Question

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.

Answer 1

Answer:

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 $F$ and for $x$. By Hooke's Law, the spring constant would be ratio of these two expressions.

Let $T$ 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 $A$ stands for the amplitude of this oscillation. In a simple harmonic oscillation, both $\mathbf{F}$ and $\mathbf{x}$ are proportional to $A$. Keep in mind that the spring constant $k$ is simply the opposite of the ratio between $\mathbf{F}$ and $\mathbf{x}$. Therefore, the exact value of $A$ 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 $t$ would be:

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

The restoring velocity at time $t$ would be:

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

The restoring acceleration at time $t$ 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 $t$ 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, $k$:

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

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

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

Before proceeding, note how $A$ was eliminated from the ratio (as expected.) Additionally, $t$ 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 $k$:

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

Side note on the unit of $k$:

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