Physics, 20.04.2020 23:18 ronnie7898

In order to study the long-term effects of weightlessness, astronauts in space must be weighed (or at least "massed"). One way in which this is done is to seat them in a chair of known mass attached to a spring of known force constant and measure the period of the oscillations of this system. The 36.4 kg chair alone oscillates with a period of 1.00 s, and the period with the astronaut sitting in the chair is 2.20 s.

Find the force constant of the spring.

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Answer from: Animallover100

Approximately $1.44\times 10^3 \; \rm N \cdot m^{-1}$ assuming that the spring has zero mass.

Explanation:

Without any external force, a piece of mass connected to an ideal spring (like the chair in this question) will undergo simple harmonic oscillation.

On the other hand, the force constant of a spring (i.e., its stiffness) can be found using Hooke's Law. If the spring exerts a restoring force $\mathbf{F}$ when its displacement is $\mathbf{x}$ , then its force constant would be:

$\displaystyle k = -\frac{\mathbf{F}}{\mathbf{x}}$ .

The goal here is to find the expressions for and for . By Hooke's Law, the spring constant would be ratio of these two expressions.

Let represent the time period of this oscillation. With the chair alone, the period of oscillation is $T = 1.00\; \rm s$ .

For a simple harmonic oscillation, the angular frequency $\omega$ can be found from the period:

$\displaystyle \omega = \frac{2\pi}{T}$ .

Let stands for the amplitude of this oscillation. In a simple harmonic oscillation, both $\mathbf{F}$ and $\mathbf{x}$ are proportional to . Keep in mind that the spring constant is simply the opposite of the ratio between $\mathbf{F}$ and $\mathbf{x}$ . Therefore, the exact value of shouldn't really affect the value of the spring constant.

In a simple harmonic motion (one that starts with maximum displacement and zero velocity,) the displacement (from equilibrium position) at time would be:

$\displaystyle \mathbf{x}(t) = A \cos(\omega \cdot t)$ .

The restoring velocity at time would be:

$\displaystyle \mathbf{v}(t) = \mathbf{x}^\prime(t) = -A\, \omega \sin(\omega\cdot t)$ .

The restoring acceleration at time would be:

$\displaystyle \mathbf{a}(t) = \mathbf{v}^\prime(t) = -A\, \omega^2 \cos(\omega\cdot t)$ .

Assume that the spring has zero mass. By Newton's Second Law of motion, the restoring force at time would be:

$\begin{aligned}& \mathbf{F}(t) \\ &= m(\text{chair}) \cdot \mathbf{a}(t) \\&= -m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)\end{aligned}$ .

Apply Hooke's Law to find the spring constant, :

$\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= -\left(\frac{-m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)}{A\cos(\omega \cdot t)}\right) \\ &= \omega^2 \cdot m(\text{chair}) \end{aligned}$ .

Again, $\omega$ stands for the angular frequency of this oscillation, where

$\displaystyle \omega = \frac{2\pi}{T}$ .

Before proceeding, note how was eliminated from the ratio (as expected.) Additionally, is also eliminated from the ratio. In other words, the spring constant is "constant" at all time. That agrees with the assumption that this spring is indeed ideal. Back to :

$\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= \cdots \\ &= \omega^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{T}\right)^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{1.00\; \rm s}\right)^2 \times 36.4\; \rm kg\end{aligned}$ .

Side note on the unit of :

$\begin{aligned} & 1\; \rm kg \cdot s^{-2} \\ &= 1\rm \; \left(kg \cdot m \cdot s^{-2}\right) \cdot m^{-1} \\ &= 1\; \rm N \cdot m^{-1}\end{aligned}$ .

Answer from: Quest

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Answer from: Quest

3. dark matter

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