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

Approximately
`1.44* 10^3 \; \rm N \cdot m^(-1)` assuming that the spring has zero mass.

Step-by-step 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 = (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 = (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((2\pi)/(T)\right)^2 \cdot m(\text{chair}) \\ &= \left((2\pi)/(1.00\; \rm s)\right)^2 * 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}`.