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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

DATA You hang various masses \(m\) from the end of a vertical, \(0.250 \mathrm{~kg}\) spring that obeys Hooke's law and is tapered, which means the diameter changes along the length of the spring. since the mass of the spring is not negligible, you must replace \(m\) in the equation \(T=2 \pi \sqrt{m / k}\) with \(m+m_{\text {eff }},\) where \(m_{\text {eff }}\) is the effective mass of the oscillating spring. (See Challenge Problem 14.93.) You vary the mass \(m\) and measure the time for 10 complete oscillations, obtaining these data: $$ \begin{array}{l|lcccc} \boldsymbol{m}(\mathbf{k g}) & 0.100 & 0.200 & 0.300 & 0.400 & 0.500 \\ \hline \text { Time (s) } & 8.7 & 10.5 & 12.2 & 13.9 & 15.1 \end{array} $$ (a) Graph the square of the period \(T\) versus the mass suspended from the spring, and find the straight line of best fit. (b) From the slope of that line, determine the force constant of the spring. (c) From the vertical intercept of the line, determine the spring's effective mass. (d) What fraction is \(m_{\text {eff }}\) of the spring's mass? (e) If a \(0.450 \mathrm{~kg}\) mass oscillates on the end of the spring, find its period, frequency, and angular frequency.

SHM of a Floating Object. An object with height \(h\)mass \(M\), and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\). (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M,\) and the cross-sectional area \(A\) of the object. You can ignore the damping due to fluid friction (see Section 14.7).

You are watching an object that is moving in SHM. When the object is displaced \(0.600 \mathrm{~m}\) to the right of its equilibrium position, it has a velocity of \(2.20 \mathrm{~m} / \mathrm{s}\) to the right and an acceleration of \(8.40 \mathrm{~m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

BIO Weighing Astronauts. This procedure has been used to "weigh" astronauts in space: A \(42.5 \mathrm{~kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes \(1.30 \mathrm{~s}\) to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes \(2.54 \mathrm{~s}\) for one cycle. What is the mass of the astronaut?

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is \(0.090 \mathrm{~m},\) it takes the block \(2.70 \mathrm{~s}\) to travel from \(x=0.090 \mathrm{~m}\) to \(x=-0.090 \mathrm{~m} .\) If the amplitude is doubled, to \(0.180 \mathrm{~m},\) how long does it take the block to travel (a) from \(x=0.180 \mathrm{~m}\) to \(x=-0.180 \mathrm{~m}\) and (b) from \(x=0.090 \mathrm{~m}\) to \(x=-0.090 \mathrm{~m} ?\)
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