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spayn [35]
3 years ago
6

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?
Physics
1 answer:
rewona [7]3 years ago
3 0

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

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