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Chapter 14: Problem 65

An object with mass \(m\) is moving in SHM. It has amplitude \(A_{1}\) and total mechanical energy \(E_{1}\) when the spring has force constant \(k_{1}\) You want to quadruple the total mechanical energy, so \(E_{2}=4 E_{1}\), and halve the amplitude, so \(A_{2}=A_{1} / 2,\) by using a different spring, one with force constant \(k_{2}\). (a) How is \(k_{2}\) related to \(k_{1}\) ? (b) What effect will the change in spring constant and amplitude have on the maximum speed of the moving object?

Short Answer

Expert verified
The spring constant \(k_{2}\) of the second spring is eight times that of the initial spring \(k_{1}\), that is \(k_{2} = 8k_{1}\). Despite quadrupling the total mechanical energy and halving the amplitude, the maximum speed of the moving object remains unaffected.

Step by step solution

01

- Understand Terms

Here, we consider an object of mass m in SHM with spring constants \(k_1\) and \(k_2\), and amplitudes \(A_1\) and \(A_2\). The total mechanical energy with the first spring is \(E_1\) and with the second spring is \(E_2=4 E_1\). The object's amplitude changes from \(A_1\) to \(A_2 = A_1 / 2\). We firstly need to identify the relationship between \(k_1\) and \(k_2\) and afterwards recognize the impact of the alterations on the maximum speed of the object.
02

- Relate \(k_2\) and \(k_1\)

The total mechanical energy of a SHM system is given by the formula: \(E_{1} = \frac{1}{2}k_{1}A_{1}^{2}\) and \(E_{2} = \frac{1}{2}k_{2}A_{2}^{2}\), but given that \(E_{2} = 4E_{1}\) and \(A_{2} = \frac{A_{1}}{2}\), substitution these gives \(\frac{1}{2}k_{2}(\frac{A_{1}}{2})^{2} = 4(\frac{1}{2}k_{1}A_{1}^{2})\). Simplifying gives the relation between \(k_{1}\) and \(k_{2}\) as \(k_{2} = 8k_{1}\)
03

- Effect on Maximum Speed

The maximum speed (V_max) in SHM can be obtained from the formula: \(V_{max} = \sqrt{k/m} * A\). With \(k_{2} = 8k_{1}\) and \(A_{2} = \frac{A_{1}}{2}\), we have \(V_{max_2} = \sqrt{8k_{1}/m} * (\frac{A_{1}}{2})\), and \(V_{max_1} = \sqrt{k_{1}/m} * A_{1}\). Cancelling out obvious equalities from both equations gives that \(V_{max_2} = V_{max_1}\). Therefore, the changes do not affect the maximum speed

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Key Concepts

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

Mechanical Energy
Mechanical energy in simple harmonic motion (SHM) is an essential concept that encapsulates both potential and kinetic energy. The total mechanical energy \(E\) of a system in SHM is a constant, assuming no energy losses due to friction or other forces. This energy can be described by the formula \(E = \frac{1}{2}kA^2\), where \(k\) is the spring constant and \(A\) is the amplitude of the motion.
  • Potential Energy: Stored in the spring when it is compressed or stretched to its amplitude.
  • Kinetic Energy: Is highest when the object passes through the equilibrium position.
Understanding mechanical energy is crucial because it allows us to predict and calculate the behavior of the oscillating system. In the exercise, when the energy is quadrupled, this indicates a significant change in the energy balance, affecting how the system can be manipulated and controlled.
Amplitude
Amplitude, represented by \(A\) in the context of simple harmonic motion, is the maximum extent of the oscillation measured from its equilibrium position. It is the distance from the rest position to the peak of the wave.
  • Physical Interpretation: Amplitude tells us the maximum displacement the oscillating object reaches from its central position.
  • Change Impact: In the problem, decreasing the amplitude by half, from \(A_1\) to \(A_2 = A_1 / 2\), significantly influences the potential energy stored in the spring.
This reduction directly affects the system's dynamics and requirements for changing the spring constant to achieve desired mechanical energy levels.
Spring Constant
The spring constant \(k\) is a measure of a spring's stiffness. It plays a pivotal role in determining the system's resonant frequency and the total mechanical energy of a harmonic oscillator. The greater the spring constant, the stiffer the spring, which means it requires more force to stretch or compress it.
  • Formula Exploration: The relation between the spring constant and energy in SHM is \(E = \frac{1}{2}kA^2\).
  • In Exercise: We found that quadrupling the total mechanical energy while halving the amplitude results in \(k_2 = 8k_1\). This means the second spring is much stiffer to accommodate the same energy conditions under a reduced amplitude.
Replacing the spring with a substantially stiffer one in light of these energy changes and amplitude reductions helps maintain proper motion characteristics.
Maximum Speed
In simple harmonic motion, maximum speed \(V_{max}\) occurs as the object passes through the equilibrium position, where its kinetic energy is at its peak while potential energy is zero. The speed can be calculated using the formula \(V_{max} = \sqrt{k/m} \cdot A\).
  • Velocity's Dependence: Directly proportional to both the amplitude and the square root of the spring constant and inversely proportional to the square root of the mass.
  • Exercise Insight: Despite changes in amplitude and spring constant, the calculated maximum speed remained the same. This occurs because the decrease in amplitude is offset by the increase in spring constant resulting in \(V_{max_2} = V_{max_1}\).
This constancy in maximum speed exemplifies the intricacy and balance of forces within harmonic motion systems.

