Question

A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation. With what period does it now oscillate?

Answer 1

Answer:

The new time period is  $T_2 = 3.8 \ s$

Explanation:

From the question we are told that

  The period of oscillation is  $T = 5 \ s$

   The  new  length is  $l_2 = 0.76 \ m$

Let assume the original length was $l_1 = 1 m$

Generally the time period is mathematically represented as

         $T = 2 \pi \sqrt{ \frac{ I }{ mgh } }$

Now  I is the moment of inertia of the stick which is mathematically represented as

           $I = \frac{m * l^2 }{12 }$

So

        $T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }$

Looking at the above equation we see that

        $T \ \ \ \alpha \ \ \ l$

=>    $\frac{ T_2 }{T_1} = \frac{l_2}{l_1}$

=>    $\frac{ T_2}{5} = \frac{0.76}{1}$

=>     $T_2 = 3.8 \ s$