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Angular Frequency Of A Pendulum

Written By Tuesday, November 1, 2022 Add Comment Edit

Learning Objectives

By the end of this section, you will be able to:

  • State the forces that act on a unproblematic pendulum
  • Decide the angular frequency, frequency, and menses of a simple pendulum in terms of the length of the pendulum and the acceleration due to gravity
  • Define the period for a concrete pendulum
  • Define the period for a torsional pendulum

Pendulums are in mutual usage. Grandfather clocks employ a pendulum to keep fourth dimension and a pendulum tin can be used to measure the dispatch due to gravity. For small displacements, a pendulum is a uncomplicated harmonic oscillator.

The Uncomplicated Pendulum

A elementary pendulum is defined to have a betoken mass, as well known equally the pendulum bob , which is suspended from a cord of length L with negligible mass ((Figure)). Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string. The mass of the string is assumed to be negligible as compared to the mass of the bob.

In the figure, a horizontal bar is shown. A string of length L extends from the bar at an angle theta counterclockwise from the vertical. The vertical direction is indicated by a dashed line extending down from where the string is attached to the bar. A circular bob of mass m is attached to the lower end of the string. The arc from the mass to the vertical is indicated by another dashed line and is a length s. A red arrow showing the time T of the oscillation of the mob is shown along the string line toward the bar. A coordinate system is shown near the bob with the positive y direction aligned with the string and pointing toward the pivot point and the positive x direction pointing tangent to the arc and away from the equilibrium position. An blue arrow from the bob toward the pivot, along the string, is labeled F sub T. A red arrow from the bob pointing down is labeled w = m g. A red arrow pointing tangent to the arc and toward equilibrium, in the minus x direction, is labeled minus m g sine theta. A red arrow at an angle theta counterclockwise from w is labeled minus m g cosine theta.

Effigy 15.20 A simple pendulum has a pocket-sized-diameter bob and a string that has a very small mass but is potent enough not to stretch appreciably. The linear displacement from equilibrium is s, the length of the arc. Too shown are the forces on the bob, which result in a net force of [latex] \text{−}mg\text{sin}\,\theta [/latex] toward the equilibrium position—that is, a restoring force.

Consider the torque on the pendulum. The force providing the restoring torque is the component of the weight of the pendulum bob that acts along the arc length. The torque is the length of the string L times the component of the net force that is perpendicular to the radius of the arc. The minus sign indicates the torque acts in the opposite direction of the athwart displacement:

[latex] \begin{array}{ccc}\hfill \tau & =\hfill & \text{−}L(mg\,\text{sin}\,\theta );\hfill \\ \hfill I\alpha & =\hfill & \text{−}Fifty(mg\,\text{sin}\,\theta );\hfill \\ \hfill I\frac{{d}^{2}\theta }{d{t}^{2}}& =\hfill & \text{−}Fifty(mg\,\text{sin}\,\theta );\hfill \\ \hfill yard{L}^{2}\frac{{d}^{2}\theta }{d{t}^{2}}& =\hfill & \text{−}50(mg\,\text{sin}\,\theta );\hfill \\ \hfill \frac{{d}^{2}\theta }{d{t}^{2}}& =\hfill & -\frac{g}{L}\text{sin}\,\theta .\hfill \end{array} [/latex]

The solution to this differential equation involves advanced calculus, and is beyond the scope of this text. But note that for modest angles (less than 15 degrees), [latex] \text{sin}\,\theta [/latex] and [latex] \theta [/latex] differ by less than one%, so we can employ the small bending approximation [latex] \text{sin}\,\theta \approx \theta . [/latex] The angle [latex] \theta [/latex] describes the position of the pendulum. Using the small bending approximation gives an approximate solution for small angles,

[latex] \frac{{d}^{2}\theta }{d{t}^{2}}=-\frac{g}{50}\theta . [/latex]

Because this equation has the same class as the equation for SHM, the solution is easy to find. The angular frequency is

[latex] \omega =\sqrt{\frac{g}{Fifty}} [/latex]

and the period is

[latex] T=2\pi \sqrt{\frac{Fifty}{g}}. [/latex]

The menstruum of a simple pendulum depends on its length and the acceleration due to gravity. The period is completely contained of other factors, such as mass and the maximum deportation. As with simple harmonic oscillators, the period T for a pendulum is nearly independent of amplitude, especially if [latex] \theta [/latex] is less than near [latex] 15\text{°}. [/latex] Even simple pendulum clocks tin be finely adjusted and remain authentic.

