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(1) A pendulum has a period of 1.35 s on Earth. What is its period on Mars, where the acceleration of gravity is about 0.37 that on Earth?

Short Answer

Expert verified
The pendulum's period on Mars is approximately 2.21 seconds.

Step by step solution

01

Understand the Pendulum Period Formula

The period of a pendulum is given 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. On Earth, \( g \approx 9.81 \, \text{m/s}^2 \).
02

Get Period on Earth

The given period of the pendulum on Earth is 1.35 seconds. This is our reference period to find the length of the pendulum and calculate the period on Mars.
03

Write the Period Ratio Equation

Since the length of the pendulum does not change, we can write the equation for the ratio of periods: \( \frac{T_{\text{Mars}}}{T_{\text{Earth}}} = \sqrt{\frac{g_{\text{Earth}}}{g_{\text{Mars}}}} \).
04

Calculate the Gravity on Mars

We know that gravity on Mars is 0.37 times that on Earth. Therefore, \( g_{\text{Mars}} = 0.37 \times 9.81 \, \text{m/s}^2 = 3.63 \, \text{m/s}^2 \).
05

Calculate the Period on Mars

Substitute the known values into the ratio equation: \( \frac{T_{\text{Mars}}}{1.35} = \sqrt{\frac{9.81}{3.63}} \). Solve for \( T_{\text{Mars}} \).
06

Solve for \( T_{\text{Mars}} \)

Calculate \( \sqrt{\frac{9.81}{3.63}} \approx 1.64 \). Therefore, \( T_{\text{Mars}} = 1.35 \times 1.64 \approx 2.21 \) seconds. The period of the pendulum on Mars is approximately 2.21 seconds.

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Key Concepts

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

Acceleration Due to Gravity
Acceleration due to gravity, often denoted as \( g \), refers to the force that pulls objects towards the center of a celestial body, such as Earth or Mars. This acceleration results from gravitational forces and is crucial in determining how fast an object will speed up while falling. On Earth, standard gravity is approximately \( 9.81 \, \text{m/s}^2 \). This value influences a pendulum's motion significantly.
Every planet has its own unique gravity depending on its mass and size. For example, Mars, which is smaller and less massive than Earth, has weaker gravity. Gravity on Mars is about 0.37 times that of Earth's, making the Martian gravity \( 3.63 \, \text{m/s}^2 \).
Understanding gravity is crucial when calculating the pendulum's period as it affects the pendulum's swing time. Less gravity means the pendulum swings more slowly, resulting in a longer period.
Pendulum Length
The length of a pendulum, represented as \( L \), is the distance from the pivot point to the center of mass of the pendulum bob. This length directly influences the pendulum's period – the time it takes to complete one full swing back and forth.
A key formula related to a pendulum's period is \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the pendulum length, and \( g \) is the acceleration due to gravity. This formula shows the relationship between these three core factors.
It's important to note that in exercises where the pendulum moves between different gravitational fields, like Earth to Mars, the length of the pendulum remains constant. This consistency allows the period's change to be calculated based solely on differences in gravity.
Mars Gravity
Mars gravity plays a vital role in understanding how objects behave on Mars compared to Earth. Mars' surface gravity is around \( 3.63 \, \text{m/s}^2 \), which is significantly lower than Earth's. This lower gravity impacts everything from the force you feel when you walk, to how fast objects fall, and even how a pendulum swings.
When considering a pendulum on Mars, the decreased gravitational acceleration leads to a longer period compared to that on Earth. As demonstrated in exercises, calculating the pendulum's period on Mars involves using the ratio of Earth's gravity to Mars' gravity. This is important in scientific studies and engineering when creating projects like Mars rovers or understanding potential human settlement experiences on the Red Planet.
  • Gravity on Mars: \( 3.63 \, \text{m/s}^2 \)
  • Effect on pendulum: Longer swing period
  • Importance: Influences scientific and exploratory calculations
Understanding Mars' gravity is essential for accurate comparisons and calculations related to motion and physics on the Martian surface.

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Most popular questions from this chapter

In some diatomic molecules, the force each atom exerts on the other can be approximated by \(F=-C / r^{2}+D / r^{3}\) where \(r\) is the atomic separation and \(C\) and \(D\) are positive constants. ( \(a\) ) Graph \(F\) vs. \(r\) from \(r=0.8 D / C\) to \(r=4 D / C\). (b) Show that equilibrium occurs at \(r=r_{0}=D / C .\) (c) Let \(\Delta r=r-r_{0}\) be a small displacement from equilibrium, where \(\Delta r \ll r_{0} .\) Show that for such small displacements, the motion is approximately simple harmonic, and \((d)\) determine the force constant. ( \(e\) ) What is the period of such motion? [Hint: Assume one atom is kept at rest.]

(II) If one oscillation has 5.0 times the energy of a second one of equal frequency and mass, what is the ratio of their amplitudes?

The amplitude of a driven harmonic oscillator reaches a value of \(23.7 F_{0} / k\) at a resonant frequency of \(382 \mathrm{~Hz}\). What is the \(Q\) value of this system?

A \(0.35-\mathrm{kg}\) mass at the end of a spring oscillates 2.5 times per second with an amplitude of \(0.15 \mathrm{~m} .\) Determine \((a)\) the velocity when it passes the equilibrium point, \((b)\) the velocity when it is \(0.10 \mathrm{~m}\) from equilibrium, \((c)\) the total energy of the system, and \((d)\) the equation describing the motion of the mass, assuming that at \(t=0, x\) was a maximum.

(II) A mass \(m\) at the end of a spring oscillates with a frequency of 0.83 Hz. When an additional \(680-g\) mass is added to \(m,\) the frequency is 0.60 \(\mathrm{Hz}\) . What is the value of \(m ?\)

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