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Chapter 6: Problem 61

Astronauts in space "weigh" themselves by oscillating on a spring. Suppose the position of an oscillating \(75 \mathrm{kg}\) astronaut is given by \(x=(0.30 \mathrm{m}) \sin ((\pi \mathrm{rad} / \mathrm{s}) \cdot t),\) where \(t\) is in \(\mathrm{s}\). What force does the spring exert on the astronaut at (a) \(t=1.0 \mathrm{s}\) and (b) 1.5 s. Note that the angle of the sine function is in radians.

Short Answer

Expert verified
The force exerted on the astronaut by the spring at \(t = 1.0s\) and \(t = 1.5s\) is 0 N.

Step by step solution

01

Find the acceleration of the astronaut

First, we'll need to find the acceleration of the astronaut by differentiating the displacement function twice. The displacement function is given by \(x=(0.30m) \sin ((\pi \mathrm{rad}/s) \cdot t)\). The first derivative with respect to \(t\) gives the velocity, and the second derivative gives the acceleration. Applying the chain rule, the first derivative is \(v = \pi(0.30 m) \cos ((\pi \mathrm{rad}/s) \cdot t)\), and the second derivative (acceleration) is \(a = -\pi^2(0.30 m) \sin ((\pi \mathrm{rad}/s) \cdot t)\).
02

Calculate the acceleration at \(t = 1.0s\) and \(t = 1.5s\)

We can substitute 1.0s and 1.5s into the acceleration equation to find the values of the acceleration at these times. For \(t = 1.0s\), \(a = -\pi^2 (0.30 m) \sin (\pi \cdot 1.0) = -\pi^2 (0.30 m) \sin (\pi) = 0 m/s^2\). For \(t = 1.5s\), \(a = -\pi^2 (0.30 m) \sin (\pi \cdot 1.5) = -\pi^2 (0.30 m) \sin (1.5\pi) = 0 m/s^2\).
03

Determine the force exerted by the spring

Using the equation of motion \(F = ma\) where \(m\) is the astronaut's mass and \(a\) is the acceleration, you can calculate the force exerted by the spring on the astronaut at \(t = 1.0s\) and \(t = 1.5s\). Since the acceleration at both of these times is 0, the spring force is also 0: \(F = 75 kg \cdot 0 = 0 N\). Thus, no force is being exerted on the astronaut by the spring at \(t = 1.0s\) and \(t = 1.5s\).

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Key Concepts

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

Oscillation
In the context of physics, oscillation refers to any kind of periodic motion that repeats itself in a regular cycle. When we talk about an astronaut oscillating on a spring, we are describing how they move back and forth in a relatively predictable way.
This movement is akin to a pendulum swinging or a weight bobbing up and down on a spring.
Purely mathematical oscillations typically involve sinusoidal functions like sine or cosine, given their inherent cyclical nature.
  • The oscillation frequency determines how often the cycles occur in a unit of time.
  • The amplitude represents the maximum extent of the oscillation away from its central, or equilibrium, position.
In this exercise, the astronaut's position is described by a sinusoidal function, highlighting its oscillatory nature. Understanding the oscillation helps us predict where the astronaut will be at a given time as they continue to move back and forth.
Spring Force
When dealing with oscillating systems, particularly those involving springs, spring force is an important concept.
Hooke's law tells us that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.
  • This is expressed mathematically as \( F = -kx \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement.
  • The negative sign indicates that the spring force acts in the opposite direction of the displacement, pulling back toward equilibrium.
In our example, the force exerted on the astronaut is calculated using the equation of motion, \( F = ma \), where \( m \) is mass and \( a \) is acceleration.
When the spring system oscillates and the sine function evaluates to zero, the acceleration is zero, leading to a zero force exerted by the spring.
Trigonometric Functions
Trigonometric functions are incredibly useful in describing oscillatory motion, such as that seen with pendulums or springs.
The sine and cosine functions naturally model these types of periodic movements due to their wave-like shapes. In physics, the argument of these functions is often given in terms of time, along with a frequency or angular velocity component.
  • In the astronaut's motion, the function \( x = (0.30 \,\text{m}) \sin((\pi \,\text{rad}/\text{s}) \cdot t) \) represents their position as a function of time.
  • The \( \pi \,\text{rad}/\text{s} \) term in the sine function is known as the angular frequency, determining how quickly the oscillations occur.
  • The amplitude, \( 0.30 \,\text{m} \), indicates the furthest point the astronaut moves from their resting position.
Trigonometric functions also simplify the calculation of velocity and acceleration through differentiation, as seen in the solution of the exercise where these functions lead to the force being zero at specific times.

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Most popular questions from this chapter

You've been called in to investigate a construction accident in which the cable broke while a crane was lifting a \(4500 \mathrm{kg} \mathrm{con}\) tainer. The steel cable is \(2.0 \mathrm{cm}\) in diameter and has a safety rating of \(50,000 \mathrm{N}\). The crane is designed not to exceed speeds of \(3.0 \mathrm{m} / \mathrm{s}\) or accelerations of \(1.0 \mathrm{m} / \mathrm{s}^{2},\) and your tests find that the crane is not defective. What is your conclusion? Did the crane operator recklessly lift too heavy a load? Or was the cable defective?

You and your friend Peter are putting new shingles on a roof pitched at \(25^{\circ} .\) You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, \(5.0 \mathrm{m}\) away, asks you for the box of nails. Rather than carry the \(2.5 \mathrm{kg}\) box of nails down to Peter, you decide to give the box a push and have it slide down to him. If the coefficient of kinetic friction between the box and the roof is \(0.55,\) with what speed should you push the box to have it gently come to rest right at the edge of the roof?

A \(2.0 \mathrm{kg}\) wood block is launched up a wooden ramp that is inclined at a \(30^{\circ}\) angle. The block's initial speed is \(10 \mathrm{m} / \mathrm{s}\) a. What vertical height does the block reach above its starting point? b. What speed does it have when it slides back down to its starting point?

An object moving in a liquid experiences a linear drag force: \(\vec{D}=(b v, \text { direction opposite the motion }),\) where \(b\) is a constant called the drag coefficient. For a sphere of radius \(R\), the drag constant can be computed as \(b=6 \pi \eta R,\) where \(\eta\) is the viscosity of the liquid. a. Find an algebraic expression for \(v_{x}(t),\) the \(x\) -component of velocity as a function of time, for a spherical particle of radius \(R\) and mass \(m\) that is shot horizontally with initial speed \(v_{0}\) through a liquid of viscosity \(\eta\) b. Water at \(20^{\circ} \mathrm{C}\) has viscosity \(\eta=1.0 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^{2} .\) Suppose a \(4.0-\mathrm{cm}\) -diameter, \(33 \mathrm{g}\) ball is shot horizontally into a tank of \(20^{\circ} \mathrm{C}\) water. How long will it take for the horizontal speed to decrease to \(50 \%\) of its initial value?

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