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What is the length of a simple pendulum whose full swing from left to right and then back again takes \(3.2 \mathrm{~s}\) ?

Short Answer

Expert verified
The pendulum is approximately 2.54 meters long.

Step by step solution

01

Understand the Pendulum Formula

The period of a simple pendulum (the time for one full set of oscillations, from one extreme to the other and back) is given by the formula: \[ T = 2\pi\sqrt{\frac{L}{g}} \]where \(T\) is the period, \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity (approximately \(9.81\, \mathrm{m/s^2}\) on Earth).
02

Identify Given Values

We are given that the full swing (back and forth) takes \(3.2\) seconds. Therefore, the period \(T = 3.2\, \mathrm{s}\). The acceleration due to gravity \(g\) is \(9.81\, \mathrm{m/s^2}\).
03

Rearrange the Pendulum Formula

Rearrange the formula to solve for the length \(L\):\[ L = \frac{gT^2}{4\pi^2} \]
04

Substitute Values Into the Formula

Substitute \(T = 3.2\, \mathrm{s}\) and \(g = 9.81\, \mathrm{m/s^2}\) into the formula:\[ L = \frac{9.81 \times (3.2)^2}{4\pi^2} \]
05

Calculate the Length

Calculate \((3.2)^2 = 10.24\), then:\[ L = \frac{9.81 \times 10.24}{39.4784} = \frac{100.4544}{39.4784} \approx 2.54 \mathrm{~m} \]
06

Conclusion

Therefore, the length of the pendulum is approximately 2.54 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pendulum Formula
The pendulum formula is a key component in understanding the behavior of simple pendulums. A simple pendulum consists of a weight, or bob, attached to the end of a string or rod of negligible mass that swings back and forth under the influence of gravity. The time it takes for a pendulum to swing from one side to the other and back is called the period, denoted as \(T\). The pendulum formula mathematically describes the relationship between the period, the length of the pendulum \(L\), and the acceleration due to gravity \(g\):
\[ T = 2\pi\sqrt{\frac{L}{g}} \]
This elegant formula reveals that the period \(T\) is proportional to the square root of the pendulum's length \(L\) and inversely proportional to the square root of gravitational acceleration \(g\). It shows that variables such as mass and amplitude (as long as they are small) do not affect the period. This universality makes the pendulum formula especially powerful in various applications, from timekeeping to scientific experiments.
Oscillation Period
The oscillation period \(T\) is an essential concept when studying pendulums. It refers to the time it takes for a pendulum to complete one full cycle of swinging from one side to the other and back. This period is measured in seconds and gives insight into the characteristics of the pendulum's motion.
The formula \( T = 2\pi\sqrt{\frac{L}{g}} \) helps calculate the period. However, if we know the period and need the length, we can rearrange it as \( L = \frac{gT^2}{4\pi^2} \).
  • As the length \(L\) of the pendulum increases, the period \(T\) also increases, meaning it takes longer for a full swing.
  • Conversely, a shorter pendulum has a shorter period.
These relationships provide a way to tailor the properties of a pendulum depending on the desired outcome, such as setting the exact time interval of a clock or understanding the motion of swings in playgrounds.
Gravity Acceleration
Gravity acceleration, often symbolized as \(g\), is a critical factor in determining the behavior of a pendulum. On Earth, gravity pulls objects at a constant acceleration of approximately \(9.81\, \mathrm{m/s^2}\). This force is responsible for returning the pendulum to its equilibrium position, thus enabling the oscillatory motion.
The interaction between gravity and the pendulum's length determines its period through the pendulum formula. In environments with different values of gravitational acceleration, such as the Moon or Mars, the period of a pendulum would notably differ. This indicates that gravity directly influences how a pendulum swings. Knowing \(g\), we see how pendulums provide valuable data not only for engineers and scientists on Earth, but also in theoretical experiments in different gravitational contexts.

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Most popular questions from this chapter

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of \(3.00\) \(\mathrm{H} z\) (a) What is the spring constant of each spring if the mass of the car is \(1450 \mathrm{~kg}\) and the mass is evenly distributed over the springs? (b) What will be the oscillation frequency if five passengers, averaging \(73.0 \mathrm{~kg}\) each, ride in the car with an even distribution of mass?

A pendulum is formed by pivoting a long thin rod about a point on the rod. In a series of experiments, the period is measured as a function of the distance \(x\) between the pivot point and the rod's center. (a) If the rod's length is \(L=2.20 \mathrm{~m}\) and its mass is \(m=22.1 \mathrm{~g}\), what is the minimum period? (b) If \(x\) is cho- sen to minimize the period and then \(L\) is increased, does the period increase, decrease, or remain the same? (c) If, instead, \(m\) is increased without \(L\) increasing, does the period increase, decrease, or remain the same?

A \(4.00 \mathrm{~kg}\) block is suspended from a spring with \(k=500 \mathrm{~N} / \mathrm{m}\). A \(50.0 \mathrm{~g}\) bullet is fired into the block from directly below with a speed of \(150 \mathrm{~m} / \mathrm{s}\) and becomes embedded in the block. (a) Find the amplitude of the resulting SHM. (b) What percentage of the original kinetic energy of the bullet is transferred to mechanical energy of the oscillator?

Although California is known for earthquakes, it has large regions dotted with precariously balanced rocks that would be easily toppled by even a mild earthquake. Apparently no major earthquakes have occurred in those regions. If an earthquake were to put such a rock into sinusoidal oscillation (parallel to the ground) with a frequency of \(2.2 \mathrm{~Hz}\), an oscillation amplitude of \(1.0\) \(\mathrm{cm}\) would cause the rock to topple. What would be the magnitude of the maximum acceleration of the oscillation, in terms of \(g\) ?

A damped harmonic oscillator consists of a block \((m=\) \(2.00 \mathrm{~kg}\) ), a spring \((k=10.0 \mathrm{~N} / \mathrm{m})\), and a damping force \((F=-b v)\). Initially, it oscillates with an amplitude of \(25.0 \mathrm{~cm}\); because of the damping, the amplitude falls to three-fourths of this initial value at the completion of four oscillations. (a) What is the value of \(b ?\) (b) How much energy has been "lost" during these four oscillations?

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