/*! 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 14 A light elastic spring \(\mathrm... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 14

A light elastic spring \(\mathrm{AB}\) of natural length \(b\) and modulus \(2 m g\) is secured to the floor at A. A light elastic string \(\mathrm{BC}\) of natural length \(4 b\) and modulus \(m g\) is attached to the spring at \(\mathrm{B}\) and to a point \(\mathrm{C}\) vertically above \(\mathrm{A}\), where \(\mathrm{AC}=5 \mathrm{~b}\). When a particle of mass \(m\) is attached at B, find: (a) the depth below \(\mathrm{C}\) of its position of equilibrium, (b) the period of its small vertical oscillations about the position of equilibrium.

Short Answer

Expert verified
The depth of equilibrium is \(-\frac{b}{7}\). The period of small oscillations is \( \frac{2\pi}{3}\sqrt{\frac{4b}{g}} \).

Step by step solution

01

Analyze Forces at Equilibrium

First, draw a free-body diagram for the particle at B in equilibrium. Since it is in equilibrium, the total force acting on it must be zero. The forces acting on the particle are the gravitational force, the tension in the spring AB, and the tension in the string BC.
02

Calculating the Stretch in Spring AB

Let the depth below C of the equilibrium position be denoted by x. The spring AB, having a natural length b and modulus 2mg, will stretch by some amount to a length b + x. The extension in the spring AB is x, and the tension in it is given by Hooke's Law: \[\text{Tension in AB} = 2mg \times \frac{x}{b}\]
03

Calculating the Stretch in String BC

The string BC has a natural length 4b and modulus mg. When the depth below C is x, the length of BC becomes 9b - x. The extension in BC is \(9b - x - 4b = 5b - x\). Thus, the tension in it is: \[\text{Tension in BC} = mg \times \frac{5b - x}{4b}\]
04

Set Up the Equilibrium Condition

At equilibrium, the upward forces must equal the downward gravitational force (mg). Therefore, setting up the equilibrium equation: \[T_{AB} + T_{BC} = mg\] Substituting the tensions, we get: \[2mg \times \frac{x}{b} + mg \times \frac{5b - x}{4b} = mg\]
05

Solve for the Depth x

Simplify the equation from the previous step to find x: \[\frac{2mg x}{b} + \frac{mg (5b - x)}{4b} = mg\] Solve this equation to find the value of x: \[\frac{8x}{4b} + \frac{5b - x}{4b} = 1\] \[8x + 5b - x = 4b\] \[7x = -b\] Therefore, \[x = -\frac{b}{7}\]
06

Determine the Oscillation Period

For vertical oscillations, the effective spring constant k is necessary. The effective spring constant is the sum of the spring constants of AB and BC: \[k = k_{AB} + k_{BC}\] where \[k_{AB} = \frac{2mg}{b}, \quad k_{BC} = \frac{mg}{4b}\] \[k = \frac{2mg}{b} + \frac{mg}{4b} = \frac{8mg + mg}{4b} = \frac{9mg}{4b}\]
07

Period of Vertical Oscillations

The period T of small vertical oscillations around the equilibrium point is given by \[T = 2\pi \sqrt{\frac{m}{k}}\] Substitute k into this equation: \[T = 2\pi \sqrt{\frac{4bm}{9mg}} = 2\pi \sqrt{\frac{4b}{9g}} = \frac{2\pi}{3}\sqrt{\frac{4b}{g}}\]

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 91影视!

