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Chapter 14: Problem 61

Object \(A\) has mass \(m_{A}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{A} .\) Object \(B\) has mass \(m_{B}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{B}\). The amplitude \(A_{A}\) for object \(A\) is twice the amplitude \(A_{B}\) for the motion of object \(B\). Also, \(m_{B}=4 m_{A}\) and \(k_{A}=9 k_{B}\). (a) What is the ratio of the maximum speeds of the two objects, \(v_{\max , A} / v_{\max , B} ?\) (b) What is the ratio of their maximum accelerations, \(a_{\max , A} / a_{\max , B} ?\)

Short Answer

Expert verified
The ratio of the maximum speeds is \(v_{\max , A} / v_{\max , B} = 3/2\) and the ratio of their maximum accelerations is \(a_{\max , A} / a_{\max , B} = 9/2\)

Step by step solution

01

Understanding Simple Harmonic Motion (SHM)

In SHM, the max speed \(v_{\max}\) can be expressed in terms of amplitude \(A\) and angular frequency \(\omega\) by \(v_{\max} = A \omega\). We can replace \(\omega\) with \(\sqrt{k/m}\), resulting in \(v_{\max} = A \sqrt{k/m}\). Similarly, the max acceleration \(a_{\max}\) can be written as \(a_{\max} = A \omega^2\), replacing \(\omega\) with \(\sqrt{k/m}\) gives us \(a_{\max} = A (k/m)\).
02

Finding Ratio of Maximum Speeds

First calculate maximum speed \(v_{\max, A}\) and \(v_{\max, B}\) for both objects using the given information. \(v_{\max, A} = A_A \sqrt{k_A/m_A}\) and \(v_{\max, B} = A_B \sqrt{k_B/m_B}\). Divide \(v_{\max, A}\) by \(v_{\max, B}\) to find the ratio of max speeds \(v_{\max , A} / v_{\max , B} \). Given that \(A_A = 2A_B\), \(k_A = 9k_B\), and \(m_B = 4m_A\), substituting these values reveals that \(v_{\max, A} = 2 \sqrt{9/4} v_{\max, B}\) which simplifies to \(v_{\max , A} / v_{\max , B} = 3/2\)
03

Finding Ratio of Maximum Accelerations

First calculate maximum acceleration \(a_{\max, A}\) and \(a_{\max, B}\) for both objects. \(a_{\max, A} = A_A k_A/m_A\) and \(a_{\max, B} = A_B k_B/m_B\). Dividing \(a_{\max, A}\) by \(a_{\max, B}\) will give us the ratio of max accelerations \(a_{\max , A} / a_{\max , B}\). Given \(A_A = 2A_B\), \(k_A = 9k_B\), and \(m_B = 4m_A\), substituting these values unveils that \(a_{\max, A} = 2 * (9/4) a_{\max, B}\) which simplifies to \(a_{\max , A} / a_{\max , B} = 9/2\)

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Key Concepts

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

Maximum Speed in SHM
When it comes to Simple Harmonic Motion (SHM), the maximum speed of an object is a point of keen interest. This maximum speed, denoted as v_max, occurs when the object passes through the equilibrium position. At this point, all the potential energy stored in the spring (for spring-mass systems) is converted into kinetic energy. The formula to determine this speed is given by v_max = A蝇, where A represents the amplitude and 蝇 (angular frequency) is the rate at which the object oscillates.

Now, angular frequency itself is connected to the physical characteristics of the system, such as the mass of the object m, and the force constant k, of the spring. It's numerically equivalent to 鈭�(办/尘). Hence, substituting for 蝇, we get v_max = A鈭�(办/尘). It's evident from the formula that for higher amplitudes and stiffer springs (high value of k), the velocity peaks at a greater value. Conversely, a greater mass effectively reduces the maximum speed.
Maximum Acceleration in SHM
The maximum acceleration in SHM, or a_max, is another critical aspect that indicates the greatest rate of change of velocity. This acceleration is felt by the object at the points of maximum displacement, where the spring's restoring force is at its peak. The mathematical relationship for maximum acceleration is expressed as a_max = A蝇虏.

