fr dt u need to to write both force equation and torque equation ....werever der is x[extn of spring] substitute it as l[theta].....solve both to find length.....