Annotation the dependence of T on m. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity, as in the following example.

Example

Measuring Dispatch due to Gravity by the Menstruation of a Pendulum

What is the acceleration due to gravity in a region where a uncomplicated pendulum having a length 75.000 cm has a period of 1.7357 s?

Strategy

Nosotros are asked to find one thousand given the catamenia T and the length 50 of a pendulum. We can solve [latex] T=2\pi \sqrt{\frac{Fifty}{1000}} [/latex] for g, bold simply that the angle of deflection is less than [latex] 15\text{°} [/latex].

Solution

  1. Foursquare [latex] T=2\pi \sqrt{\frac{50}{1000}} [/latex] and solve for 1000:

    [latex] k=iv{\pi }^{2}\frac{L}{{T}^{2}}. [/latex]

  2. Substitute known values into the new equation:

    [latex] g=4{\pi }^{2}\frac{0.75000\,\text{yard}}{{(i.7357\,\text{southward})}^{2}}. [/latex]

  3. Summate to find g:

    [latex] g=ix.8281{\,\text{1000/s}}^{2}. [/latex]

Significance

This method for determining g can be very authentic, which is why length and period are given to five digits in this instance. For the precision of the approximation [latex] \text{sin}\,\theta \approx \theta [/latex] to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about [latex] 0.5\text{°} [/latex].

Cheque Your Understanding

An engineer builds two simple pendulums. Both are suspended from pocket-size wires secured to the ceiling of a room. Each pendulum hovers 2 cm in a higher place the flooring. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendulums will differ if the bobs are both displaced past [latex] 12\text{°} [/latex].

The move of the pendulums volition not differ at all because the mass of the bob has no result on the motion of a simple pendulum. The pendulums are only affected by the period (which is related to the pendulum'due south length) and past the acceleration due to gravity.

Concrete Pendulum

Whatsoever object can oscillate like a pendulum. Consider a coffee mug hanging on a hook in the pantry. If the mug gets knocked, it oscillates back and forth like a pendulum until the oscillations die out. We accept described a simple pendulum as a indicate mass and a cord. A physical pendulum is any object whose oscillations are like to those of the simple pendulum, but cannot be modeled as a point mass on a cord, and the mass distribution must be included into the equation of motion.

Equally for the simple pendulum, the restoring force of the physical pendulum is the strength of gravity. With the simple pendulum, the force of gravity acts on the center of the pendulum bob. In the case of the physical pendulum, the forcefulness of gravity acts on the centre of mass (CM) of an object. The object oscillates near a point O. Consider an object of a generic shape as shown in (Figure).

A drawing of a physical pendulum. In the figure, the pendulum is an irregularly shaped object. The center of mass, C M, is a distance L from the pivot point, O. The center of mass traces a circular arc, centered at O. The line from O to L makes an angle theta counterclockwise from the vertical. Three forces are depicted by red arrows at the center of mass. The force m g points down. Its components are minus m g sine theta which points tangent to the arc traced by the center of mass, and m g cosine theta which points radially outward.

Effigy 15.21 A physical pendulum is whatsoever object that oscillates as a pendulum, but cannot be modeled equally a signal mass on a string. The force of gravity acts on the eye of mass (CM) and provides the restoring force that causes the object to oscillate. The minus sign on the component of the weight that provides the restoring force is present because the force acts in the opposite direction of the increasing angle [latex] \theta [/latex].

When a physical pendulum is hanging from a point but is free to rotate, it rotates because of the torque practical at the CM, produced by the component of the object'southward weight that acts tangent to the movement of the CM. Taking the counterclockwise management to exist positive, the component of the gravitational forcefulness that acts tangent to the motion is [latex] \text{−}mg\,\text{sin}\,\theta [/latex]. The minus sign is the result of the restoring force interim in the opposite management of the increasing angle. Recall that the torque is equal to [latex] \overset{\to }{\tau }=\overset{\to }{r}\,×\,\overset{\to }{F} [/latex]. The magnitude of the torque is equal to the length of the radius arm times the tangential component of the force applied, [latex] |\tau |=rF\text{sin}\,\theta [/latex]. Hither, the length L of the radius arm is the distance between the indicate of rotation and the CM. To clarify the movement, beginning with the net torque. Like the simple pendulum, consider only small angles so that [latex] \text{sin}\,\theta \approx \theta [/latex]. Recollect from Fixed-Axis Rotation on rotation that the net torque is equal to the moment of inertia [latex] I=\int {r}^{2}dm [/latex] times the angular dispatch [latex] \alpha , [/latex] where [latex] \blastoff =\frac{{d}^{2}\theta }{d{t}^{2}} [/latex]:

