Question

A common device for entertaining a toddler is a jump seat that hangs from the horizontal portion of a doorframe via elastic cords (Fig. 15-63). Assume that only one cord is on each side in spite of the more realistic arrangement shown. When a child is placed in the seat, they both descend by a distance ${\mathbf{d}}_{\mathbf{s}}$as the cords stretch (treat them as springs). Then the seat is pulled down an extra distance ${\mathbf{d}}_{\mathbf{m}}$and released, so that the child oscillates vertically, like a block on the end of a spring. Suppose you are the safety engineer for the manufacturer of the seat. You do not want the magnitude of the child’s acceleration to exceed 0.20 gfor fear of hurting the child’s neck. If ${\mathbf{d}}_{\mathbf{m}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{cm}$, what value of ${\mathbf{d}}_{\mathbf{s}}$corresponds to that acceleration magnitude?

Short answer

The distance ${\mathrm{d}}_{\mathrm{s}}$corresponding to the magnitude of the child’s acceleration is 0.5 m .

Step-by-step solution

The given data

Stretched distance, $d_m=10\mathrm{cm}$.

Understanding the concept of simple harmonic motion

Using the force of spring and force of gravity on a child, we can find the relation between distance and angular velocity. From this relationship, we can distance corresponding to the magnitude of the child’s acceleration when stretched to 10 cm.

Formulae:

The force due to gravity, $F_g=mg$ (i)

The force due to Hooke’s Law, $F_s=-K{d}_3$ (ii)

The angular frequency of a body in SHM, $\omega =\sqrt{k/m}$ (iii)

The acceleration of a body in SHM, $a={\omega }^{2}x$ (iv)

Calculation of the distance that corresponds to magnitude of acceleration

At equilibrium, force of spring is equal to force of gravity on the block.

Therefore, using equations (i) and (ii), we get

$$\begin{gathered} -k{d}_s=mg \\ {d}_s=-\frac{mg}{k} \\ =-\frac{g}{{\omega }^2} \end{gathered}$$

When the spring is stretched,distance$d_m$, acceleration on child is

$a_g=0.20g$

Since

$a_g=0.20g,x=10\mathrm{cm}=0.1\mathrm{m}$

Therefore,angular frequency of oscillation at this acceleration using equation (iv)will be;

$$\begin{gathered} {\omega }^2=\frac{a_g}{x} \\ =\frac{0.20\mathrm{g}}{0.1\mathrm{m}} \\ =2{\mathrm{gm}}^{-1} \end{gathered}$$

Therefore, substituting this above value in equation (a), we get the distance at which the magnitude of acceleration is attained as:

$$\begin{gathered} d_s=\frac{g}{2g{m}^{-1}} \\ =\frac{1}{2}\mathrm{m} \\ =0.5\mathrm{m} \end{gathered}$$

Therefore, the distance role="math" localid="1657365398809" ${\mathrm{d}}_{\mathrm{s}}$corresponding to the magnitude of the child’s acceleration is 0.5 m .