/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 An apple weighs 1.00 \(\mathrm{N... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An apple weighs 1.00 \(\mathrm{N}\) . When you hang it from the end of a long spring of force constant 1.50 \(\mathrm{N} / \mathrm{m}\) and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back- and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed?

Short Answer

Expert verified
The unstretched length of the spring is 2.00 meters.

Step by step solution

01

Determine the spring displacement

First, we need to calculate how much the spring stretches due to the weight of the apple. The force exerted by the apple is equal to its weight, which is given by \( F = 1.00 \, \mathrm{N} \). Using Hooke's Law, \( F = kx \), where \( k = 1.50 \, \mathrm{N}/\mathrm{m} \) is the spring constant, we can solve for the displacement \( x \): \( x = \frac{F}{k} = \frac{1.00}{1.50} = 0.67 \, \mathrm{m} \).
02

Calculate the frequency of bouncing (SHM)

The frequency of simple harmonic motion (SHM) for a mass-spring system is given by \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). We know the weight of the apple \( W = 1.00 \, \mathrm{N} \) and \( W = mg \), so \( m = \frac{W}{g} = \frac{1.00}{9.8} \approx 0.102 \, \mathrm{kg} \). Substitute \( k = 1.50 \, \mathrm{N}/\mathrm{m} \) and \( m \): \[ f = \frac{1}{2\pi} \sqrt{\frac{1.50}{0.102}} \approx 0.61 \, \mathrm{Hz} \].
03

Determine the pendulum frequency relation

The problem states the pendulum frequency is half of the bounce frequency (SHM frequency). Therefore, the frequency of the pendulum, \( f_p \), is:\[ f_p = \frac{f}{2} = \frac{0.61}{2} \approx 0.305 \, \mathrm{Hz} \].
04

Calculate the effective length of the pendulum

The frequency formula for a simple pendulum is \( f_p = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \). We want to find \( L \), the effective length:\[ f_p = \frac{1}{2\pi} \sqrt{\frac{9.8}{L}} = 0.305 \].Solving for \( L \), we square both sides, \[ (2\pi \times 0.305)^2 = \frac{9.8}{L} \].Calculate:\[ L = \frac{9.8}{(2\pi \times 0.305)^2} \approx 2.67 \, \mathrm{m} \].
05

Determine the unstretched length of the spring

The effective length of the pendulum (i.e., \( L = 2.67 \, \mathrm{m} \)) corresponds to the full stretched length of the spring. We previously calculated the spring displacement as \( 0.67 \, \mathrm{m} \). The unstretched length, \( L_0 \), is thus:\[ L_0 = L - x = 2.67 - 0.67 = 2.00 \, \mathrm{m} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hooke's Law
Hooke's Law is a fundamental principle in physics that explains how springs work. It states that the force needed to extend or compress a spring is directly proportional to the distance it is stretched or compressed. This relationship can be written as \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position.

This principle helps us understand why a spring stretches when a weight is hung from it. In the original exercise with the apple, the spring stretches due to the weight of the apple it holds, consistent with Hooke's Law. By using the equation \( x = \frac{F}{k} \), we can calculate that the apple, exerting a force of 1.00 N, stretches the spring by 0.67 m. This understanding is crucial for solving problems that involve spring displacement due to weight or force.
Pendulum
A pendulum is a simple mechanical system consisting of a mass (called a bob) hanging from a pivot. It swings back and forth under the influence of gravity. When the pendulum swings through small angles, its motion closely resembles that of simple harmonic motion (SHM). This similarity makes it useful for exploring concepts in physics like periodic motion and frequency.

For the apple in the exercise, when it stops bouncing and swings side to side, it moves like a simple pendulum. In such a setup, its frequency of swinging is determined by the length of the pendulum and acceleration due to gravity. Interestingly, the frequency of a pendulum is half of the bouncing frequency from the spring motion, as specified in the exercise. Understanding this relationship is important for calculating the pendulum's effective length.
Spring Constant
The spring constant, denoted by \( k \), describes the stiffness of a spring. It tells us how resistant a spring is to being compressed or stretched. The larger the spring constant, the stiffer the spring, and more force is required to stretch or compress it by a given amount.

A critical part of the exercise is knowing the spring constant to predict how much the spring will stretch under a specific force. Here, with a spring constant of 1.50 N/m, the spring stretches 0.67 m when a 1.00 N force from the apple is applied. For any given spring system, knowing \( k \) allows us to calculate forces and displacements accurately, and is essential to problems dealing with springs.
Frequency Calculation
Frequency calculation is part of understanding oscillatory motions, such as those seen in springs and pendulums. It tells us how many oscillations occur in a given time period, normally interpreted as hertz (Hz), which means cycles per second.

In the context of the exercise, two types of frequencies are crucial: the frequency of the spring bouncing (SHM) and the pendulum swinging frequency. To find the spring system's frequency, the exercise uses the formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). After identifying that the mass of the apple is approximately 0.102 kg, it is plugged into this formula along with the spring constant to determine the frequency of 0.61 Hz for the bouncing spring.

For the pendulum, the frequency is half of the spring frequency, calculated to be around 0.305 Hz, leveraging its physical pendulum formula \( f_p = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \). Both calculations demonstrate important applications of frequency determination for different oscillating systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a physics lab, you attach a \(0.200-\mathrm{kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring's force constant.

The point of the needle of a sewing machine moves in SHM along the \(x\) -axis with a frequency of 2.5 Hz. At \(t=0\) its position and velocity components are \(+1.1 \mathrm{cm}\) and \(-15 \mathrm{cm} / \mathrm{s},\) respectively. (a) Find the acceleration component of the needle at \(t=0\) . (b) Write equations giving the position, velocity, and acceleration components of the point as a function of time.

BIO Weighing a Virus. In February \(2004,\) scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 30 \(\mathrm{nm}\) long with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{v}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by the formula \(\frac{f_{\mathrm{S}+\mathrm{V}}}{f_{\mathrm{S}}}=\frac{1}{\sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)}}\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?

A 1.50-kg ball and a 2.00 -kg ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant \(165 \mathrm{N} / \mathrm{m},\) and the system is vibrating vertically with amplitude 15.0 \(\mathrm{cm} .\) The glue connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20 \(\mathrm{m} / \mathrm{s}\) to the right and an acceleration of 8.40 \(\mathrm{m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.