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Chapter 10: Problem 23

A vertical spring (spring constant \(=112 \mathrm{N} / \mathrm{m}\) ) is mounted on the floor. A 0.400-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

Short Answer

Expert verified
(a) Frequency: 2.66 Hz. (b) Amplitude: 0.035 m.

Step by step solution

01

Understanding the Problem

Given a vertical spring with a spring constant \( k = 112 \, \text{N/m} \) and a block of mass \( m = 0.400 \, \text{kg} \) placed on it, we are required to find the frequency of oscillation and the amplitude at which the block loses contact with the spring.
02

Calculate Frequency of the Motion

The frequency of simple harmonic motion can be determined using the formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). Plug in the given values: \( k = 112 \, \text{N/m} \) and \( m = 0.400 \, \text{kg} \). Calculate:\[f = \frac{1}{2\pi} \sqrt{\frac{112}{0.400}} \approx \frac{1}{2\pi} \times 16.73 \approx 2.66 \, \text{Hz}\]
03

Determine Amplitude for Block Losing Contact

The block loses contact with the spring when the spring force equals the gravitational force, at the highest point of oscillation. The block loses contact when the acceleration \( a_m = g \), where \( g = 9.81 \, \text{m/s}^2 \). The maximum acceleration in SHM is \( a_m = \omega^2 A \).First, find \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{112}{0.400}} = 16.73 \, \text{rad/s} \).Using \( a_m = g \), set \( \omega^2 A = g \) and solve for \( A \):\[16.73^2 A = 9.81 \implies A = \frac{9.81}{280} \approx 0.035 \, \text{m}\]
04

Conclusion

Based on the calculations: (a) The frequency of the oscillating block is approximately 2.66 Hz. (b) The amplitude at which the block loses contact with the spring is approximately 0.035 m.

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Key Concepts

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

Frequency Calculation
When dealing with simple harmonic motion (SHM) of a block-spring system, calculating the frequency is an essential step. Frequency tells us how many oscillations occur in one second and is measured in hertz (Hz).
For our vertical spring system, we use the formula for the frequency of a block-spring system as:
  • \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
Here, \(k\) is the spring constant, which indicates how stiff the spring is, and \(m\) is the mass of the object causing the oscillation.
In this exercise:
  • The spring constant \(k\) is given as 112 N/m, and the block's mass \(m\) is 0.400 kg.
  • Substituting these values into the formula, we get:\[ f = \frac{1}{2\pi} \sqrt{\frac{112}{0.400}} \approx 2.66 \text{ Hz} \]
This calculation means that within a single second, the block completes approximately 2.66 oscillations. Understanding this concept helps us predict the motion of similar harmonic systems.
Spring Constant
The spring constant, often represented as \(k\), is a measure of a spring's stiffness. The higher the spring constant, the stiffer the spring. It is measured in newtons per meter (N/m).
In SHM, the spring constant is crucial because:
  • It affects how much force is needed to stretch or compress the spring by a certain distance.
  • It determines the frequency of oscillation of attached objects, along with the mass of the object.
In our exercise, the spring constant is \(k = 112 \) N/m. This means that 112 newtons of force are needed to stretch the spring by 1 meter.
This stiffness impacts both how quickly the system oscillates and how far the block will travel during its motion. More importantly, understanding the spring constant allows us to explore how different springs will behave under similar conditions.
Amplitude
Amplitude in SHM refers to the maximum distance from the equilibrium position that the oscillating object, like our block, can reach. It is measured in meters (m) and is a vital parameter showing the extent of the block's motion.
For a block losing contact with the spring, the critical amplitude is reached when the spring force no longer supports the block against gravity.
  • This point is determined by setting the maximum acceleration of the block equal to gravitational acceleration (\( g = 9.81 \text{ m/s}^2 \)).
For the exercise:
  • We equate \( a_m = \omega^2 A = g \) to find the time at which the block loses contact.
  • With \(\omega\) calculated as \(16.73 \text{ rad/s}\), solve for \(A\):\[ A = \frac{9.81}{280} \approx 0.035 \text{ m} \]
This tells us that the block will lose contact if the amplitude exceeds 0.035 meters, which aids in underscoring how gravitational forces interplay with oscillatory motion.

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Most popular questions from this chapter

The femur is a bone in the leg whose minimum cross-sectional area is about \(4.0 \times 10^{-4} \mathrm{m}^{2} .\) A compressional force in excess of \(6.8 \times 10^{4} \mathrm{N}\) will fracture this bone. (a) Find the maximum stress that this bone can with- stand. (b) What is the strain that exists under a maximum-stress condition?

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.

In a room that is \(2.44 \mathrm{m}\) high, a spring (unstrained length \(=0.30 \mathrm{m}\) ) hangs from the ceiling. A board whose length is \(1.98 \mathrm{m}\) is attached to the free end of the spring. The board hangs straight down, so that its \(1.98-\mathrm{m}\) length is perpendicular to the floor. The weight of the board \((104 \mathrm{N})\) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?

A vertical ideal spring is mounted on the floor and has a spring constant of \(170 \mathrm{N} / \mathrm{m} .\) A \(0.64-\mathrm{kg}\) block is placed on the spring in two different ways. (a) In one case, the block is placed on the spring and not released until it rests stationary on the spring in its equilibrium position. Determine the amount (magnitude only) by which the spring is compressed. (b) In a second situation, the block is released from rest immediately after being placed on the spring and falls downward until it comes to a momentary halt. Determine the amount (magnitude only) by which the spring is now compressed.

A spring stretches by \(0.018 \mathrm{m}\) when a \(2.8-\mathrm{kg}\) object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0 \mathrm{Hz} ?\)
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