Q38E - University Physics with Modern Physics

Question

A proud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.180 m. (a) Find the force constant of the spring. The fish is now pulled down 5.00 cm and released. (b) What is the period of oscillation of the fish? (c) What is the maximum speed it will reach?

Step-by-Step Solution

Answer

1. The force constant of the spring is $\mathit{k}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{54}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{ }\mathit{N}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$.
2. The period of oscillation of the fish is T = 0.85 s.

           3. The maximum speed it will reach is ${\mathbf{V}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathbf{037}\mathbf{ }\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$.

Step 1: Write the given information and determine the force constant

Here, given that the mass of the fish is m = 65 kg.

The spring stretched by the fish is x = 0.18 m.

Now as the fish is pulled down 5.00 cm and released it means the amplitude of the SHM is A = 5 cm = 5 x ${10}^{-2}$m.

As, the fish is hanging with the spring, the magnitude of the force applied on the fish by the spring is,  $F_S=kx$ is balanced by the gravitational force $F_G$= mg.

So,

$\begin{array}{l}{F}_s=F_G\\ kx=mg\end{array}$

Where, k is spring force constant, x is the displacement, m is the mass of the object, and g = 9.8m $s^{-2}$ is the acceleration due to gravity.

Therefore, by rearranging the above equation, you get,

$\begin{array}{r}k=\frac{mg}{x} \\ =\frac{65\mathrm{kg}\times 9.8{\mathrm{ms}}^{-2}}{0.18\mathrm{m}}\\ =3538.9\mathrm{Nm}\\ =3.54\times {10}^3{\mathrm{Nm}}^{-1}\end{array}$ 

Therefore, the force constant of the spring is $\mathit{k}\mathbf{=}\mathbf{3}\mathbf{.54}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathit{N}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$.

Step 2: Calculate the time period

The time period in SHM of the fish having the mass m and spring force constant k is,

$\begin{array}{r}T=2\pi \sqrt{\frac{m}{k}} \\ =2\pi \sqrt{\frac{65\mathrm{kg}}{3.54\times {10}^3{\mathrm{Nm}}^{-1}}}\\ =0.85\mathrm{s} \end{array}$

So, the period of oscillation of the fish is T = 0.85 s.

Step 3: Calculate the maximum velocity of the fish

As the velocity $v_x$ of the fish at a given displacement is,

$v_x=\pm \sqrt{\frac{k}{m}}\sqrt{A^2-x^2}$

Since the block's maximum speed occurs when it is moving through x = 0,

$\begin{array}{r}{v}_x=\sqrt{\frac{k}{m}}\sqrt{A^2-0^2} \\ =\sqrt{\frac{k}{m}}A \\ =\sqrt{\frac{3.54\times {10}^3{\mathrm{Nm}}^{-1}}{65\mathrm{kg}}}\times 5\times {10}^{-2}\mathrm{m}\\ =0.37{\mathrm{ms}}^{-1} \end{array}$

Hence, the maximum speed it will reach is \({v_{max}} = 0.37\;m\;{s^{ - 1}}\).