/*! 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 17 BIO Weighing Astronauts. This pr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

BIO Weighing Astronauts. This procedure has 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 \(1.30 \mathrm{~s}\) to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes \(2.54 \mathrm{~s}\) for one cycle. What is the mass of the astronaut?

Short Answer

Expert verified
The astronaut's mass is approximately \( 130 \mathrm{~kg} \).

Step by step solution

01

Determine the spring constant using the chair without the astronaut

We will first determine the spring constant using the chair's mass and oscillation period when empty. Using the formula \( T = 2 \pi \sqrt{\frac{m}{k}} \), we can solve for the spring constant k, which should be \( k = \frac{m}{(T/2 \pi)^2} \). Substituting the given values, \( m = 42.5 \mathrm{~kg} \) and \( T = 1.3 \mathrm{~s} \), we find \( k \approx 13.6 \mathrm{~N/m} \).
02

Determine the mass of the astronaut

Now we can use the spring constant to find the mass of the astronaut. Rearranging the same formula for m and using the period of oscillation when the astronaut is sitting yields \( m = (T/2 \pi)^2 \cdot k \). Substituting the given values \( k \approx 13.6 \mathrm{~N/m} \) and \( T = 2.54 \mathrm{~s} \), we calculate the mass of the astronaut to be approximately \( 130 \mathrm{~kg} \).
03

Final check

Ensure that your final mass makes sense and is reasonable. Keeping in mind the mass of a typical astronaut, our result seems reasonable.

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.

Spring Constant
To understand the concept of spring constant, we begin with Hooke's Law. This fundamental law in physics states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, this can be expressed as \[ F = -kx \] where:
  • \( F \) is the force applied to the spring,
  • \( k \) is the spring constant, and
  • \( x \) is the displacement of the spring from its natural length.
The spring constant \( k \) is a measure of the stiffness of a spring. A higher \( k \) indicates a stiffer spring that requires more force to achieve the same displacement. In the context of the given problem, we calculated the spring constant by first using the empty chair's mass and oscillation period. Knowing \( m = 42.5 \, \text{kg} \) and \( T = 1.3 \, \text{s} \), we solved the equation: \[ T = 2 \pi \sqrt{\frac{m}{k}} \] Rearranging for \( k \), the spring constant is approximately \( 13.6 \, \text{N/m} \). This helped us proceed to find the astronaut's mass.
Simple Harmonic Motion
Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement. It is one of the most important principles for understanding oscillations. The detailed features of SHM include:
  • The motion is repetitive with a fixed rhythm.
  • The force driving the object towards the center is proportional to the displacement.
  • The energy in SHM is conserved and continuously switches between kinetic and potential forms.
In the given problem, the oscillations of the chair and astronaut follow simple harmonic motion. The essential formula used is the period of oscillation \( T \), given by: \[ T = 2 \pi \sqrt{\frac{m}{k}} \] This signifies that the period is dependent on both the mass \( m \) and the spring constant \( k \). When the chair is occupied by the astronaut, the longer period implies increased effective mass, which tells us about the mass of the astronaut when solved.
Mass Measurement in Space
In space, conventional methods of weighing do not work due to the lack of gravity. Instead, astronauts’ mass is measured using principles like those involving oscillations. The key idea is that a known spring constant can be used to determine an unknown mass based on the period of oscillation.Using the known values of the empty chair, the spring constant was first identified. As this same constant is applicable when the astronaut joins the system, her mass was calculated from the altered period. The derived formula for finding the astronaut's mass \( m \) is \[ m = \left( \frac{T}{2 \pi} \right)^2 \cdot k \] where \( T \) is the new period with the astronaut. Upon performing the calculations, the astronaut’s mass works out to approximately \( 130 \, \text{kg} \), reflecting the simplicity and efficiency of using oscillations for mass measurement in a zero-gravity environment.

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

When an object of unknown mass is attached to an ideal spring with force constant \(120 \mathrm{~N} / \mathrm{m},\) it is found to vibrate with a frequency of \(6.00 \mathrm{~Hz}\). Find (a) the period of the motion; (b) the angular frequency; (c) the mass of the object.

SHM of a Floating Object. An object with height \(h\)mass \(M\), and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\). (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M,\) and the cross-sectional area \(A\) of the object. You can ignore the damping due to fluid friction (see Section 14.7).

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced \(0.120 \mathrm{~m}\) from its equilibrium position and released with zero initial speed, then after \(0.800 \mathrm{~s}\) its displacement is found to be \(0.120 \mathrm{~m}\) on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

A rifle bullet with mass \(8.00 \mathrm{~g}\) and initial horizontal velocity \(280 \mathrm{~m} / \mathrm{s}\) strikes and embeds itself in a block with mass \(0.992 \mathrm{~kg}\) that rests on a friction less surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of \(15.0 \mathrm{~cm} .\) After the impact, the block moves in SHM. Calculate the period of this motion.

A machine part is undergoing SHM with a frequency of \(4.00 \mathrm{~Hz}\) and amplitude \(1.80 \mathrm{~cm} .\) How long does it take the part to go from \(x=0\) to \(x=-1.80 \mathrm{~cm} ?\)

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.