Chapter 10, Problem 46P

To determine

The frequency of oscillation of the spring without anybody sitting on the wooden horse.

Answer to Problem 46P

The frequency of oscillation of the spring without the child is $2.5\text{Hz}$.

Explanation of Solution

The mass of the child is $24\text{kg}$, the compression of the spring is $28\text{cm}$ and the frequency of oscillation with the child sitting on the horse is $0.88\text{Hz}$.

Write the expression for total force in vertical y-direction on the spring when the child is at equilibrium position. (Refer figure 1)

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ ky-mg=0\end{array}$     

Here, $k$ is the spring constant, $m$ is the mass of the child, $y$ is the compressed distance of the spring and $g$ is the acceleration due to gravity.

Rearranging the above equation for $k$

$k=\frac{mg}{y}$                                                                          (I)

Write the expression for the frequency when there’s nobody on the horse

$f_h=\frac{1}{2\pi }\sqrt{\frac{k}{m_h}}$                                                                           (II)

Here, $f_h$ is the frequency of the spring without the child and $m_h$ is the mass of the horse.

Write the expression for force constant of the spring

$\begin{array}{c}k=m{\omega }_c^2\\ =\left(m_c+m_h\right){\omega }_c^2\end{array}$

Here, $m$ is the total mass on the spring, $m_c$ is the mass of the child and ${\omega }_c$ is the angular frequency of the spring with the child.

Rearranging the above equation for $m_h$

$m_h=\frac{k}{{\omega }_c^2}-m_c$                                                                          (III)

Substitute (III) in (II)

$f_h=\frac{1}{2\pi }\sqrt{\frac{k}{k/{\omega }_c^2-m_c}}$                                                                            (IV)

Substitute $24\text{kg}$ for $m$, $9.80{\text{ms}}^{-}{}^{\text{2}}$ for $g$ and $28\text{cm}$ for $y$ in (I) to find $k$.

$\begin{array}{c}k=\frac{\left(24\text{kg}\right)\left(9.80{\text{ms}}^{-}{}^{\text{2}}\right)}{28\text{ cm}}\\ =\frac{\left(24\text{kg}\right)\left(9.80{\text{ms}}^{-}{}^{\text{2}}\right)}{28\times {10}^{-2}\text{ m}}\\ =840{\text{Nm}}^{-1}\end{array}$

Substitute $840{\text{Nm}}^{-1}$ for $k$ and $0.88\text{Hz}$ for ${\omega }_c$ and $24\text{kg}$ for $m_c$ in (IV) to find $f_h$

$\begin{array}{c}{f}_h=\frac{1}{2\pi }\sqrt{\frac{840{\text{Nm}}^{-1}}{\left({840{\text{Nm}}^{-1}/\left(0.88\text{Hz}\right)}^2\right)-24\text{kg}}}\\ =\frac{1}{2\pi }\sqrt{\frac{840{\text{Nm}}^{-1}}{\left({840{\text{kgms}}^{-2}\cdot {\text{m}}^{-1}/\left(0.88\text{Hz}\times 2\pi \text{}\text{ rad}\cdot {\text{s}}^{-1}\right)}^2\right)-24\text{kg}}}\\ =2.5\text{Hz}\end{array}$

Thus, the frequency of oscillation of the spring without the child is $2.5\text{Hz}$.