[latex] I\blastoff ={\tau }_{\text{net}}=Fifty(\text{−}mg)\text{sin}\,\theta . [/latex]

Using the small angle approximation and rearranging:

[latex] \begin{array}{ccc}\hfill I\alpha & =\hfill & \text{−}50(mg)\theta ;\hfill \\ \hfill I\frac{{d}^{2}\theta }{d{t}^{2}}& =\hfill & \text{−}Fifty(mg)\theta ;\hfill \\ \hfill \frac{{d}^{2}\theta }{d{t}^{2}}& =\hfill & \text{−}(\frac{mgL}{I})\theta .\hfill \cease{array} [/latex]

In one case again, the equation says that the second fourth dimension derivative of the position (in this example, the bending) equals minus a constant [latex] (-\frac{mgL}{I}) [/latex] times the position. The solution is

[latex] \theta (t)=\text{Θ}\text{cos}(\omega t+\varphi ), [/latex]

where [latex] \text{Θ} [/latex] is the maximum angular displacement. The angular frequency is

[latex] \omega =\sqrt{\frac{mgL}{I}}. [/latex]

The menses is therefore

[latex] T=two\pi \sqrt{\frac{I}{mgL}}. [/latex]

Note that for a uncomplicated pendulum, the moment of inertia is [latex] I=\int {r}^{ii}dm=g{Fifty}^{2} [/latex] and the period reduces to [latex] T=2\pi \sqrt{\frac{L}{g}} [/latex].

Example

Reducing the Swaying of a Skyscraper

In extreme conditions, skyscrapers can sway up to ii meters with a frequency of upwardly to xx.00 Hz due to high winds or seismic activity. Several companies have developed physical pendulums that are placed on the pinnacle of the skyscrapers. Equally the skyscraper sways to the right, the pendulum swings to the left, reducing the sway. Assuming the oscillations take a frequency of 0.50 Hz, design a pendulum that consists of a long beam, of constant density, with a mass of 100 metric tons and a pivot point at 1 end of the beam. What should be the length of the axle?

The figure depicts a tall building with a column on its roof and a long rod of length L that swings on a pivot point near the top of the column.

Strategy

We are asked to find the length of the physical pendulum with a known mass. We showtime demand to find the moment of inertia of the beam. Nosotros tin can then use the equation for the period of a physical pendulum to detect the length.

Solution

  1. Find the moment of inertia for the CM:
  2. Use the parallel axis theorem to observe the moment of inertia about the point of rotation:

    [latex] I={I}_{\text{CM}}+{\frac{L}{4}}^{two}One thousand=\frac{1}{12}G{L}^{2}+\frac{ane}{4}M{Fifty}^{two}=\frac{1}{3}M{L}^{ii}. [/latex]

  3. The flow of a concrete pendulum has a period of [latex] T=2\pi \sqrt{\frac{I}{mgL}} [/latex]. Utilize the moment of inertia to solve for the length Fifty:

    [latex] \begin{array}{ccc}\hfill T& =\hfill & 2\pi \sqrt{\frac{I}{MgL}}=two\pi \sqrt{\frac{\frac{1}{3}Thou{L}^{two}}{MgL}}=two\pi \sqrt{\frac{L}{3g}};\hfill \\ \hfill L& =\hfill & 3g{(\frac{T}{ii\pi })}^{2}=three(ix.eight\frac{\text{m}}{{\text{south}}^{2}}){(\frac{2\,\text{s}}{2\pi })}^{2}=2.98\,\text{chiliad}\text{.}\hfill \end{array} [/latex]

Significance

In that location are many ways to reduce the oscillations, including modifying the shape of the skyscrapers, using multiple physical pendulums, and using tuned-mass dampers.

