A thrill seeking cat of mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillated vertically in SHM.

The amplitude is 0.50 m, and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of all these three energies when the cat is

a) at its highest point

b) at its lowest point

c) at its equilibrium position.

(Exercise 14.39, 14th edition or 15th edition)

To solve this problem, we need to analyze the energy at different points in the oscillation of the cat. We will consider elastic potential energy (${U}_{}U\_\{\backslash text\{elastic\}\}$), kinetic energy ($KK$), and gravitational potential energy (${U}_{g}U\_g$). The total mechanical energy ($EE$) in simple harmonic motion is conserved and is given by the sum of these three energies.

Given:

First, let’s establish the spring constant $kk$k. At the equilibrium position, the spring is stretched by the weight of the cat. This stretch ${x}_{}x\_\{\backslash text\{eq\}\}$xeq can be found using Hooke’s law and the equilibrium condition:

$mg=k{x}_{}mg\; =\; k\; x\_\{\backslash text\{eq\}\}$

At equilibrium: ${x}_{}=\frac{mg}{k}x\_\{\backslash text\{eq\}\}\; =\; \backslash frac\{mg\}\{k\}$

The total mechanical energy in the system is constant and can be expressed as:

$E=\frac{1}{2}k{A}^{2}E\; =\; \backslash frac\{1\}\{2\}kA^2$

The three positions to analyze are:

Highest point(where the spring is unstretched, and the cat is at $y=Ay\; =\; A$y=A)Lowest point(where the spring is most stretched, and the cat is at $y=-Ay\; =\; -A$y=−A)Equilibrium position(where the spring is stretched by ${x}_{}x\_\{\backslash text\{eq\}\}$xeq, and the cat is at $y=0y\; =\; 0$y=0)## 1. Highest Point ( $y=Ay\; =\; A$ )

At the highest point:

${U}_{g}=mg({y}_{}-{y}_{})=mg\left(2A\right)=4\cdot 9.8\cdot 2\cdot 0.5=39.2\hspace{0.17em}JU\_g\; =\; mg(y\_\{\backslash text\{max\}\}\; \u2013\; y\_\{\backslash text\{min\}\})\; =\; mg(2A)\; =\; 4\; \backslash cdot\; 9.8\; \backslash cdot\; 2\; \backslash cdot\; 0.5\; =\; 39.2\; \backslash ,\; \backslash text\{J\}$

Total energy: $E={U}_{}+K+{U}_{g}=0+0+39.2=39.2\hspace{0.17em}JE\; =\; U\_\{\backslash text\{elastic\}\}\; +\; K\; +\; U\_g\; =\; 0\; +\; 0\; +\; 39.2\; =\; 39.2\; \backslash ,\; \backslash text\{J\}$

## 2. Lowest Point ( $y=-Ay\; =\; -A$ )

At the lowest point:

${U}_{}=\frac{1}{2}k(2A{)}^{2}=2k{A}^{2}U\_\{\backslash text\{elastic\}\}\; =\; \backslash frac\{1\}\{2\}k(2A)^2\; =\; 2kA^2$

Total energy: $E={U}_{}+K+{U}_{g}=2k{A}^{2}+0+0=2\left(\frac{1}{2}k{A}^{2}\right)=k{A}^{2}=39.2\hspace{0.17em}JE\; =\; U\_\{\backslash text\{elastic\}\}\; +\; K\; +\; U\_g\; =\; 2kA^2\; +\; 0\; +\; 0\; =\; 2\; \backslash left(\; \backslash frac\{1\}\{2\}\; k\; A^2\; \backslash right)\; =\; kA^2\; =\; 39.2\; \backslash ,\; \backslash text\{J\}$

## 3. Equilibrium Position ( $y=0y\; =\; 0$ )

At the equilibrium position:

${x}_{}=\frac{mg}{k}x\_\{\backslash text\{eq\}\}\; =\; \backslash frac\{mg\}\{k\}$

Thus the elastic potential energy:

${U}_{}=\frac{1}{2}k{x}_{2}^{}=\frac{1}{2}k{\left(\frac{mg}{k}\right)}^{2}=\frac{1}{2}\frac{{m}^{2}{g}^{2}}{k}U\_\{\backslash text\{elastic\}\}\; =\; \backslash frac\{1\}\{2\}\; k\; x\_\{\backslash text\{eq\}\}^2\; =\; \backslash frac\{1\}\{2\}\; k\; \backslash left(\; \backslash frac\{mg\}\{k\}\; \backslash right)^2\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{m^2\; g^2\}\{k\}$

${U}_{g}=mgAU\_g\; =\; mgA$

Using conservation of mechanical energy: $E=\frac{1}{2}k{A}^{2}={U}_{}+K+{U}_{g}E\; =\; \backslash frac\{1\}\{2\}\; k\; A^2\; =\; U\_\{\backslash text\{elastic\}\}\; +\; K\; +\; U\_g$

Substituting:

$39.2=\frac{1}{2}\frac{{m}^{2}{g}^{2}}{k}+K+mgA39.2\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{m^2\; g^2\}\{k\}\; +\; K\; +\; mgA$

Solving for $KK$K:

$K=39.2-\frac{1}{2}\frac{(4\cdot 9.8{)}^{2}}{k}-4\cdot 9.8\cdot 0.5=39.2-0.5\cdot 19.6\cdot 4-19.6K\; =\; 39.2\; \u2013\; \backslash frac\{1\}\{2\}\; \backslash frac\{(4\; \backslash cdot\; 9.8)^2\}\{k\}\; \u2013\; 4\; \backslash cdot\; 9.8\; \backslash cdot\; 0.5\; =\; 39.2\; \u2013\; 0.5\; \backslash cdot\; 19.6\; \backslash cdot\; 4\; \u2013\; 19.6$

This gets us back to our starting total mechanical energy: $E=39.2\hspace{0.17em}JE\; =\; 39.2\; \backslash ,\; \backslash text\{J\}$

Thus, all the individual calculations should maintain the total energy value $E=39.2\hspace{0.17em}JE\; =\; 39.2\; \backslash ,\; \backslash text\{J\}$ in each state for consistency.