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2 years ago

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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?

1 answer:

rewona [7]2 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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Answer with Explanation:

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1 year ago

A rock is thrown straight up with an initial velocity of 19.6 m/s. What time interval elapses between the rock’s being thrown an

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Answer:

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2 years ago

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