Torsional Pendulum

A torsional pendulum consists of a rigid torso suspended by a light wire or bound ((Figure)). When the body is twisted some small maximum bending [latex] (\text{Θ}) [/latex] and released from residual, the body oscillates between [latex] (\theta =+\text{Θ}) [/latex] and [latex] (\theta =\text{−}\,\text{Θ}) [/latex]. The restoring torque is supplied by the shearing of the string or wire.

A torsional pendulum is illustrated in this figure. The pendulum consists of a horizontal disk that hangs by a string from the ceiling. The string attaches to the disk at its center, at point O. The disk and string can oscillate in a horizontal plane between angles plus Theta and minus Theta. The equilibrium position is between these, at theta = 0.

Figure fifteen.22 A torsional pendulum consists of a rigid body suspended by a string or wire. The rigid torso oscillates betwixt [latex] \theta =+\text{Θ} [/latex] and [latex] \theta =\text{−}\text{Θ} [/latex].

The restoring torque can exist modeled as being proportional to the angle:

[latex] \tau =\text{−}\kappa \theta . [/latex]

The variable kappa [latex] (\kappa ) [/latex] is known as the torsion abiding of the wire or string. The minus sign shows that the restoring torque acts in the opposite direction to increasing angular deportation. The cyberspace torque is equal to the moment of inertia times the angular acceleration:

[latex] \begin{assortment}{}\\ I\frac{{d}^{ii}\theta }{d{t}^{ii}}=\text{−}\kappa \theta ;\hfill \\ \frac{{d}^{two}\theta }{d{t}^{2}}=-\frac{\kappa }{I}\theta .\hfill \end{array} [/latex]

This equation says that the second time derivative of the position (in this case, the bending) equals a negative constant times the position. This looks very like to the equation of movement for the SHM [latex] \frac{{d}^{2}ten}{d{t}^{2}}=-\frac{chiliad}{g}x [/latex], where the period was plant to be [latex] T=two\pi \sqrt{\frac{m}{yard}} [/latex]. Therefore, the period of the torsional pendulum tin be found using

[latex] T=2\pi \sqrt{\frac{I}{\kappa }}. [/latex]

The units for the torsion constant are [latex] [\kappa ]=\text{Northward-thou}=(\text{kg}\frac{\text{k}}{{\text{s}}^{2}})\text{m}=\text{kg}\,\frac{{\text{m}}^{2}}{{\text{due south}}^{2}} [/latex] and the units for the moment of inertial are [latex] [I]={\text{kg-m}}^{ii}, [/latex] which show that the unit for the period is the second.

Example

Measuring the Torsion Constant of a Cord

A rod has a length of [latex] 50=0.30\,\text{g} [/latex] and a mass of 4.00 kg. A cord is fastened to the CM of the rod and the organization is hung from the ceiling ((Effigy)). The rod is displaced 10 degrees from the equilibrium position and released from rest. The rod oscillates with a period of 0.5 s. What is the torsion constant [latex] \kappa [/latex]?

Figure a shows a horizontal rod, length 30.0 centimeters and mass 4.00 kilograms, hanging by a string from the ceiling. The string attaches to the middle of the rod. The rod rotates with the string in the horizontal plane. Figure b shows the rod with the details needed for finding its moment of inertia. The rod's length, end to end, is L and its total mass is M. It has linear mass density lambda equals d m d x which also equals M over L. A small segment of the rod that has length d x at a distance x from the center of the rod is highlighted. The string is attached to the rod at the center of the rod.

Figure 15.23 (a) A rod suspended by a cord from the ceiling. (b) Finding the rod'south moment of inertia.

Strategy

We are asked to observe the torsion abiding of the string. We start demand to find the moment of inertia.

Solution

  1. Find the moment of inertia for the CM:

    [latex] {I}_{\text{CM}}=\int {ten}^{2}dm={\int }_{\text{−}L\text{/}ii}^{+50\text{/}2}{x}^{ii}\lambda dx=\lambda {[\frac{{ten}^{3}}{3}]}_{\text{−}L\text{/}ii}^{+Fifty\text{/}2}=\lambda \frac{2{L}^{iii}}{24}=(\frac{M}{L})\frac{2{50}^{3}}{24}=\frac{1}{12}M{L}^{ii}. [/latex]

