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Chapter 10: Problem 43

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
0.40 seconds before the pendulum reaches its greatest speed.

Step by step solution

01

Understanding the Problem

We have a simple pendulum with a length of 0.65 meters, and we need to find the time it takes for the pendulum to reach its maximum speed after being released from rest.
02

Identify the Greatest Speed Condition

The pendulum reaches its greatest speed at its lowest point, where all the potential energy has been converted to kinetic energy. For a simple pendulum, this occurs at the midpoint of its oscillation.
03

Use the Pendulum Period Formula

The period of a simple pendulum (the time for one complete oscillation back and forth) is given by the formula: \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity, which is approximately 9.81 m/s虏.
04

Calculate the Period

Substitute the given length into the formula: \( T = 2\pi \sqrt{\frac{0.65}{9.81}} \). Calculate this to find the total period of the pendulum's oscillation.
05

Calculate Time to Maximum Speed

Since the maximum speed is at the lowest point, which is one-quarter of the period from the start, the time elapsed to reach this point is \( \frac{T}{4} \). Calculate this value.
06

Final Calculation

Compute \( T = 2\pi \sqrt{\frac{0.65}{9.81}} \), then divide by 4 to find \( \frac{T}{4} \). Perform the arithmetic to find the time elapsed.

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Key Concepts

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

Pendulum Period
The period of a simple pendulum is a crucial concept that tells us how long it takes for the pendulum to swing back and forth once. It is determined by the formula \[T = 2\pi \sqrt{\frac{L}{g}}\]where:
  • \( T \) is the period of oscillation.
  • \( L \) is the length of the pendulum.
  • \( g \) is the acceleration due to gravity, roughly 9.81 m/s虏 on Earth.
The period is directly proportional to the square root of the pendulum's length, meaning longer pendulums take more time to complete one oscillation.
Interestingly, the mass of the pendulum doesn't affect the period, assuming the angle is small.
Potential Energy
Potential energy in a pendulum is highest at the endpoints of its swing. At these points, it is momentarily at rest. This energy is stored due to the pendulum's position above its lowest point. The potential energy \[U = mgh\]where:
  • \( m \) is the mass of the pendulum.
  • \( g \) is the acceleration due to gravity.
  • \( h \) is the height from the lowest point.
As the pendulum swings downward, its potential energy decreases while its kinetic energy increases, showing the conversion of energy.
Kinetic Energy
Kinetic energy is greatest when the pendulum passes through its lowest point. Here, its speed is highest, and all potential energy has been converted into kinetic energy.
The kinetic energy of the pendulum is given by the formula:\[K = \frac{1}{2}mv^2\]where:
  • \( m \) is the mass.
  • \( v \) is the velocity of the pendulum.
The transition of energy in a pendulum between potential and kinetic is a classical example of conservation of energy.
Oscillation
An oscillation in a simple pendulum includes swinging from one side to the other and returning to the starting position. It is marked by periodic motion that repeats at regular intervals.
Each complete movement back and forth is one oscillation. The time taken for the pendulum to return to its initial point is known as the period.
The keys to understanding pendulum oscillations include:
  • The symmetry of motion: each swing has equal time durations.
  • Energy exchanges: potential and kinetic energy transition smoothly.
  • The consistent period: calculated using the pendulum period formula.
Understanding oscillation helps to comprehend how pendulums behave predictably.

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Most popular questions from this chapter

One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned so as to stretch the wire. The piano wire is made from steel \(\left(Y=2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2}\right) .\) It has a radius of \(0.80 \mathrm{mm}\) and an unstrained length of \(0.76 \mathrm{m}\). The radius of the tuning peg is \(1.8 \mathrm{mm} .\) Initially, there is no tension in the wire, but when the tuning peg is turned, tension develops. Find the tension in the wire when the tuning peg is turned through two revolutions. Ignore the radius of the wire compared to the radius of the tuning peg.

A 6.8-kg bowling ball is attached to the end of a nylon cord with a cross- sectional area of \(3.4 \times 10^{-5} \mathrm{m}^{2} .\) The other end of the cord is fixed to the ceiling. When the bowling ball is pulled to one side and released from rest, it swings downward in a circular arc. At the instant it reaches its lowest point, the bowling ball is \(1.4 \mathrm{m}\) lower than the point from which it was released, and the cord is stretched \(2.7 \times 10^{-3} \mathrm{m}\) from its unstrained length. What is the unstrained length of the cord? (Hint: When calculating any quantity other than the strain, ignore the increase in the length of the cord.\()\)

A heavy-duty stapling gun uses a 0.140 -kg metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

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 person who weighs \(670 \mathrm{N}\) steps onto a spring scale in the bathroom, and the spring compresses by \(0.79 \mathrm{cm} .\) (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by \(0.34 \mathrm{cm} ?\)
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