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A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where \(g =\) 3.71 m/s\(^2\)?

Short Answer

Expert verified
The period on Mars is approximately 2.60 seconds.

Step by step solution

01

Understand the formula for the period of a pendulum

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

Express the period in terms of gravity

Since we are comparing periods on Earth and Mars, we can express the periods in terms of their respective gravities. Let \( T_\text{Earth} \) be the period on Earth and \( T_\text{Mars} \) be the period on Mars. Then, from the formula, \( \frac{T_\text{Mars}^2}{T_\text{Earth}^2} = \frac{g_\text{Earth}}{g_\text{Mars}} \).
03

Substitute the known values

We know that \( T_\text{Earth} = 1.60 \) s and \( g_\text{Earth} = 9.81 \) m/s². Substitute these values along with \( g_\text{Mars} = 3.71 \) m/s² into the equation: \[ \frac{T_\text{Mars}^2}{(1.60)^2} = \frac{9.81}{3.71} \].
04

Solve for the period on Mars

First, calculate the right side of the equation: \( \frac{9.81}{3.71} \approx 2.643 \).Now solve for \( T_\text{Mars}^2 \):\( T_\text{Mars}^2 = 1.60^2 \times 2.643 \).Calculate \( 1.60^2 \approx 2.56 \). Thus, \( T_\text{Mars}^2 = 2.56 \times 2.643 \approx 6.76 \).Take the square root to find \( T_\text{Mars} \):\( T_\text{Mars} \approx \sqrt{6.76} \approx 2.60 \) s.

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Key Concepts

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

Period of a Pendulum
The period of a pendulum is the time it takes to complete one full swing back and forth. To understand this better, think of a swing at the playground. The time it takes to go from the starting point, to the highest point on the other side, and back to the start is the period. For a simple pendulum, this period is determined by both the length of the pendulum and the acceleration due to gravity. The formula to calculate the period, expressed as \( T \), is:\[ T = 2\pi \sqrt{\frac{L}{g}} \]where:
  • \( T \) is the period
  • \( L \) is the length of the pendulum
  • \( g \) is the acceleration due to gravity
The longer the pendulum, the longer the period will be. Similarly, stronger gravity results in a shorter period. This formula shows how the physical properties around the pendulum, especially gravity, influence its behavior.
Acceleration Due to Gravity
Gravity is an attractive force pulling objects towards each other. On Earth, gravity gives us a pull of approximately 9.81 m/s². This means, in one second, an object speeds up by 9.81 meters per second when falling freely. Acceleration due to gravity varies depending on the planet or celestial body you are on. For Earth, we have this fixed value, but on Mars, gravity is weaker, with \( g = 3.71 \text{ m/s}^2 \). This means things fall slower on Mars than on Earth, impacting the motion of a pendulum by making its swings slower. Consider:
  • Higher gravity: shorter period (faster swings)
  • Lower gravity: longer period (slower swings)
Understanding gravity’s role is crucial when comparing physical phenomena across different planetary bodies.
Pendulum on Different Planets
A pendulum's behavior isn't constant; it changes if we move from one planet to another because gravity differs. When you take a pendulum from Earth to Mars, its period adjusts due to the differences in gravitational pull.Using the equation for the period, you can compare how a pendulum performs on different planets. On Earth, a pendulum with a period of \( 1.60 \text{ s} \) will swing more slowly as you travel to a planet like Mars, where \( g = 3.71 \text{ m/s}^2 \). To find the new period on Mars \( (T_\text{Mars}) \), you use:\[ \frac{T_\text{Mars}^2}{T_\text{Earth}^2} = \frac{g_\text{Earth}}{g_\text{Mars}} \]Substituting the known values gives us a longer period because the gravitational pull is weaker on Mars. This calculation results in a period of approximately \( 2.60 \text{ s} \) on Mars, showing the direct influence of gravity on pendulum motion. Traveling across planets helps illustrate how fundamental forces, like gravity, change the behavior of objects.

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Most popular questions from this chapter

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.250 m and the period is 3.20 s. What are the speed and acceleration of the block when \(x =\) 0.160 m?

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 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 \((f_S+\,_V)\) to the frequency without the virus \((f_S)\) is given by \(f_S+\,_V/f_S = 1\sqrt{ 1 + (m_V/m_S) }\), where \(m_V\) is the mass of the virus and \(m_S\) is the mass of the silicon sliver. Notice that it 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\(^{-16}\) g and a frequency of 2.00 \(\times\) 10\(^{15}\) Hz without the virus and 2.87 \(\times\) 10\(^{14}\) Hz with the virus. What is the mass of the virus, in grams and in femtograms?

A block with mass \(m =\) 0.300 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The other end of the spring is attached to a wall. When the block is at \(x = +\)0.240 m, its acceleration is \(a_x = -\)12.0 m/s\(^{2}\) and its velocity is \(v_x = +\)4.00 m/s. What are (a) the spring's force constant \(k\); (b) the amplitude of the motion; (c) the maximum speed of the block during its motion; and (d) the maximum magnitude of the block's acceleration during its motion?

You want to find the moment of inertia of a complicated machine part about an axis through its center of mass. You suspend it from a wire along this axis. The wire has a torsion constant of 0.450 N \(\cdot\) m/rad. You twist the part a small amount about this axis and let it go, timing 165 oscillations in 265 s. What is its moment of inertia?

A spring of negligible mass and force constant \(k =\) 400 N/m is hung vertically, and a 0.200-kg pan is suspended from its lower end. A butcher drops a 2.2-kg steak onto the pan from a height of 0.40 m. The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

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