Question

An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.
(a) Calculate the amplitude of the motion.
_____ m
(b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]
______ m/s

Answer 1

Concepts and reason

Use the concept of simple harmonic motion and conservation of energy to solve this problem.

First, apply the conservation of energy to the spring-mass system to obtain the amplitude of motion.

Later, find the maximum velocity attained by the object using the conservation of energy concept.

Fundamentals

The formula for the potential energy of the spring is as follows:

$PE = \frac{1}{2}k{x^2}$ …… (1)

Here, k is the spring constant, x is the displacement.

The kinetic energy of the mass attached to a spring is as follows:

$KE = \frac{1}{2}m{v^2}$ …… (2)

Here, m is the mass of the object, and v is the velocity of the mass.

The expression for the maximum potential energy of the spring is as follows:

${\left( {P{E_{Spring}}} \right)_{\max }} = \frac{1}{2}kx_{\max }^2$ …… (3)

Here, ${\left( {P{E_{Spring}}} \right)_{\max }}$ is the maximum potential energy of the spring, and ${x_{\max }}$ is the maximum displacement.

(a)

From the conservation of energy;

${\left( {P{E_{Spring}}} \right)_{\max }} = PE + KE$

Substitute the equations (1), (2), and (3) in the above equation.

$\frac{1}{2}kx_{\max }^2 = \frac{1}{2}k{x^2} + \frac{1}{2}m{v^2}$

Rearrange the above equation for ${x_{\max }}$ as follows:

$\begin{array}{c}\\x_{\max }^2 = \frac{{k{x^2} + m{v^2}}}{k}\\\\{x_{\max }} = \sqrt {\frac{{k{x^2} + m{v^2}}}{k}} \\\end{array}$

Substitute 270 N/m for k, 0.020 m for x, 3.5 kg for m, and 0.55 m/s for v.

$\begin{array}{c}\\{x_{\max }} = \sqrt {\frac{{\left( {270\;{\rm{N/m}}} \right){{\left( {0.020\;{\rm{m}}} \right)}^2} + \left( {3.5\;{\rm{kg}}} \right){{\left( {0.55\;{\rm{m/s}}} \right)}^2}}}{{\left( {270\;{\rm{N/m}}} \right)}}} \\\\ = 0.066\;{\rm{m}}\\\end{array}$

(b)

From the conservation of energy;

$\begin{array}{c}\\\frac{1}{2}k{x^2} = \frac{1}{2}m{v^2}\\\\k{x^2} = m{v^2}\\\end{array}$

Substitute ${v_{\max }}$ for v and ${x_{\max }}$ for x.

$kx_{\max }^2 = mv_{\max }^2$

Here, ${v_{\max }}$ is the maximum velocity and ${x_{\max }}$ is the maximum displacement.

Rearrange the above equation for ${v_{\max }}$ , to obtain the maximum velocity gained by the object.

${v_{\max }} = \sqrt {\frac{{kx_{\max }^2}}{m}}$

Substitute 270 N/m for k, 0.066 m for ${x_{\max }}$ , and 3.5 kg for m.

$\begin{array}{c}\\{v_{\max }} = \sqrt {\frac{{\left( {270\;{\rm{N/m}}} \right){{\left( {0.066\;{\rm{m}}} \right)}^2}}}{{3.5\;{\rm{kg}}}}} \\\\ = 0.58\;{\rm{m/s}}\\\end{array}$

Ans: Part a

The amplitude of the motion of the spring is 0.066 m.

Part b

The maximum velocity attained by the object is 0.58 m/s.