Problem 27 - College Physics

Question

A toy is undergoing SHM on the end of a horizontal spring with force constant \(300.0 \mathrm{~N} / \mathrm{m} .\) When the toy is \(0.120 \mathrm{~m}\) from its equilibrium position, it is observed to have a speed of \(3 \mathrm{~m} / \mathrm{s}\) and a total energy of \(4.4 \mathrm{~J}\). Find (a) the mass of the toy, (b) the amplitude of the motion, and (c) the maximum speed attained by the toy during its motion.

Step-by-Step Solution

Answer

(a) Mass = 0.498 kg, (b) Amplitude = 0.171 m, (c) Max speed = 4.2 m/s.

Step 1: Understand the Given Information

We are given a toy oscillating in simple harmonic motion (SHM) with a spring constant \( k = 300.0 \, \text{N/m} \). The toy is displaced \( x = 0.120 \, \text{m} \) from its equilibrium with a speed of \( v = 3.0 \, \text{m/s} \) and total mechanical energy \( E = 4.4 \, \text{J} \).

Step 2: Calculate the Potential Energy

The potential energy \( U \) in SHM for a spring is given by \( U = \frac{1}{2}kx^2 \). Substituting the given values: \[ U = \frac{1}{2} \times 300.0 \, \text{N/m} \times (0.120 \, \text{m})^2 = 2.16 \, \text{J} \].

Step 3: Calculate the Kinetic Energy

The total energy \( E \) is the sum of potential energy \( U \) and kinetic energy \( KE \). Solve for \( KE \) using \( KE = E - U \): \[ KE = 4.4 \, \text{J} - 2.16 \, \text{J} = 2.24 \, \text{J} \].

Step 4: Determine the Mass of the Toy

The kinetic energy in SHM is also given by \( KE = \frac{1}{2}mv^2 \). Solve for mass \( m \) using the known kinetic energy: \[ 2.24 = \frac{1}{2}m(3)^2 \] \[ m = \frac{2 \times 2.24}{3^2} = 0.498 \, \text{kg} \].

Step 5: Calculate the Amplitude of the Motion

The amplitude \( A \) is the maximum displacement where all energy is potential. Using total energy \( E = \frac{1}{2}kA^2 \), solve for \( A \): \[ 4.4 = \frac{1}{2} \times 300.0 \times A^2 \] \[ A = \sqrt{\frac{2 \times 4.4}{300.0}} = 0.171 \, \text{m} \].

Step 6: Determine the Maximum Speed

The maximum speed \( v_{\text{max}} \) occurs when all the energy is kinetic. Using \( KE_{\text{max}} = \frac{1}{2}mv_{\text{max}}^2 = 4.4 \, \text{J} \), solve for \( v_{\text{max}} \): \[ 4.4 = \frac{1}{2} \times 0.498 \times v_{\text{max}}^2 \] \[ v_{\text{max}} = \sqrt{\frac{2 \times 4.4}{0.498}} = 4.2 \, \text{m/s} \].

Key Concepts

Mass Calculation

To find the mass of the toy in simple harmonic motion, we need to use the formula for kinetic energy. The kinetic energy in such a system can be given by \( KE = \frac{1}{2}mv^2 \),where \( m \) is mass and \( v \) is velocity.
In this problem, we already calculated the kinetic energy as \( 2.24 \, \text{J} \) when the toy is moving with a speed of \( 3.0 \, \text{m/s} \).Since the kinetic energy and velocity are known, solving for the mass \( m \) becomes straightforward:\[m = \frac{2 \times KE}{v^2}\]Plug the known values:\[m = \frac{2 \times 2.24}{(3.0)^2} = 0.498 \, \text{kg}\]This calculation shows the importance of understanding relationships in equations in SHM.

Amplitude Determination

Determining the amplitude of motion is crucial in understanding simple harmonic motion.Amplitude \( A \) is the maximum displacement from the equilibrium.At this point, all the energy is potential, expressed as:\( E = \frac{1}{2}kA^2 \).
Here, \( E \) is the total mechanical energy and \( k \) is the spring constant.We know that the total energy \( E \) is \( 4.4 \, \text{J} \), and the spring constant \( k = 300.0 \, \text{N/m} \).Plug these values into the potential energy formula:\[A = \sqrt{\frac{2 \times E}{k}}\]\[A = \sqrt{\frac{2 \times 4.4}{300.0}} = 0.171 \, \text{m}\]This amplitude tells us how far from equilibrium the toy can get in its motion.

Maximum Speed Calculation

In simple harmonic motion, maximum speed is achieved when the potential energy is zero, and all the energy is kinetic.Again, kinetic energy in this scenario is given by:\( KE_{\text{max}} = \frac{1}{2}mv_{\text{max}}^2 \),where \( v_{\text{max}} \) is the maximum velocity.
Knowing the total mechanical energy of the toy is \( 4.4 \, \text{J} \), we can solve for the maximum speed:\[v_{\text{max}} = \sqrt{\frac{2 \times E}{m}}\]\[v_{\text{max}} = \sqrt{\frac{2 \times 4.4}{0.498}} = 4.2 \, \text{m/s}\]Achieving this speed means the toy reaches its highest kinetic phase during oscillation.

Kinetic Energy

Kinetic energy in simple harmonic motion is the energy due to the toy's motion.It's calculated from the toy's mass and velocity using the formula:\( KE = \frac{1}{2}mv^2 \).
In the given exercise, at a displacement of \( 0.120 \, \text{m} \), the kinetic energy was found to be \( 2.24 \, \text{J} \).This is calculated by subtracting the potential energy from the total energy:\[KE = E - U \KE = 4.4 \, \text{J} - 2.16 \, \text{J} = 2.24 \, \text{J}\]This energy indicates how the toy is partitioning its energy between potential and kinetic throughout its oscillation.

Potential Energy

Potential energy in simple harmonic motion is the energy stored in the system when the toy is displaced.It's given by:\( U = \frac{1}{2}kx^2 \),where \( k \) is the spring constant and \( x \) is the displacement.
For the toy at \( 0.120 \, \text{m} \) from equilibrium:\[U = \frac{1}{2} \times 300.0 \, \text{N/m} \times (0.120 \, \text{m})^2 = 2.16 \, \text{J}\]This potential energy is part of the mechanical energy of the system, demonstrating the dynamic exchange between kinetic and potential energies that characterize simple harmonic motion.