91Ó°ÊÓ

You are given the task of opening an antiquated "light lock," which is unlocked by shining red light pulses of a certain frequency for a long duration of time into a light sensor on the lock. You are given a red laser pointer, a spring of unstretched length \(L=15.0 \mathrm{cm}\) and spring constant \(k=7.20 \mathrm{N} / \mathrm{m},\) a sheet of steel \(\left(\rho=7.60 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is 0.125 inches thick, and some tools. You come up with the idea to take a piece of the steel sheet (of mass \(m\) ), cut a slot in it, and hang it from the spring. If you shine the laser through the slot and onto the sensor, and then stretch the spring and let it go, the steel plate will oscillate and cause the beam to pass through the slot periodically. (a) Assuming the beam is passing through the slot (and onto the lock's sensor) when the spring-mass system is in equilibrium, how is the frequency at which the light pulses hit the sensor related to the frequency of the spring/mass (i.e., steel plate) system. (b) Based on your answer for (a), what should be the frequency of the spring/mass system if the unlocking frequency is \(3.50 \mathrm{Hz} ?\) (c) What should be the mass \(m\) of the steel plate? (d) Calculate some reasonable dimensions for the steel plate (i.e., they should be consistent with the mass that is required for the spring-mass system). You may assume the material removed to make the slot in the steel plate is of negligible mass.

Step by step solution

01

Analyze the Problem

We are dealing with a spring-mass system where the steel plate oscillates vertically, activating the light pulses. The equilibrium position is when the steel plate is at rest, and the laser passes through the slot consistently.

02

Determine Frequency Relationship

For a basic harmonic oscillator, the frequency of oscillation is given by the equation \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \) where \( k \) is the spring constant and \( m \) is the mass of the steel plate. The frequency of the light pulses is the same as the oscillation frequency of the system due to this periodic motion.

03

Calculate Required Oscillation Frequency

The desired frequency of light pulses to unlock the lock is given as \( 3.50 \, \text{Hz} \). Thus, the oscillation frequency of the spring-mass system, \( f \), must also be \( 3.50 \, \text{Hz} \).

04

Solve for Mass of the Steel Plate

Using the frequency formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), we solve for mass \( m \):\[m = \frac{k}{(2\pi f)^2}\]Substituting \( k = 7.20 \, \text{N/m} \) and \( f = 3.50 \, \text{Hz} \), we find:\[m = \frac{7.20}{(2\pi \times 3.50)^2} \approx 0.09354 \, \text{kg}\]

05

Calculate Dimensions of Steel Plate

The steel sheet's density is \( \rho = 7.60 \, \text{g/cm}^3 \) converted to \( 7600 \, \text{kg/m}^3 \). Knowing \( m = 0.09354 \, \text{kg} \) and the thickness is \( 0.125 \, \text{inches} \approx 0.003175 \, \text{m} \), we use the volume-mass relationship, \( V = \frac{m}{\rho} \), to find surface area:\[V = \frac{0.09354}{7600} = 1.2313 \times 10^{-5} \, \text{m}^3\]Thus, area \( A = \frac{V}{\text{thickness}} = \frac{1.2313 \times 10^{-5}}{0.003175} \approx 0.003878 \, \text{m}^2 \). Therefore, a reasonable dimension could be a square of about \( \sqrt{0.003878} = 0.0623 \, \text{m} \approx 6.23 \, \text{cm} \) on each side.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant

In the context of the problem, the spring constant (\( k \)) is a measure of how stiff the spring is. A spring constant describes the spring's ability to resist deformation under a force. In simple terms, it tells us how much force is needed to stretch or compress the spring by a certain distance.

For our spring-mass system, the given spring constant is \( 7.20 \, \text{N/m} \), which indicates that the spring is moderately stiff. When a spring constant is higher, it means the spring is stiffer, requiring more force to stretch or compress it by the same amount compared to a spring with a lower spring constant.

Understanding the spring constant is essential in predicting the behavior of a harmonic oscillator, including how quickly the system can oscillate. This directly influences the oscillation frequency of the spring when a mass is attached to it.

Oscillation Frequency

Oscillation frequency is a critical concept in understanding how frequently an oscillating system, like the spring-mass setup in this exercise, completes one full cycle of motion. It's measured in hertz (Hz), where one hertz equals one cycle per second.

In a harmonic oscillator like our spring-mass system, the oscillation frequency is given by the formula:

• \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)

This expression links the spring constant (\( k \)) and the mass of the object (\( m \)) on the spring to the frequency of oscillation. A key aspect of this formula is that as the mass increases, the frequency decreases, and as the spring constant increases, the frequency also increases.

For this problem, we are tasked with tuning the system to have an oscillation frequency of \( 3.50 \, \text{Hz} \), which matches the lock's unlocking frequency.

Mass

Mass in the context of a harmonic oscillator refers to the amount of matter in the object attached to the spring. In our exercise, this is the steel plate cut from the sheet. The mass (\( m \)) affects the oscillation frequency of the system based on \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \).

To achieve the necessary frequency of \( 3.50 \, \text{Hz} \), we need to calculate the precise mass using the given spring constant and desired frequency. By rearranging the frequency formula, \( m = \frac{k}{(2\pi f)^2} \), allows us to determine the mass of the steel plate.

For our system, using \( k = 7.20 \text{N/m} \) and \( f = 3.50 \text{Hz} \), the mass is calculated to be approximately \( 0.09354 \text{kg} \), making it crucial to cut just the right size of steel to meet this requirement.

Density

Density is a material property that represents mass per unit volume. For the steel sheet given in the problem, its density (\( \rho \)) is \( 7.60 \, \text{g/cm}^3 \), which converts to \( 7600 \, \text{kg/m}^3 \).

In this exercise, understanding the density of the steel helps in determining the dimensions of the steel plate that must be cut. Knowing the volume and thickness, we can calculate the required surface area to ensure the correct mass for our spring-mass system.

The relationship of interest is \( V = \frac{m}{\rho} \), where \( V \) is the volume. This relationship helps us calculate that for our specific mass, the volume must be \( 1.2313 \times 10^{-5} \, \text{m}^3 \). Considering the thickness of the steel, we deduced the surface area needed, allowing us to cut a piece of steel with appropriate dimensions to maintain the desired mass and achieve the correct oscillation frequency.