/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Weighing Astronauts. This proced... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Weighing Astronauts. This procedure has actually 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 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?

Short Answer

Expert verified
The mass of the astronaut is approximately 214.81 kg.

Step by step solution

01

Understand the Formulas

To find the astronaut's mass, we use the formula for the period of a simple harmonic oscillator: \[ T = 2\pi \sqrt{\frac{m}{k}} \]where \( T \) is the period, \( m \) is the mass on the spring, and \( k \) is the spring constant. We will have two equations: one for the chair alone, and one for the chair with the astronaut.
02

Solve for the Spring Constant

Using the first case where the chair is empty and has a period of 1.30 seconds:\[ T_{chair} = 2\pi \sqrt{\frac{m_{chair}}{k}} \]Substitute \( m_{chair} = 42.5 \, \text{kg} \) and \( T_{chair} = 1.30 \, \text{s} \) into the equation:\[ 1.30 = 2\pi \sqrt{\frac{42.5}{k}} \]Solving for \( k \) gives:\[ k = \frac{4\pi^2 \times 42.5}{1.30^2} \approx 1577.53 \, \text{N/m} \]
03

Formulate the Equation with the Astronaut

With the astronaut, the period is 2.54 seconds:\[ T_{total} = 2\pi \sqrt{\frac{m_{total}}{k}} \]Substitute \( T_{total} = 2.54 \, \text{s} \) and \( k = 1577.53 \, \text{N/m} \) , and let \( m_{total} = m_{chair} + m_{astronaut} \):\[ 2.54 = 2\pi \sqrt{\frac{42.5 + m_{astronaut}}{1577.53}} \]
04

Solve for Astronaut's Mass

Re-arrange and solve for \( m_{astronaut} \):\[ \frac{2.54}{2\pi} = \sqrt{\frac{42.5 + m_{astronaut}}{1577.53}} \]\[ \left(\frac{2.54}{2\pi}\right)^2 = \frac{42.5 + m_{astronaut}}{1577.53} \]Calculating the left side gives approximately 0.16313:\[ 0.16313 = \frac{42.5 + m_{astronaut}}{1577.53} \]\[ 42.5 + m_{astronaut} = 0.16313 \times 1577.53 \]\[ m_{astronaut} = 257.31 - 42.5 \approx 214.81 \, \text{kg} \]
05

Final Answer

The mass of the astronaut is approximately 214.81 kg. This concludes the problem by using the data provided and the oscillation equations to find the astronaut's mass in space.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mass Determination
In the realm of physics, determining mass using oscillations is a fascinating process. When we talk about mass determination using a harmonic oscillator, it involves understanding how objects attached to a spring oscillate. To determine an object's mass via oscillations, we measure the period of these oscillations. This method is not only accurate but also practical, especially in environments like space, where traditional weighing scales are not feasible. In our example, the chair's known mass is an essential factor in calculating the astronaut's mass by differentiating between the two states of oscillation: with and without the astronaut.
Oscillation Period
The term "oscillation period" refers to the time it takes for one full cycle of motion in an oscillating system, such as a spring-mass system like our chair.
  • When the chair is empty, the period is 1.30 seconds.
  • When the astronaut sits in the chair, the period increases to 2.54 seconds.
This change in the oscillation period is due to the added mass of the astronaut, which affects the system's motion. The period of oscillation is directly related to both the mass attached to the spring and the spring constant through the formula: \[ T = 2\pi \sqrt{\frac{m}{k}} \]Thus, observing the alteration in the oscillation period is a crucial diagnostic tool to determine unknown masses, such as the mass of the astronaut in this problem.
Spring Constant
The spring constant, denoted as \(k\), is a measure of a spring's stiffness and plays a significant role in harmonic motion. It essentially defines how much force is required to compress or extend the spring by a certain amount. The spring constant is determined using the formula: \[ k = \frac{4\pi^2 \times m}{T^2} \]In our scenario, the spring constant is derived from the oscillation of the chair when it's empty. Given the chair's mass (42.5 kg) and its period (1.30 seconds), we calculated:
  • \(k \approx 1577.53\, \text{N/m}\)
This value remains constant in both cases—whether the chair is empty or occupied by the astronaut—thus serving as a key factor in determining the astronaut’s mass based on the observed oscillation period.
Astronaut Weight Measurement
Measuring weight in space is tricky, as standard scales relying on gravitational pulls don't work. Instead, oscillation methods provide a robust and innovative solution. In this context, weight measurement involves determining mass using the oscillation characteristics of the system.The astronaut's mass causes a change in the system's oscillation period. Using the known spring constant and the initial conditions of the chair's oscillation when empty, we can precisely calculate the additional mass added by the astronaut:
  • The formula adjustment for determining the astronaut's mass is: \[ \left(\frac{T_{total}}{2\pi}\right)^2 \times k = m_{total} \]
Such precise calculations conclude the mass of the astronaut, emphasizing the usefulness of harmonic oscillators in space for tasks that ground-based scales usually fulfill.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A block with mass \(M\) rests on a frictionless surface and is connected to a horizontal spring of force constant \(k\) . The other end of the spring is attached to a wall (Fig. 13.36\() .\) A second block with mass \(m\) rests on top of the first block. The coefficient of static friction between the blocks is \(\mu_{s}\) Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.

After landing on an unfamiliar planet, a space explorer con- structs a simple pendulum of length 50.0 \(\mathrm{cm} .\) She finds that the pendulum makes 100 complete swings in 136 \(\mathrm{s}\) . What is the value of \(g\) on this planet?

A uniform, solid metal disk of mass 6.50 \(\mathrm{kg}\) and diameter 24.0 \(\mathrm{cm}\) hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.23 \(\mathrm{N}\) tangent to the rim of the disk to turn it by \(3.34^{\circ},\) thus twisting the wire. You now remove this force and release the disk from rest. (a) What is the torsion constant for the metal wire? (b) What are the frequency and period of the torsional oscillations of the disk? (c) Write the equation of motion for \(\theta(t)\) for the disk.

SHM in a Car Engine. The motion of the piston of an automobile engine (Fig. 13.35\()\) is approximately simple harmonic. (a) If the stroke of an engine (twice the amplitude) is 0.100 m and the engine runs at 3500 rev /min, compute the acceleration of the piston at the endpoint of its stroke. \((b)\) If the piston has mass 0.450 \(\mathrm{kg}\) , what net force must be exerted on it at this point? (c) What are the speed and kinetic energy of the piston at the midpoint of its stroke? (d) What average power is required to accelerate the piston from rest to the speed found in part (c)? (e) If the engine runs at 7000 rev/min, what are the answers to parts (b), (c), and (d)?

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

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.