Chapter 6, Problem 4SP

Suppose that a 300-g mass (0.30 kg) is oscillating at the end of a spring upon a horizontal surface that is essentially friction-free. The spring can be both stretched and compressed and has a spring constant of 400 N/m. It was originally stretched a distance of 12 cm (0.12 m) from its equilibrium (unstretched) position prior to release.

1. a. What is its initial potential energy?
2. b. What is the maximum velocity that the mass will reach in its oscillation? Where in the motion is this maximum reached?
3. c. Ignoring friction, what are the values of the potential energy, kinetic energy, and velocity of the mass when the mass is 6 cm from the equilibrium position?
4. d. How does the value of velocity computed in part c compare to that computed in part b? (What is the ratio of the values? This is interesting because even though you are midway between the equilibrium point and the maximum displacement, the velocity is much closer to the maximum value you found in part b.)

To determine

The initial potential energy of the spring.

Answer to Problem 4SP

The initial potential energy of the spring-mass system is $2.88\text{J}$.

Explanation of Solution

Given info: The spring constant is $400\text{N/m}$ and the stretched distance is $0.12\text{m}$.

Write the expression for the elastic potential energy of a spring.

$PE=\frac{1}{2}k{x}^2$

Here,

$PE$ is the elastic potential energy

$k$ is the spring constant

$x$ is the stretched distance

Substitute $400\text{N/m}$ for $k$ and $0.12\text{m}$ for $x$ to find the potential energy $PE$.

$\begin{array}{c}PE=\frac{1}{2}\left(400\text{N/m}\right){\left(0.12\text{m}\right)}^2\\ =2.88\text{J}\end{array}$

Conclusion:

Therefore, the initial potential energy of the spring-mass system is $2.88\text{J}$.

To determine

The maximum velocity that the mass will reach in its oscillation and the position where then system achieves maximum velocity.

Answer to Problem 4SP

The maximum velocity that the mass will reach in its oscillation is $4.38\text{m/s}$ and this velocity occurs as the mass moves through the equilibrium position.

Explanation of Solution

Given info: The mass hanged attached to the spring is $0.30\text{kg}$ and the initial potential energy of the spring-mass system is $2.88\text{J}$.

The maximum velocity will be at a point where the kinetic energy of the system is maximum. The maximum kinetic energy is obtained when all the potential energy stored in the state, is completely converted to the kinetic energy. Since the initial potential energy is $2.88\text{J}$, the maximum kinetic energy that the system can achieve is $2.88\text{J}$.

$KE=2.88\text{J}$

Write the expression for kinetic energy of an object.

$KE=\frac{1}{2}m{v}^2$

Here,

$m$ is the mass of the object

$v$ is the velocity of the object

Solve for $v$.

$v=\sqrt{\frac{2\left(KE\right)}{m}}$

Substitute $2.88\text{J}$ for $KE$ and $0.30\text{kg}$ for $m$ to find the maximum velocity $v$.

$\begin{array}{c}v=\sqrt{\frac{2\left(2.88\text{J}\right)}{0.30\text{kg}}}\\ =4.38\text{m/s}\end{array}$

The potential energy of the spring-mass system converts completely to kinetic energy when the system moves through the equilibrium position. Thus, the maximum velocity occurs as the mass moves through the equilibrium position.

Conclusion:

Therefore, the maximum velocity that the mass will reach in its oscillation is $4.38\text{m/s}$ and this velocity occurs as the mass moves through the equilibrium position.

To determine

The potential energy, kinetic energy and the velocity of the mass when the mass is $6\text{cm}$ from the equilibrium position.

Answer to Problem 4SP

When the mass is $6\text{cm}$ from the equilibrium position, the potential energy is $0.72\text{J}$, kinetic energy is $2.16\text{J}$ and the velocity is $3.79\text{J}$.

Explanation of Solution

Given info: The mass attached to spring is $0.30\text{kg}$, the maximum kinetic energy is $2.88\text{J}$, the spring constant is $400\text{N/m}$ and the distance from the equilibrium position is $6\text{cm}$.

Write the expression for the elastic potential energy of a spring.

$PE=\frac{1}{2}k{x}^2$

Substitute $400\text{N/m}$ for $k$ and $6\text{cm}$ for $x$ to find the potential energy $PE$.

$\begin{array}{c}PE=\frac{1}{2}\left(400\text{N/m}\right){\left(6\text{cm}\right)}^2\\ =\frac{1}{2}\left(400\text{N/m}\right){\left(6\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\\ =0.72\text{J}\end{array}$

At a particular distance away from the equilibrium position, the system acquires some potential energy. Hence the kinetic energy at a particular position is obtained by subtracting the potential energy at that point from the maximum kinetic energy.

Thus, the kinetic energy at the given point is obtained as,

$\begin{array}{c}KE=2.88\text{J}-0.72\text{J}\\ =2.16\text{J}\end{array}$

Write the expression for velocity from the expression for kinetic energy.

$v=\sqrt{\frac{2\left(KE\right)}{m}}$

Substitute $2.16\text{J}$ for $KE$ and $0.30\text{kg}$ for $m$ to find the velocity at the given position.

$\begin{array}{c}v=\sqrt{\frac{2\left(2.16\text{J}\right)}{0.30\text{kg}}}\\ =3.79\text{m/s}\end{array}$

Conclusion:

Therefore, when the mass is $6\text{cm}$ from the equilibrium position, the potential energy is $0.72\text{J}$, kinetic energy is $2.16\text{J}$ and the velocity is $3.79\text{J}$.

To determine

The ratio of the values of velocity of the mass at the equilibrium position and the position $6\text{cm}$ away from the equilibrium position.

Answer to Problem 4SP

The ratio of the values of velocity of the mass at the position $6\text{cm}$ away from the equilibrium position and that at equilibrium position is $0.87$.

Explanation of Solution

Given info: The velocity of the mass at the equilibrium position is $4.38\text{m/s}$ and that at the position $6\text{cm}$ away from the equilibrium position is $3.79\text{m/s}$.

The ratio of the values of velocity of the mass at the position $6\text{cm}$ away from the equilibrium position and that at equilibrium position is obtained as;

$\frac{3.79}{4.38}=0.87$

Thus, the velocity of the mass at the position $6\text{cm}$ away from the equilibrium position is nearly $87\%$ of that at equilibrium position.

Conclusion:

Therefore, the ratio of the values of velocity of the mass at the position $6\text{cm}$ away from the equilibrium position and that at equilibrium position is $0.87$.