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A pendulum on Mars. A certain simple pendulum has a period on earth of \(1.60 \mathrm{~s}\). What is its period on the surface of Mars, where the acceleration due to gravity is \(3.71 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
The period on Mars is approximately 2.60 seconds.

Step by step solution

01

Understand the relationship of period and gravity

The period of a simple pendulum on Earth is given by the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length, and \( g \) is the acceleration due to gravity. The period on Mars can be found using a similar relationship.
02

Identify given values and equations

On Earth, the period \( T_{E} = 1.60 \, \mathrm{s} \) and the gravity \( g_{E} = 9.81 \, \mathrm{m/s^2} \). On Mars, \( g_{M} = 3.71 \, \mathrm{m/s^2} \). We are asked to find the period \( T_{M} \) on Mars.
03

Relate periods on Earth and Mars

Using the period formula, we have \( T_{E}^2 = 4\pi^2 \frac{L}{g_{E}} \) and \( T_{M}^2 = 4\pi^2 \frac{L}{g_{M}} \). Since the length \( L \) is constant for the pendulum, we can write \( \frac{T_{M}^2}{T_{E}^2} = \frac{g_{E}}{g_{M}} \).
04

Solve for the period on Mars

Rearrange the relationship to solve for \( T_{M} \):\[ T_{M} = T_{E} \sqrt{\frac{g_{E}}{g_{M}}} \]Substitute the known values:\[ T_{M} = 1.60 \times \sqrt{\frac{9.81}{3.71}} \]
05

Calculate the period on Mars

Perform the calculation:\[ T_{M} = 1.60 \times \sqrt{2.644] \]\[ T_{M} \approx 1.60 \times 1.626 \approx 2.60 \, \mathrm{s} \]

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Key Concepts

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

Simple Harmonic Motion
Simple Harmonic Motion, or SHM, is a type of periodic motion where an object moves back and forth over the same path. It is characterized by its restoring force proportional to the displacement from its equilibrium position. In simpler terms, imagine a pendulum swinging back and forth. This to-and-fro motion is a classic example of simple harmonic motion.

When studying a pendulum, the periodic motion is caused by the gravitational force pulling it toward its resting position. This creates a continuous, smooth, oscillating motion that can be easily predicted using specific formulas. It's all about balance and rhythm in motion.

SHM forms the basis for understanding many natural phenomena and engineering applications, like clocks and musical instruments. So, when we consider the pendulum, we're not only talking about a weight on a string, but also about a fundamental physical concept.
Gravitational Acceleration on Mars
Gravitational acceleration is a key factor influencing the pendulum's motion, as it affects the period or the time it takes for a complete cycle. On Earth, the acceleration due to gravity is approximately 9.81 m/s².

However, planets are not all the same regarding gravity. Mars, for example, has a different gravitational pull due to its smaller size and mass, with gravity measured at approximately 3.71 m/s². This difference greatly impacts how things move on Mars compared to Earth.

The smaller gravitational acceleration on Mars means that the force pulling the pendulum towards its equilibrium is weaker. This results in a slower swing, which is why the period of the pendulum is longer on Mars compared to Earth.
Period of a Pendulum
The period of a pendulum is the duration it takes to complete one full swing, or oscillation, back to its starting point. The formula 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 of the pendulum, and \( g \) is the acceleration due to gravity.

This relationship shows that the period is dependent on both the gravitational force and the length of the pendulum. Importantly, the period does not depend on the mass of the pendulum or the amplitude of the swing as long as the oscillations are small.

Understanding this relationship helps in solving physics problems related to pendulums, whether on Earth, Mars, or elsewhere in the universe.
Physics Problem Solving
Physics problem solving often involves breaking down the problem into more manageable parts and using known formulas to find unknown quantities. Here's a simple strategy:

  • Identify what you know and what you need to find out.
  • Write down the relevant formulas.
  • Plug in the known values.
  • Solve for the unknown.
  • Check your work for consistency.


In the case of the pendulum problem on Mars, we were given the period on Earth and gravitational forces for Earth and Mars. By understanding how these relate through the pendulum period formula, we calculated the period on Mars.

This approach ensures that even complex problems become a step-by-step process that is easier to handle. Physics isn't just about memorizing formulas; it's about applying them logically to find solutions to real-world questions.

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Most popular questions from this chapter

An astronaut notices that a pendulum that took \(2.50 \mathrm{~s}\) for a complete cycle of swing when the rocket was waiting on the launch pad takes \(1.25 \mathrm{~s}\) for the same cycle of swing during liftoff. What is the acceleration of the rocket? (Hint: Inside the rocket, it appears that \(g\) has increased.)

A science museum has asked you to design a simple pendulum that will make 25.0 complete swings in \(85.0 \mathrm{~s}\). What length should you specify for this pendulum?

"Seeing" surfaces at the nanoscale. One technique for making images of surfaces at the nanometer scale, including membranes and biomolecules, is dynamic atomic force microscopy. In this technique, a small tip is attached to a cantilever, which is a flexible, rectangular slab supported at one end, like a diving board (Figure 11.40 ). The cantilever vibrates, so the tip moves up and down in simple harmonic motion. In one mode of operation, the resonant frequency for a cantilever with force constant \(k=1000 \mathrm{~N} / \mathrm{m}\) is \(100 \mathrm{kHz}\). As the oscillating tip is brought within a few nanometers of the surface of a sample (as shown in the figure), it experiences an attractive force from the surface. For an oscillation with a small amplitude (typically \(0.050 \mathrm{nm}),\) the force \(F\) that the sample surface exerts on the tip varies linearly with the displacement \(x\) of the tip, \(|F|=k_{\text {surf }} x,\) where \(k_{\text {surf }}\) is the effective force constant for this force. The net force on the tip is therefore \(\left(k+k_{\text {surf }}\right) x,\) and the frequency of the oscillation changes slightly due to the interaction with the surface. Measurements of the frequency as the tip moves over different parts of the sample's surface can provide information about the sample. If we model the vibrating system as a mass on a spring, what is the mass necessary to achieve the desired resonant frequency when the tip is not interacting with the surface? A. \(25 \mathrm{ng}\) B. \(100 \mathrm{ng}\) C. \(25 \mu \mathrm{g}\) D. \(100 \mu \mathrm{g}\)

An object suspended from a spring vibrates with simple harmonic motion. At an instant when the displacement of the object is equal to one-half the amplitude, what fraction of the total energy of the system is kinetic and what fraction is potential?

A pendulum consisting of a \(0.5 \mathrm{~kg}\) mass tied to a \(0.1 \mathrm{~m}\) string is set into oscillation at the same moment that a stone is dropped from a 44.1 -m-tall building. How many cycles of oscillation will the pendulum go through before the stone hits the ground?

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