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A simple pendulum is made from a 0.65-m-long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest speed?

Short Answer

Expert verified
The time elapsed is approximately 0.81 seconds.

Step by step solution

01

Understanding the Pendulum Motion

A simple pendulum is a mass (or ball) connected to a string, swinging back and forth. When pulled and released, it swings to the other side until it reaches its highest speed at the lowest point (bottom of the swing).
02

Identifying Half of the Period

The ball attains its greatest speed at the lowest point of its motion, which is halfway through its first complete swing from one side to the other and back again. This time is half the oscillation period.
03

Calculating the Pendulum's Period

The period of a simple pendulum can be found using the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( L = 0.65 \) m is the length of the pendulum, and \( g = 9.81 \) m/s² is the acceleration due to gravity.
04

Solving for the Period

Substitute \( L = 0.65 \) and \( g = 9.81 \) into the formula: \[ T = 2\pi \sqrt{\frac{0.65}{9.81}} \approx 2\pi \sqrt{0.0662} \approx 2\pi \times 0.2574 \approx 1.62 \text{ seconds}. \]
05

Calculating the Time to Greatest Speed

Since the greatest speed is achieved halfway through a period, the time taken is half of the period: \[ \text{Time} = \frac{T}{2} = \frac{1.62}{2} \approx 0.81 \text{ seconds}. \]

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Key Concepts

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

Simple Pendulum
A simple pendulum is a fundamental concept in physics, illustrating motion and forces in a straightforward way. It's composed of two main parts:
The string and the weight (usually referred to as the "bob"). The string is connected to a pivot point, allowing the pendulum to swing freely under the influence of gravity. As the bob moves, it traces an arc, swinging back and forth in a regular, periodic motion.
When you pull the bob to one side and release it, this sets the pendulum into motion. Its path will resemble a curve as it swings due to gravitational forces acting on it.
  • Initially, the bob moves faster as it approaches the lowest point due to gravity.
  • At the lowest point, it reaches its maximum speed.
  • As it ascends, it slows down until it momentarily stops at the peak of its swing.
    From this point, the bob will reverse direction and the process will repeat.
This movement is regular and predictable, making the simple pendulum an excellent model for understanding oscillatory motion.
Oscillation Period
The oscillation period is a crucial aspect of pendulum motion. It's the time it takes for the pendulum to complete one full back-and-forth swing. Understanding this helps to predict how quickly or slowly a pendulum swings. For a simple pendulum,
the period is dependent on two key factors:
- The length of the string (L)
- The acceleration due to gravity (g)
  • The formula used to determine the period ( T ) of a pendulum is: \( T = 2\pi \sqrt{\frac{L}{g}} \).
  • This tells us the period increases as the length of the pendulum string increases.
  • Conversely, a stronger gravitational force (higher g) results in a shorter period.
So, when the pendulum's period is known, you can predict half of this period to find the time taken for the pendulum to reach its greatest speed, which is halfway through its swing.
Acceleration due to Gravity
Acceleration due to gravity, often denoted as g, is a vital parameter when dealing with pendulum motion. On Earth, g is approximately 9.81 m/s², influencing how fast the pendulum swings back and forth. This acceleration pulls the pendulum bob towards the Earth, enabling the pendulum to continue its swing.
  • This force affects the speed at which the pendulum reaches its lowest point.
  • It is also crucial in determining the oscillation period, as seen in the formula \( T = 2\pi \sqrt{\frac{L}{g}} \).
The importance of g is evident when conducting experiments or making predictions involving pendulum motion.
In different environments, such as on other planets where g values differ, the pendulum's behavior will alter due to varying gravitational pulls. This concept demonstrates how changes in gravity can impact motion, which is essential in both educational and real-world applications.

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Most popular questions from this chapter

Ssm A hand exerciser utilizes a coiled spring. A force of \(89.0 \mathrm{N}\) is required to compress the spring by \(0.0191 \mathrm{m} .\) Determine the force needed to compress the spring by \(0.0508 \mathrm{m}\).

\(\mathrm{A}\) square plate is \(1.0 \times 10^{-2} \mathrm{m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{m}\) on a side, and has a mass of \(7.2 \times 10^{-2} \mathrm{kg} .\) The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is 0.91. A force is applied to the top of the plate, as in Figure 10.29 a. Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10.29 b\) ) that can be created by the applied force just before the plate begins to move.

In preparation for shooting a ball in a pinball machine, a spring \((k=675 \mathrm{N} / \mathrm{m})\) is compressed by \(0.0650 \mathrm{m}\) relative to its unstrained length. The ball \((m=0.0585 \mathrm{kg})\) is at rest against the spring at point A. When the spring is released, the ball slides (without rolling). It leaves the spring and arrives at point \(\mathrm{B},\) which is \(0.300 \mathrm{m}\) higher than point A. Ignore friction, and find the ball's speed at point B.

A 1.1-kg object is suspended from a vertical spring whose spring constant is \(120 \mathrm{N} / \mathrm{m}\). (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of \(0.20 \mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.

A \(1.00 \times 10^{-2}\) -kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is \(124 \mathrm{N} / \mathrm{m} .\) The block is shoved parallel to the spring axis and is given an initial speed of \(8.00 \mathrm{m} / \mathrm{s},\) while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

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