Key Concepts

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

equilibrium position
In the context of physics, the equilibrium position is where all forces acting on a particle or system are balanced, resulting in no net force. For the particle at B in our problem, gravity pulls it downward while tensions in the spring and string pull upward. At equilibrium, these forces equalize.
To find this position, we balance the gravitational force (mg) with the tensions in the spring (AB) and string (BC). Based on Hooke's Law, these tensions relate to their respective extensions. Importantly, the equilibrium happens at the point where the forces sum to zero, known as the system's resting state.
Hooke's Law
Hooke's Law is a principle in physics that states the force needed to extend or compress a spring by some distance (x) is proportional to that distance. Mathematically, it's expressed as:
\[ F = -kx \]
where:
  • \( F \) is the force applied
  • \( k \) is the spring constant
  • \( x \) is the extension or compression of the spring
A spring's or string's tension arises from this law, which we use in our exercise to calculate tensions in spring AB and string BC.
Using the formulas derived from Hooke's Law, we find the contributions of each to the equilibrium condition.
oscillation period
The oscillation period (T) is the time a system takes to complete one full cycle of oscillation. For a mass-spring system like ours, the period depends on the mass (m) and the effective spring constant (k). It's given by:
\[T = 2\pi \sqrt{\frac{m}{k}}\]
To determine T, we first identify the effective spring constant which, in our exercise, combines the constants for AB and BC. Once we have it, we plug values into the formula. The periodicity reveals how often and quickly the system returns to its equilibrium position, showcasing oscillatory motion's predictability.
spring constant
The spring constant (k) defines a spring's stiffness, indicating how much force is needed for a unit displacement. Higher values of k represent stiffer springs.
For our problem, we calculate the constants for AB and BC as follows:
  • \( k_{AB} = \frac{2mg}{b} \)
  • \( k_{BC} = \frac{mg}{4b} \)
Summing these gives the effective spring constant:
\( k = \frac{9mg}{4b} \)
Understanding k helps predict the system's response to forces. In oscillations, it combines with the mass to affect the period, encoding the system's intrinsic properties.
tension in spring
The tension in a spring is a force that acts to restore the spring to its natural length. For spring AB in our exercise, the tension is calculated using Hooke's Law:
\[ \text{Tension in AB} = 2mg \frac{x}{b} \]
Similarly, for string BC:
\[ \text{Tension in BC} = mg \frac{5b - x}{4b} \]
These tensions arise due to extensions in AB and BC from natural lengths. Their proper calculation is crucial for determining the equilibrium and, subsequently, the system's behavior under oscillations.
When summed, these tensions balance out the gravitational pull, aligning perfectly to establish equilibrium and leading us to deeper insights into the system mechanics.

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

A light elastic spring, of modulus \(8 m g\) and natural length \(l\), has one end attached to a ceiling and carries a scale pan of mass \(m\) at the other end. The scale pan is given a vertical displacement from its equilibrium position and released to oscillate with period \(T\). Prove that $$ T=2 \pi \sqrt{\left(\frac{l}{8 g}\right)} $$ A weight of mass \(k m\) is placed in the scale pan and from the new equilibrium position the procedure is repeated. The period of oscillation is now \(2 T\). Find the value of \(k\). Find also the maximum amplitude of the latter oscillations if the weight and the scale pan do not separate during the motion.

One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(k m g\), is attached at a fixed point on a frictionless plane inclined at an angle \(\theta\) to the horizontal. A heavy particle is atiached to the other end of the string. The particle is at rest on the plane with the string along a line of greatest slope and extended by a length \(b\). The particle is then pulled down a distance \(d\) in the line of the string and released. Show that the period of the simple harmonic motion with which the particle starts to move is independent of \(\theta\). If \(d=2 b\), find the time from release to the string going slack and find also the speed of the particle at the instant when the string goes slack.

A particle P moves with an acceleration which is proportional to its distance from a fixed point \(C\), and is always directed towards \(C\). If the particle starts from rest at a point A distant \(d\) from \(C\), where the magnitude of its acceleration. is \(\lambda^{2} d\), find an expression for the velocity of \(\mathrm{P}\) when it is distant \(x\) from \(\mathrm{C}\). After what time does P next come to instantaneous rest? In what time does \(\mathrm{P}\) travel from \(\mathrm{A}\) to a point \(\mathrm{Q}\) where \(\mathrm{AQ}=\frac{d}{2} ?\)

A particle which is oscillating is not necessarily performing S.H.M.

A particle P describes S.H.M. of amplitude \(1 \mathrm{~m}\). In performing one complete oscillation, P travels a distance: (a) \(2 \mathrm{~m}\) (b) 0 (c) \(4 \mathrm{~m}\) (d) \(-2 \mathrm{~m}\).
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation
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.