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Chapter 14: Problem 47

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length \(50.0 \mathrm{~cm} .\) She finds that the pendulum makes 100 complete swings in 136 s. What is the value of \(g\) on this planet?

Short Answer

Expert verified
The value of \(g\) on the planet is approximately \(6.75 m/s^2\).

Step by step solution

01

Understand the Problem and Gather Information

From the problem, length of the pendulum \(L = 50.0 cm = 0.5 m\). The pendulum makes 100 complete swings in 136 s, the time for 1 swing, which is the period \(T\), can be calculated by \(T = 136 s / 100 swings = 1.36 s\). We are asked to find the value of gravitational acceleration \(g\) on this planet.
02

Apply the formula for the period of a simple pendulum

The period of a simple pendulum is given by \(T = 2\pi\sqrt{L / g}\) where \(T\) is the period, \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity. Here we know \(T\) and \(L\), we need to rearrange the formula to solve for \(g\).
03

Rearrange the formula to solve for \(g\)

The formula for \(g\) from the given equation becomes \(g = 4\pi^2L / T^2\). We substitute the given values into the formula.
04

Substitute the values and solve the problem

Now, put \(L = 0.5 m\) and \(T = 1.36 s\) into the equation. The value of \(g\) becomes \(g = 4\pi^2 * 0.5 /1.36^2 = 6.75 m/s^2\)

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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 consists of a weight suspended from a pivot point so that it can swing freely back and forth.
It is often used in physics to study motion and gravity because its simple design provides a clear example of harmonic oscillation.
In a simple pendulum, the only forces acting on the pendulum bob are the tension in the string and the gravitational force. This simplicity makes it ideal for understanding basic principles.
The main components of a simple pendulum include:
  • a mass, often called the bob
  • a string or rod, which is inextensible and has negligible mass
  • a fixed pivot point that allows the bob to swing
When the bob is displaced sideways and then released, it will swing back due to gravity, creating periodic motion. This simple construct can become the basis for exploring more complex physical concepts such as damping and resonance.
Pendulum Period
The period of a pendulum is the time it takes for the pendulum to complete one full swing back and forth.
In a simple pendulum, this period depends on the length of the pendulum and the local gravitational field, but not on the mass of the pendulum bob.
The mathematical formula used to calculate the period of a simple pendulum is: \[ T = 2\pi\sqrt{\frac{L}{g}} \] Where:
  • \(T\) is the period,
  • \(L\) is the length,
  • \(g\) is the acceleration due to gravity.
This formula tells us that the period is proportional to the square root of the length, meaning that a longer pendulum will swing more slowly.
The period is inversely proportional to the square root of the gravitational acceleration, meaning stronger gravity makes pendulums swing faster.
Understanding this relationship helps us determine the gravitational force in different environments, such as other planets.
Planetary Gravity
Gravity is the force that attracts bodies toward the center of any celestial object.
On Earth, the average gravitational acceleration is about 9.81 m/s虏, but this varies slightly depending on location.
When exploring other planets, understanding gravitational acceleration is crucial as it affects how objects fall and how pendulums behave.
In the exercise, the space explorer uses a pendulum to measure gravity on an unknown planet. By observing the pendulum's period and knowing its length, she calculates the gravitational acceleration using the formula:
  • \( g = \frac{4\pi^2L}{T^2} \)
This formula arises from rearranging the pendulum period equation.
Substituting the known values (length and period) allows her to find the gravity of the planet.
Such experiments illustrate how simple tools can provide profound insights about planetary characteristics without complex technology.

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Most popular questions from this chapter

A rifle bullet with mass \(8.00 \mathrm{~g}\) and initial horizontal velocity \(280 \mathrm{~m} / \mathrm{s}\) strikes and embeds itself in a block with mass \(0.992 \mathrm{~kg}\) that rests on a friction less surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of \(15.0 \mathrm{~cm} .\) After the impact, the block moves in SHM. Calculate the period of this motion.

\(\mathrm{A}\) mass is oscillating with amplitude \(A\) at the end of a spring. How far (in terms of \(A\) ) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

The Silently Ringing Bell. A large, \(34.0 \mathrm{~kg}\) bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of mass is \(0.60 \mathrm{~m}\) below the pivot. The bell's moment of inertia about an axis at the pivot is \(18.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The clapper is a small, \(1.8 \mathrm{~kg}\) mass attached to one end of a slender rod of length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently - that is, for the period of oscillation for the bell to equal that of the clapper?

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\theta / 2\) from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

A \(1.50 \mathrm{~kg}\) mass on a spring has displacement as a function of time given by $$ x(t)=(7.40 \mathrm{~cm}) \cos [(4.16 \mathrm{rad} / \mathrm{s}) t-2.42] $$ Find (a) the time for one complete vibration; (b) the force constant of the spring; (c) the maximum speed of the mass; (d) the maximum force on the mass; (e) the position, speed, and acceleration of the mass at \(t=1.00 \mathrm{~s} ;\) (f) the force on the mass at that time.
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