A thrill seeking cat of mass 4.00 kg

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:

• Mass of the cat, $m=4.00\hspace{0.17em}kgm\; =\; 4.00\; \backslash ,\; \backslash text\{kg\}$
  
• Amplitude of oscillation, $A=0.50\hspace{0.17em}mA\; =\; 0.50\; \backslash ,\; \backslash text\{m\}$
  
• At the highest point, the spring is unstretched, which means the spring is at its natural length.
• Spring constant, $kk$, needs to be determined.

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:

1. Highest point (where the spring is unstretched, and the cat is at $y=Ay\; =\; A$y=A)
2. Lowest point (where the spring is most stretched, and the cat is at $y=-Ay\; =\; -A$y=−A)
3. 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:

• The spring is unstretched, so $U_{}=0U\_\{\backslash text\{elastic\}\}\; =\; 0$Uelastic​=0
• The kinetic energy $K=0K\; =\; 0$K=0 because the velocity is zero at the turning point.
• The gravitational potential energy $U_{g}U\_g$Ug​ relative to the lowest point:

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

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:

• The spring is stretched by $2A2A$, so the elastic potential energy:

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

• The kinetic energy $K=0K\; =\; 0$K=0 because the velocity is zero at the turning point.
• The gravitational potential energy $U_g=0U\_g\; =\; 0$Ug​=0 because it’s the reference point.

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:

• The spring is stretched by $x_{}x\_\{\backslash text\{eq\}\}$​, and:

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

Thus the elastic potential energy:

$U_{}=\frac{1}{2}k{x}_2^{}=\frac{1}{2}k{\left(\frac{mg}{k}\right)}^2=\frac{1}{2}\frac{m^2{g}^2}{k}U\_\{\backslash text\{elastic\}\}\; =\; \backslash frac\{1\}\{2\}\; k\; x\_\{\backslash text\{eq\}\}^2\; =\; \backslash frac\{1\}\{2\}\; k\; \backslash left(\; \backslash frac\{mg\}\{k\}\; \backslash right)^2\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{m^2\; g^2\}\{k\}$​

• The kinetic energy is at its maximum because the potential energy is shared between gravitational and elastic forms.
• The gravitational potential energy:

$U_g=mgAU\_g\; =\; mgA$

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:

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

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.