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(II) A meter stick is hung at its center from a thin wire (Fig. 35 a). It is twisted and oscillates with a period of 5.0 $\mathrm{s}$ . The meter stick is sawed off to a length of 70.0 $\mathrm{cm} .$ This piece is again balanced at its center and set in oscillation (Fig, 35b). With what period does it oscillate?

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$$3.0 \mathrm{s}$$

03:51

Shital Rijal

Physics 101 Mechanics

Chapter 14

Oscillators

Motion Along a Straight Line

Motion in 2d or 3d

Periodic Motion

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

02:18

In physics, an oscillation is the repetitive variation, typically in time, of some measure about a central value or between two or more different states. The oscillation may be periodic or aperiodic.

03:52

A very light rigid rod wit…

04:52

03:10

A uniform thin rod of leng…

02:32

A meter stick is suspended…

01:34

05:03

A physical pendulum consis…

07:10

A very light rigid rod of …

02:00

A thin, straight, uniform …

03:23

05:39

for this problem. On the topic of oscillators, we have a meter stick which is hung at its center from it than wire. The media stick is twisted and begins to oscillate with a period of five seconds. It is then sort of to a length of 70 cm and the new piece is balanced at its center once again and set into oscillation. Again. You want to know the new period with which it oscillates. Now this is called a torsion pendulum and the angular frequency for a torsion pendulum, omega is equal to the square root of K. Over I where K. Is the torch in constant, which is constant in this problem. And I is the moment of inertia. Now the moment of financial of a rod about center I is 1/12 times the mass of the rod, M times the length of the road squared. And so the angular frequency of oscillations, omega is equal to the square root of K. Over I we know make us equal to two pi over the period of oscillations. T. And so from here we can see the period of oscillations. T. is two pi times the square root I. Of K. So if we take the ratio of periods a new period to the original period T. Over T. Note, This is equal to two pi times the square root of I. Over K. Over the original period two pi times the square root of the original moment of inertia are not over K, which is constant. So this simply becomes the square root. Oh I a new moment of inertia over the original moment of inertia high note. And so we can right this in terms of the rotational moment of inertia for a rod about its center is one of 12 um L squared Divided by the original amount of Inertia 1/12 am not are not squared. And so if we put our values into this equation has become the square root of zero point seven AM, not Times 0.7. Well, not since we saw the meter rule to 70 cm and that's squid all divided by I am not no squared. So this ratio of periods is zero point 58 566 And so the new period of oscillations is 0.58566 times the original period, not Which is 0.58566 Times the original period of five seconds. This gives us the new period to be too 0.9 seconds.

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