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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 this planet is approximately 5.34 \( \text{m/s}^2 \).

Step by step solution

01

Identify the pendulum formula

The period of a simple pendulum is related to the acceleration due to gravity by the formula \( T = 2\pi\sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.
02

Calculate the period of the pendulum

Since the pendulum makes 100 complete swings in 136 seconds, we first calculate the period of one complete swing: \( T = \frac{136\, \text{s}}{100} = 1.36\, \text{s} \).
03

Rearrange the formula to solve for g

Rearrange the pendulum formula to solve for \( g \): \( g = \frac{4\pi^2L}{T^2} \).
04

Substitute the known values

Substitute \( L = 0.50 \) cm and \( T = 1.36 \) s into the formula: \( g = \frac{4\pi^2 \times 0.50}{(1.36)^2} \).
05

Calculate g

Calculate \( g \) using the substituted values: \[ g \approx \frac{4\pi^2 \times 0.50}{1.8496} \approx \frac{19.7392 \times 0.50}{1.8496} \approx \frac{9.8696}{1.8496} \approx 5.34\, \text{m/s}^2 \].

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Key Concepts

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

Acceleration Due to Gravity
The acceleration due to gravity, symbolized as \(g\), is a crucial concept in physics that describes how fast objects accelerate towards the center of a planet due to gravity. On Earth, \(g\) is approximately \(9.81 \, \text{m/s}^2\). It means that an object will increase its velocity by \(9.81 \, \text{m/s}\) every second, a principle first described by Sir Isaac Newton.

This value of \(g\) can vary depending on where you are in the universe. For instance, on the Moon or Mars, \(g\) is lower because these celestial bodies are smaller and exert a weaker gravitational force. When you measure \(g\) on an unfamiliar planet, you essentially want to understand how strong gravity is on that planet compared to Earth.

In experiments like the one done by our space explorer, \(g\) is calculated by observing a pendulum's movement, which provides a reliable way to gauge the planet's gravitational pull.
Period of a Pendulum
The period of a pendulum, denoted as \(T\), refers to the time it takes for the pendulum to complete one full back-and-forth swing. Understanding the period is essential because it links directly to the pendulum's length and the gravity affecting it, through the pendulum formula. It's a concept that helps us observe how gravity influences time and motion.

To find the period, you calculate the time it takes for multiple swings and then divide this by the number of swings. In our explorer's experiment, the pendulum made 100 complete swings in 136 seconds. Thus, the period \(T\) is \(1.36 \, \text{seconds per swing}\).

A pendulum's period reveals not only its speed of swinging but also the gravitational force at play. This period will be longer if gravity is weaker or the pendulum is longer, and shorter with stronger gravity or a shorter pendulum.
Pendulum Formula
The pendulum formula is a mathematical expression that gives us insight into the relationship between a pendulum's period \(T\), its length \(L\), and gravity \(g\). The formula is expressed as \( T = 2\pi\sqrt{\frac{L}{g}} \).

This equation is derived from Newtonian mechanics and shows that the period of a pendulum is:
  • Directly proportional to the square root of its length \(L\).
  • Inversely proportional to the square root of the acceleration due to gravity \(g\).
The simpler interpretation of this is that a longer pendulum or lower gravity results in a longer period, and vice versa.

In practical applications, if you know the period of a pendulum and its length, you can rearrange this formula to solve for \(g\), thereby finding the gravitational acceleration, just as the explorer did. This is done using the rearranged formula: \(g = \frac{4\pi^2L}{T^2}\). By incorporating known values of \(L\) and \(T\), one can effectively calculate \(g\) and gain insights into the gravitational characteristics of an unfamiliar environment.

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Most popular questions from this chapter

BIO Weighing a Virus. In February \(2004,\) scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 30 \(\mathrm{nm}\) long with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{v}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by the formula \(\frac{f_{\mathrm{S}+\mathrm{V}}}{f_{\mathrm{S}}}=\frac{1}{\sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)}}\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 \(\mathrm{cm}\) and a frequency of 0.850 \(\mathrm{Hz}\) . Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom- pom's coordinate is \(x=+9.0 \mathrm{cm} ;(\mathrm{c})\) the time required to move from the equilibrium position directly to a point 12.0 \(\mathrm{cm}\) away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found using the energy approach used in Section \(14.3,\) and which cannot? Explain.

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