  2. Calculate the torsion abiding using the equation for the period:

    [latex] \brainstorm{array}{ccc}\hfill T& =\hfill & 2\pi \sqrt{\frac{I}{\kappa }};\hfill \\ \hfill \kappa & =\hfill & I{(\frac{two\pi }{T})}^{2}=(\frac{1}{12}M{L}^{2}){(\frac{2\pi }{T})}^{2};\hfill \\ & =\hfill & (\frac{1}{12}(four.00\,\text{kg}){(0.30\,\text{m})}^{two}){(\frac{2\pi }{0.50\,\text{s}})}^{2}=4.73\,\text{North}·\text{thou}\text{.}\hfill \end{array} [/latex]

Significance

Similar the force constant of the system of a block and a spring, the larger the torsion constant, the shorter the menstruation.

Summary

  • A mass yard suspended past a wire of length L and negligible mass is a simple pendulum and undergoes SHM for amplitudes less than almost [latex] xv\text{°} [/latex]. The menstruum of a unproblematic pendulum is [latex] T=2\pi \sqrt{\frac{L}{g}} [/latex], where L is the length of the string and g is the dispatch due to gravity.
  • The menstruation of a concrete pendulum [latex] T=2\pi \sqrt{\frac{I}{mgL}} [/latex] can exist found if the moment of inertia is known. The length between the point of rotation and the center of mass is Fifty.
  • The catamenia of a torsional pendulum [latex] T=2\pi \sqrt{\frac{I}{\kappa }} [/latex] can be found if the moment of inertia and torsion constant are known.

Conceptual Questions

Pendulum clocks are made to run at the correct charge per unit past adjusting the pendulum's length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will y'all have to lengthen or shorten the pendulum to keep the right fourth dimension, other factors remaining constant? Explain your answer.

A pendulum clock works past measuring the period of a pendulum. In the springtime the clock runs with perfect time, but in the summer and wintertime the length of the pendulum changes. When most materials are heated, they expand. Does the clock run besides fast or too tedious in the summer? What most the winter?

The period of the pendulum is [latex] T=2\pi \sqrt{Fifty\text{/}1000}. [/latex] In summer, the length increases, and the menses increases. If the period should be 1 2d, only menstruation is longer than one second in the summer, it will oscillate fewer than 60 times a minute and clock will run slow. In the winter information technology volition run fast.

With the use of a stage shift, the position of an object may be modeled as a cosine or sine role. If given the choice, which office would you choose? Assuming that the phase shift is nada, what are the initial conditions of function; that is, the initial position, velocity, and acceleration, when using a sine function? How about when a cosine function is used?

Problems

What is the length of a pendulum that has a period of 0.500 south?

Some people think a pendulum with a period of 1.00 s tin be driven with "mental energy" or psycho kinetically, because its menstruation is the same every bit an average heartbeat. True or not, what is the length of such a pendulum?

What is the period of a i.00-m-long pendulum?

How long does it take a child on a swing to consummate i swing if her center of gravity is 4.00 m beneath the pivot?

The pendulum on a cuckoo clock is 5.00-cm long. What is its frequency?

Two parakeets sit on a swing with their combined CMs 10.0 cm beneath the pivot. At what frequency do they swing?

(a) A pendulum that has a period of 3.00000 s and that is located where the dispatch due to gravity is [latex] 9.79\,{\text{m/due south}}^{2} [/latex] is moved to a location where the acceleration due to gravity is [latex] ix.82\,{\text{m/s}}^{2} [/latex]. What is its new menses? (b) Explicate why and then many digits are needed in the value for the catamenia, based on the relation between the period and the acceleration due to gravity.

A pendulum with a period of 2.00000 south in i location ([latex] one thousand=9.80{\text{chiliad/s}}^{two} [/latex]) is moved to a new location where the menses is at present i.99796 south. What is the dispatch due to gravity at its new location?

[latex] 9.82002\,{\text{m/s}}^{2} [/latex]

(a) What is the effect on the menstruation of a pendulum if you double its length? (b) What is the effect on the flow of a pendulum if you decrease its length by v.00%?

Glossary

physical pendulum
any extended object that swings like a pendulum
simple pendulum
point mass, called a pendulum bob, attached to a most massless string
torsional pendulum
any suspended object that oscillates by twisting its intermission

Angular Frequency Of A Pendulum,

Source: https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/15-4-pendulums/

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