Again deriving from the angular frequency, we can rewrite 蝇 as 鈭�(办/尘). Consequently, the formula gets transformed into a_max = A(k/m). The interpretation here is straightforward: an object with a larger amplitude and attached to a stiffer spring experiences a greater maximum acceleration. Notably, increasing the mass of the object, while keeping other factors constant, will lessen the maximum acceleration. The interplay of these variables determines how violently or gently the system oscillates.
Angular Frequency
Angular frequency, denoted by the Greek letter 蝇, is a vital term in describing the motion of SHM. It represents how quickly the object undergoes cyclic motion and is related to the time period T and frequency f of oscillation via the relationships 蝇 = 2蟺/T or 蝇 = 2蟺f. In SHM, the spring force constant k and the mass m define the angular frequency with the equation 蝇 = 鈭�(办/尘).

Stepping into more specific territory, a higher force constant k or a lower mass m means a quicker oscillation, as the object has a higher angular frequency. This concept is central to many phenomena outside of springs and masses 鈥� for example, pendulum clocks, vibrating molecules, and even the frequency of light waves can all be related to angular frequency.
Force Constant
The force constant, symbolized as k, is pivotal in the realm of SHM as it dictates the 'springiness', or rather, the stiffness of a spring. In physical terms, it's the restoring force exerted per unit displacement from equilibrium. In equation form, it is utilized as F = -kx, where F speaks to the restoring force and x is the displacement.

In the specific context of SHM exercises, knowing the force constant allows us to calculate the angular frequency and, as a result, other dynamics of the motion such as maximum speed and acceleration. The larger the force constant, the more force is needed to stretch or compress the spring by some length. This property also serves as a bridge to connect the concepts of energy in the system since the potential energy stored in a compressed or stretched spring is given by PE = (1/2)kx虏. Therefore, a stiffer spring not only causes the object to oscillate faster but also stores more potential energy for a given displacement.

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Most popular questions from this chapter

A \(0.500 \mathrm{~kg}\) glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{~N} / \mathrm{m},\) undergoes \(\mathrm{SHM}\) with an amplitude of \(0.040 \mathrm{~m}\). Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x=-0.015 \mathrm{~m} ;\) (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x=-0.015 \mathrm{~m} ;\) (e) the total mechanical energy of the glider at any point in its motion.

An object is moving in damped SHM, and the damping constant can be varied. If the angular frequency of the motion is \(\omega\) when the damping constant is zero, what is the angular frequency, expressed in terms of \(\omega\), when the damping constant is one-half the critical damping value?

Quantum mechanics is used to describe the vibrational motion of molecules, but analysis using classical physics gives some useful insight. In a classical model the vibrational motion can be treated as SHM of the atoms connected by a spring. The two atoms in a diatomic molecule vibrate about their center of mass, but in the molecule HI, where one atom is much more massive than the other, we can treat the hydrogen atom as oscillating in SHM while the iodine atom remains at rest. (a) A classical estimate of the vibrational frequency is \(f=7 \times 10^{13} \mathrm{~Hz}\). The mass of a hydrogen atom differs little from the mass of a proton. If the HI molecule is modeled as two atoms connected by a spring, what is the force constant of the spring? (b) The vibrational energy of the molecule is measured to be about \(5 \times 10^{-20} \mathrm{~J}\). In the classical model, what is the maximum speed of the H atom during its SHM? (c) What is the amplitude of the vibrational motion? How does your result compare to the equilibrium distance between the two atoms in the HI molecule, which is about \(1.6 \times 10^{-10} \mathrm{~m} ?\)

A block with mass \(M\) rests on a friction less surface and is connected to a horizontal spring of force constant \(k .\) The other end of the spring is attached to a wall (Fig. \(\mathbf{P 1 4 . 6 6 )} .\) A second block with mass \(m\) rests on top of the first block. The coefficient of static friction between the blocks is \(\mu_{\mathrm{s}}\). Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.

A uniform, solid metal disk of mass \(6.50 \mathrm{~kg}\) and diameter \(24.0 \mathrm{~cm}\) hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of \(4.23 \mathrm{~N}\) tangent to the rim of the disk to turn it by \(3.34^{\circ},\) thus twisting the wire. You now remove this force and release the disk from rest. (a) What is the torsion constant for the metal wire? (b) What are the frequency and period of the torsional oscillations of the disk? (c) Write the equation of motion for \(\theta(t)\) for the disk.
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