91Ó°ÊÓ

Given, that the mass of the cat is m = 4 kg.

The amplitude of the SHM is A = 0.05 m.

The elastic potential energy of the spring having a spring force constant k is,

$U_S=\frac{1}{2}k{x}^2\left(1\right)$

As, the kinetic energy of a body of mass m and having a speed v is,

$E_K=\frac{1}{2}m{v}^2\left(2\right)$

The gravitational potential energy is,

$U_G=mgy$ (3)

Where, m is the mass of the body, g is the acceleration due to gravity, and y is the height of the body.

So, the total energy is,

role="math" localid="1668145837816" $E_T=U_S+E_K+U_G$ (4)

When the cat is at its highest point, then at the highest point of the motion the spring has its natural unstretched length, x = 0, so from equation (1), the potential energy of the spring is,

$U_S=0$ (5)

At the highest point, the speed is also zero, so from equation (2) the kinetic energy is,

$E_K=0$ (6)

At the highest point, y = 2A, from equation (3), the gravitational potential energy is,

$\begin{array}{r}{U}_G=mg\left(2A\right)\\ =4\mathrm{kg}×9.8{\mathrm{ms}}^{−2}×2×0.05\mathrm{m}\\ =3.92\mathrm{J}\end{array}$

Equate equations (4), (5), (6), and (7), you get,

$$\begin{gathered} E_T=0+0+3.92J \\ =3.92J \end{gathered}$$

Therefore, when the cat is at its highest point the elastic potential energy of the spring, kinetic energy of the cat, the gravitational potential energy of the system and sum of the energies are ${\mathbf{U}}_{\mathbf{S}}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{J}\mathbf{,}\mathbf{ }{\mathbf{E}}_{\mathbf{K}}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{J}\mathbf{,}\mathbf{ }{\mathbf{U}}_{\mathbf{G}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{92}\mathbf{ }\mathbf{J}\mathbf{,}\mathbf{ }{\mathbf{E}}_{\mathbf{T}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{92}\mathbf{ }\mathbf{J}$.

When the cat is at its lowest point, then at the lowest point of the motion the spring has its maximum stretched length, x = 2A, from equation (1), the elastic potential energy of the spring is,

$$\begin{gathered} U_S=\frac{1}{2}k{\left(2A\right)}^2 \\ =2k{A}^2 \end{gathered}$$

but,

$$\begin{gathered} mg=k \\ =2A \end{gathered}$$

From the above, you get,

$k=\frac{mg}{2A}$

SO, the equation (8) becomes,

$\begin{array}{r}{U}_s=2\frac{mg}{A}{A}^2\\ =mgA\\ =4\mathrm{kg}×9.8{\mathrm{ms}}^{−2}×2×0.05\mathrm{m}\\ =3.92\mathrm{J}\end{array}$

At the lowest point, the speed is also zero, from equation (2) the kinetic energy is,

$E_K=0$ (10)

At the lowest point, y = 0, from equation (3), the gravitational potential energy is,

$U_G=0$ (11)

Hence from equations (4), (9), (10), and (11), you get,

$\begin{array}{l}{E}_T=3.92+0+0J\\ =3.92J\end{array}\left(12\right)$

So, when the cat is at its lowest point the elastic potential energy of the spring, kinetic energy of the cat, the gravitational potential energy of the system and sum of the energies are ${\mathbf{U}}_{\mathbf{S}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{92}\mathbf{ }\mathbf{J}\mathbf{,}\mathbf{ }{\mathbf{E}}_{\mathbf{K}}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{J}\mathbf{,}\mathbf{ }{\mathbf{U}}_{\mathbf{G}}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{J}\mathbf{,}\mathbf{ }{\mathbf{E}}_{\mathbf{T}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{92}\mathbf{ }\mathbf{J}$.

When the cat is at its equilibrium position, then at the equilibrium position of the motion the spring has its maximum stretched length is, x = A, again from equation (1), the elastic potential energy of the spring is,

$U_S=\frac{1}{2}k{A}^2$

But here we know

$\begin{array}{l}mg=kx\\ \begin{array}{l}mg=k\\ k=\frac{mg}{A}\end{array}\end{array}$

equation (12) becomes,

$\begin{array}{r}{U}_S=\frac{mg}{2A}{A}^2\\ =\frac{1}{2}×4\mathrm{kg}×9.8{\mathrm{ms}}^{−2}×0.05\mathrm{m}\\ =0.98\mathrm{J}\end{array}$

At the equilibrium position, the speed,

$\begin{array}{r}V=V_{max}\\ =\sqrt{\frac{k}{m}}A\\ =\sqrt{\frac{mg}{mA}}A\\ =\sqrt{gA}\end{array}$

Hence from equation (2) the kinetic energy is,

$\begin{array}{r}{E}_K=\frac{1}{2}m(\sqrt{gA}{)}^2\\ =\frac{1}{2}mgA\\ =\frac{1}{2}×4kg×9.8{\mathrm{ms}}^{−2}×0.05\mathrm{m}\\ =0.98\mathrm{J}\end{array}$

At the lowest point, y = A, so from equation (3), the gravitational potential energy IS,

$\begin{array}{r}{U}_G=mgA\\ =4\mathrm{kg}×9.8{\mathrm{ms}}^{−2}×0.05\mathrm{m}\\ =1.96\mathrm{J}\end{array}$

Hence from equations (4), (13), (14), and (15), you get,

role="math" localid="1668146341810" $$\begin{gathered} E_T=0.98+0.88+1.96J \\ =3.92J \end{gathered}$$

Hence, when the cat is at its equilibrium point the elastic potential energy of the spring, kinetic energy of the cat, the gravitational potential energy of the system and sum of the energies are role="math" localid="1668146363981" ${\mathbf{U}}_{\mathbf{S}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{98}\mathbf{ }\mathbf{J}\mathbf{,}\mathbf{ }{\mathbf{E}}_{\mathbf{K}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{98}\mathbf{ }\mathbf{J}\mathbf{,}\mathbf{ }{\mathbf{U}}_{\mathbf{G}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{96}\mathbf{ }\mathbf{J}\mathbf{,}\mathbf{ }{\mathbf{E}}_{\mathbf{T}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{92}\mathbf{ }\mathbf{J}$.