/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 Concept Simulation 10.2 at allow... [FREE SOLUTION] | 91Ó°ÊÓ

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Concept Simulation 10.2 at allows you to explore the effect of the acceleration due to gravity on pendulum motion, which is the focus of this problem. Astronauts on a distant planet set up a simple pendulum of length \(1.2 \mathrm{~m}\). The pendulum executes simple harmonic motion and makes 100 complete vibrations in \(280 \mathrm{~s}\). What is the acceleration due to gravity?

Short Answer

Expert verified
The acceleration due to gravity is approximately 6.04 m/s².

Step by step solution

01

Understand the Problem

Read the problem carefully to identify what is being asked. We need to find the acceleration due to gravity on a distant planet using a pendulum that makes 100 complete vibrations in 280 seconds, with a length of 1.2 meters.
02

Define Relevant Formula

The formula for the period of a simple pendulum is \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.
03

Calculate the Period

To find the period \( T \), divide the total time by the number of vibrations: \( T = \frac{280}{100} = 2.8 \) seconds per vibration.
04

Rearrange the Formula to Solve for g

Rearrange the formula \( T = 2\pi \sqrt{\frac{L}{g}} \) to solve for \( g \):\[ g = \frac{4\pi^2 L}{T^2} \].
05

Substitute the Known Values into the Formula

Substitute the known values for \( L \) and \( T \) into the formula: \( g = \frac{4\pi^2 \times 1.2}{2.8^2} \).
06

Calculate the Value of g

Calculate the right-hand side of the equation: \( 4\pi^2 = 39.4784 \), then \( 2.8^2 = 7.84 \). So, \[ g = \frac{39.4784 \times 1.2}{7.84} \approx 6.04 \text{ m/s}^2 \].

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Key Concepts

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

Simple Harmonic Motion
Simple harmonic motion is a type of periodic, oscillatory movement. It occurs when an object moves back and forth along the same path, like a pendulum. In this motion, the restoring force that brings the object back to its central position is directly proportional to its displacement from that point. This results in an elegant, wave-like motion that is incredibly predictable.
One common example of simple harmonic motion is a pendulum swinging. The motion of a pendulum can be described using precise mathematical equations, making it a fantastic tool for experiments. When a pendulum swings, it reaches a certain maximum height on either side, and the force due to gravity pulls it back to its lowest point. This creates a continuous back-and-forth motion that is characteristic of simple harmonic motion.
  • Predictable and regular motion.
  • Can be described using sine or cosine functions.
  • Restoring force is proportional to displacement.
Acceleration Due to Gravity
The acceleration due to gravity is an important concept in the physics of motion. It represents the gravitational force that pulls objects towards the center of a celestial body, such as a planet. This force is experienced by objects in free fall and is a crucial variable in pendulum motion.
On Earth, the acceleration due to gravity is approximately 9.81 m/s². However, this value changes on other planets. In the given problem, astronauts aimed to discover the acceleration due to gravity on a distant planet by using a pendulum.
By measuring the pendulum's period on this new planet, they can work backward using a formula that relates the pendulum's period and length to the acceleration due to gravity. Knowing these measurements allows them to solve for the unknown gravitational acceleration, which is different from that on Earth.
  • Force pulling objects toward planet centers.
  • Varies with the planet or celestial body.
  • Affects pendulum motion and simple harmonic motion.
Period of a Pendulum
The period of a pendulum is the time it takes for the pendulum to make one complete swing from one side to the other and back. The period is a fundamental characteristic of the pendulum's motion and is directly related to both the length of the pendulum and the acceleration due to gravity.
Calculating the period involves dividing the total time by the number of complete oscillations. For a pendulum, the formula used is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \] This formula tells us that:
  • The period, \( T \), increases with the square root of the length, \( L \), of the pendulum.
  • It decreases with the square root of the acceleration due to gravity, \( g \).
When solving problems involving pendulum motion, adjusting the known formula to solve for one of the factors (like \( g \) in the case of the distant planet) can help students apply their understanding in a meaningful way. In the exercise, by knowing the period and the length of the pendulum, it's possible to determine the gravitational pull on the pendulum, revealing important properties of the environment.

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Most popular questions from this chapter

An archer, about to shoot an arrow, is applying a force of \(+240 \mathrm{~N}\) to a drawn bowstring. The bow behaves like an ideal spring whose spring constant is \(480 \mathrm{~N} / \mathrm{m}\). What is the displacement of the bowstring?

A rifle fires a \(2.10 \times 10^{-2}\) kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by \(9.10 \times 10^{-2} \mathrm{~m}\) from its unstrained length. The pellet rises to a maximum height of \(6.10 \mathrm{~m}\) above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

An archer pulls the bowstring back for a distance of \(0.470 \mathrm{~m}\) before releasing the arrow. The bow and string act like a spring whose spring constant is \(425 \mathrm{~N} / \mathrm{m}\). (a) What is the elastic potential energy of the drawn bow? (b) The arrow has a mass of \(0.0300 \mathrm{~kg}\). How fast is it traveling when it leaves the bow?

Multiple-Concept Example 6 presents a model for solving this problem. As far as vertical oscillations are concerned, a certain automobile can be considered to be mounted on four identical springs, each having a spring constant of \(1.30 \times 10^{5} \mathrm{~N} / \mathrm{m}\). Four identical passengers sit down inside the car, and it is set into a vertical oscillation that has a period of \(0.370 \mathrm{~s}\). If the mass of the empty car is \(1560 \mathrm{~kg}\), determine the mass of each passenger. Assume that the mass of the car and its passengers is distributed evenly over the springs.

Interactive Solution \(10.9\) at discusses a method used to solve this problem. To measure the static friction coefficient between a \(1.6-\mathrm{kg}\) block and a vertical wall, the setup shown in the drawing is used. A spring (spring constant \(=510 \mathrm{~N} / \mathrm{m}\) ) is attached to the block. Someone pushes on the end of the spring in a direction perpendicular to the wall until the block does not slip downward. If the spring in such a setup is compressed by \(0.039 \mathrm{~m}\), what is the coefficient of static friction?

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