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Most popular questions from this chapter

An object is undergoing SHM with period \(1.200 \mathrm{~s}\) and amplitude \(0.600 \mathrm{~m} .\) At \(t=0\) the object is at \(x=0\) and is moving in the negative \(x\) -direction. How far is the object from the equilibrium position when \(t=0.480 \mathrm{~s} ?\)

A sinusoidally varying driving force is applied to a damped harmonic oscillator of force constant \(k\) and mass \(m .\) If the damping constant has a value \(b_{1},\) the amplitude is \(A_{1}\) when the driving angular frequency equals \(\sqrt{k / m}\). In terms of \(A_{1}\), what is the amplitude for the same driving frequency and the same driving force amplitude \(F_{\max },\) if the damping constant is (a) \(3 b_{1}\) and (b) \(b_{1} / 2 ?\)

BIO "Seeing" Surfaces at the Nanoscale. One technique for making images of surfaces at the nanometer scale, including membranes and biomolecules, is dynamic atomic force microscopy. In this technique, a small tip is attached to a cantilever, which is a flexible, rectangular slab supported at one end, like a diving board. The cantilever vibrates, so the tip moves up and down in simple harmonic motion. In one operating mode, the resonant frequency for a cantilever with force constant \(k=1000 \mathrm{~N} / \mathrm{m}\) is \(100 \mathrm{kHz}\). As the oscillating tip is brought within a few nanometers of the surface of a sample (as shown in the figure), it experiences an attractive force from the surface. For an oscillation with a small amplitude (typically, \(0.050 \mathrm{nm}),\) the force \(F\) that the sample surface exerts on the tip varies linearly with the displacement \(x\) of the tip, \(|F|=k_{\text {surf }} x,\) where \(k_{\text {surf }}\) is the effective force constant for this force. The net force on the tip is therefore \(\left(k+k_{\text {surf }}\right) x\), and the frequency of the oscillation changes slightly due to the interaction with the surface. Measurements of the frequency as the tip moves over different parts of the sample's surface can provide information about the sample.

A block with mass \(m=0.300 \mathrm{~kg}\) is attached to one end of an ideal spring and moves on a horizontal friction less surface. The other end of the spring is attached to a wall. When the block is at \(x=+0.240 \mathrm{~m},\) its acceleration is \(a_{x}=-12.0 \mathrm{~m} / \mathrm{s}^{2}\) and its velocity is \(v_{x}=+4.00 \mathrm{~m} / \mathrm{s} .\) What are (a) the spring's force constant \(k ;\) (b) the amplitude of the motion; (c) the maximum speed of the block during its motion; and (d) the maximum magnitude of the block's acceleration during its motion?

The Silently Ringing Bell. A large, \(34.0 \mathrm{~kg}\) bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of mass is \(0.60 \mathrm{~m}\) below the pivot. The bell's moment of inertia about an axis at the pivot is \(18.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The clapper is a small, \(1.8 \mathrm{~kg}\) mass attached to one end of a slender rod of length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently - that is, for the period of oscillation for the bell to equal that of the clapper?
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