91Ó°ÊÓ

CALC A Spring with Mass. The preceding problems in this chapter have assumed that the springs had negligible mass. But of course no spring is completely massless. To find the effect of the spring's mass, consider a spring with mass \(M\) , equilibrium length \(L_{0}\), and spring constant \(k\). When stretched or compressed to a length \(L,\) the potential energy is \(\frac{1}{2} k x^{2},\) where \(x=L-L_{0-}\) (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\) . Assume that the speed of points along the length of the spring varies linearly with distance \(I\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v .\) (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L\) . The result is not \(\frac{1}{2} M v^{2}\) , since not all of the spring moves with the same speed.) (b) Take the time derivative of the conservation of energy equation, Eq. (14.21), for a mass \(m\) moving on the end of a massless. By comparing your results to Eq. (14.8), which defines \(\omega,\) show that the angular frequency of oscillation is \(\omega=\sqrt{k / m}\) (c) Apply the procedure of part (b) to obtain the angular frequency of oscillation \(\omega\) of the spring considered in part (a). If the effective mass \(M^{\prime}\) of the spring is defined by \(\omega=\sqrt{k / M^{\prime}},\) what is \(M^{\prime}\) in terms of \(M ?\)

Step by step solution

01

Set Up the Problem

We need to calculate the kinetic energy of a spring with mass. Given that the speed of points along the spring varies linearly, we will divide the spring into small segments and sum their kinetic energies.

02

Describe Velocity of a Segment

Consider a small segment of the spring at a distance \(l\) from the fixed end. The speed \(v_l\) of this segment can be expressed as \(v_l = \frac{v}{L} l\), where \(v\) is the speed of the free end.

03

Compute Mass of a Segment

The mass of this small segment \(dm\) can be expressed as \(dm = \frac{M}{L} dl\), where \(M\) is the total mass of the spring.

04

Integrate to Find Total Kinetic Energy

The kinetic energy of the small segment is \(dKE = \frac{1}{2} dm \cdot v_l^2\). Substituting expressions for \(v_l\) and \(dm\), we have \[dKE = \frac{1}{2} \cdot \frac{M}{L} \cdot dl \cdot \left(\frac{v}{L} l\right)^2 = \frac{M v^2}{2L^3} l^2 dl\]. Integrate this expression from 0 to \(L\):\[KE = \int_0^L \frac{M v^2}{2L^3} l^2 dl = \frac{M v^2}{2L^3} \cdot \left[ \frac{l^3}{3} \right]_0^L = \frac{M v^2}{2L^3} \cdot \frac{L^3}{3} = \frac{M v^2}{6}.\]

05

Conservation of Energy and Angular Frequency

For a mass \(m\) with a massless spring, the angular frequency is given by \(\omega = \sqrt{\frac{k}{m}}\). In this case, considering a spring with mass, the conservation of energy approach confirms relation to previous cases.

06

Effective Mass of Spring

Given the relations from step 5, if the effective mass \(M'\) satisfies \(\omega = \sqrt{\frac{k}{M'}}\), then by setting equal to kinetic energy expression, we determine \(M' = \frac{M}{3}\).

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.

Kinetic Energy

Kinetic energy in a spring with mass is an interesting topic. Unlike a massless spring, where kinetic energy is straightforward to calculate, a spring with mass requires special considerations. When a spring stretches or compresses, its speed varies along its length. The speed is highest at the free end and zero at the fixed end. This linear speed distribution affects the kinetic energy calculation. To find the kinetic energy, we divide the spring into tiny segments.

• The speed of each segment is determined by its position along the spring.
• The mass of each segment is calculated based on the total mass and length of the spring.

By integrating these small kinetic energy contributions over the spring's length, we find that the total kinetic energy is \[\text{KE} = \frac{M v^2}{6}\], which is less than the kinetic energy for a point mass at speed \(v\). This demonstrates that not all parts of the spring move at the same speed, impacting the kinetic energy.

Angular Frequency

Angular frequency is key when studying oscillations. It defines how quickly an object oscillates around its equilibrium position. For a traditional mass-spring system, the angular frequency \(\omega\) is determined by the spring constant \(k\) and the mass \(m\). The standard formula is:\[\omega = \sqrt{\frac{k}{m}}\]When considering a spring with mass, this formula adapts because the spring itself contributes to the system's inertia. In practical terms:

• Angular frequency can be thought of as the number of oscillation cycles per unit time.
• Incorporating the mass of the spring modifies the traditional mass-spring system calculations.

Calculating angular frequency for a spring with mass involves using the effective mass, as it accounts for the spring’s distribution of mass.

Effective Mass

The effective mass in a spring-mass system is a concept that aids in calculating oscillations. It's an adjustment in the system for when a spring itself has mass, not just the object at the end. Normally, we consider only the mass attached to the spring. However, a massive spring itself also oscillates, behaving as if it contributes additional mass to the system.To determine the effective mass \(M'\), we use the relation for angular frequency with a massive spring:\[\omega = \sqrt{\frac{k}{M'}}\]By comparing it to the system with just an attached mass, it's found that \[M' = \frac{M}{3}\]. Here are some points on effective mass:

• It represents how the mass of the spring influences its oscillation.
• It makes calculations for oscillation dynamics more accurate.

The effective mass helps us better understand the inertia of the combined mass-spring system.

Oscillation

Oscillation refers to the motion repeated back and forth within a system, such as a mass attached to a spring. These oscillations can be affected by various factors, including the spring's own mass. In a spring-mass system:

• The mass at the end moves up and down or back and forth, repeating in cycles.
• The time for one full cycle of motion is called the period.
• Oscillations are typically described by their frequency and amplitude.

Adding the spring’s mass into the equation influences these properties, causing changes in the dynamic behavior of the spring-mass system. These influence how the spring-mass system returns to its initial state after its motion is interrupted. Understanding oscillation in this context is crucial for accurately predicting system behavior.