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A \(1.35 \mathrm{~kg}\) object is attached to a horizontal spring of force constant \(2.5 \mathrm{~N} / \mathrm{cm}\) and is started oscillating by pulling it \(6.0 \mathrm{~cm}\) from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only \(3.5 \mathrm{~cm}\). (a) How much energy has this system lost to damping during these eight cycles? (b) Where did the "lost" energy go? Explain physically how the system could have lost energy.

Short Answer

Expert verified
Approximately 0.30 J of energy is lost to damping, likely as heat due to resistive forces.

Step by step solution

01

Calculating Initial Potential Energy

To calculate the initial potential energy when the object is pulled to 6.0 cm from its equilibrium, we use the formula for the potential energy stored in a spring: \( U = \frac{1}{2} k x^2 \). First, convert the spring constant from N/cm to N/m: \( k = 2.5 \text{ N/cm} = 250 \text{ N/m} \). The initial displacement \( x = 6.0 \text{ cm} = 0.06 \text{ m} \). Therefore, the initial energy \( U_0 = \frac{1}{2} \times 250 \times 0.06^2 = 0.45 \text{ J} \).
02

Calculating Final Potential Energy

Similarly, calculate the potential energy after eight cycles when the maximum displacement reduces to 3.5 cm. Convert this displacement to meters: \( x = 3.5 \text{ cm} = 0.035 \text{ m} \). The final potential energy \( U_f = \frac{1}{2} \times 250 \times 0.035^2 = 0.153125 \text{ J} \).
03

Calculating Energy Lost

The energy lost due to damping during the eight cycles is the difference between the initial and final potential energies. Therefore, the energy lost \( \Delta U = U_0 - U_f = 0.45 \text{ J} - 0.153125 \text{ J} = 0.296875 \text{ J} \approx 0.30 \text{ J} \).
04

Understanding Energy Loss Mechanics

In a damped oscillating system, energy is not conserved as it is gradually transferred to other forms such as thermal energy due to air resistance or internal friction. The lost energy as calculated, \( 0.30 \text{ J} \), is likely dissipated as heat due to such resistive forces acting against motion.

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Key Concepts

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

Spring Constant
The spring constant, often denoted by the symbol \( k \), is a crucial aspect in the study of oscillations and springs. It defines the stiffness of a spring. Essentially, it tells us how much force is needed to stretch or compress a spring by a unit of length. In the given exercise, the spring constant has been provided as \( 2.5 \text{ N/cm} \). For ease of calculations in the international system of units, we convert it to \( 250 \text{ N/m} \) by recognizing that 1 cm is equivalent to 0.01 m.

This transformed value explains how rigid or easily stretchable the spring is. A larger spring constant implies a stiffer spring that requires more force for the same amount of displacement compared to a spring with a smaller constant. Understanding \( k \) is foundational to predicting how a spring will behave under different loads and computing the potential energy stored in it.
Potential Energy Calculation
Potential energy in the context of a spring, often denoted as \( U \), is the energy stored due to its position or configuration. In mechanical terms, it's the energy held when the spring is either stretched or compressed from its equilibrium position. The formula to calculate this potential energy is: \( U = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

Initially, when the object attached to the spring is pulled 6.0 cm (or 0.06 m) away from the equilibrium, the potential energy is calculated using the spring constant \( 250 \text{ N/m} \). This results in an initial potential energy of \( 0.45 \text{ J} \).

After eight cycles of oscillation, the maximum displacement decreases to 3.5 cm (or 0.035 m), recalculating the potential energy to \( 0.153125 \text{ J} \). The observed decrease in potential energy hints at the system's energy loss over the cycles due to damping.
Energy Dissipation
Energy dissipation in a system refers to how energy is transformed from one form into another lost form, commonly thermal energy through mechanisms like friction or air resistance. In the phenomenon of a damped harmonic oscillation, energy is not fully conserved. Over time, some of it is transferred from mechanical forms to heat, mainly due to the resistive forces working against the motion, such as air drag or intrinsic friction in materials.

In our system, the computation shows an initial potential energy of \( 0.45 \text{ J} \) reducing to \( 0.153125 \text{ J} \) after eight oscillation cycles. This results in an energy loss of approximately \( 0.30 \text{ J} \). Physically, this lost energy likely turns into heat due to air friction or internal damping within the spring and mass setup, illustrating a real-world example of energy conservation principles not applying exactly due to non-ideal conditions.
Oscillation Cycles
Oscillation cycles represent the repeated movements back and forth around an equilibrium position, typical to systems that exhibit harmonic motion. Each cycle comprises a complete to-and-fro movement. In a damped harmonic oscillator, despite cycles occurring consistently, the amplitude reduces gradually with each cycle due to energy being siphoned off, mostly as heat.

In the given scenario, over eight cycles, the oscillation's maximum displacement was observed to shrink from 6.0 cm to 3.5 cm. This reduction conclusively indicates energy dissipation within the oscillating system. The concept of cycles is not only pivotal in defining the period and frequency of the oscillation but also helps discern the rate of energy loss due to damping forces acting over time.

Understanding oscillation cycles helps in grasping the dynamics of different systems—be it mechanical, electrical, or even biological—that resonate and evolve similarly over time.

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Most popular questions from this chapter

Weighing astronauts. In order to study the long-term effects of weightlessness, astronauts in space must be weighed (or at least "massed"). One way in which this is done is to seat them in a chair of known mass attached to a spring of known force constant and measure the period of the oscillations of this system. If the \(35.4 \mathrm{~kg}\) chair alone oscillates with a period of \(1.25 \mathrm{~s}\), and the period with the astronaut sitting in the chair is \(2.23 \mathrm{~s},\) find (a) the force constant of the spring and (b) the mass of the astronaut.

You've made a simple pendulum with a length of \(1.55 \mathrm{~m}\), and you also have a (very light) spring with force constant \(2.45 \mathrm{~N} / \mathrm{m} .\) What mass should you add to the spring so that its period will be the same as that of your pendulum?

Effect of diving on blood. It is reasonable to assume that the bulk modulus of blood is about the same as that of water ( \(2.2 \mathrm{GPa}\) ). As one goes deeper and deeper in the ocean, the pressure increases by \(1.0 \times 10^{4} \mathrm{~Pa}\) for every meter below the surface. (a) If a diver goes down \(33 \mathrm{~m}\) (a bit over \(100 \mathrm{ft}\) ) in the ocean, by how much does each cubic centimeter of her blood change in volume? (b) How deep must a diver go so that each drop of blood compresses to half its volume at the surface? Is the ocean deep enough to have this effect on the diver?

If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose it is displaced \(0.120 \mathrm{~m}\) from its equilibrium position and released with zero initial speed. After \(0.800 \mathrm{~s}\), its displacement is found to be \(0.120 \mathrm{~m}\) on the opposite side and it has passed the equilibrium position once during this interval. Find (a) the amplitude, (b) the period, and (c) the frequency of the motion.

A \(0.500 \mathrm{~kg}\) glider on an air track is attached to the end of an ideal spring with force constant \(450 \mathrm{~N} / \mathrm{m} ;\) it undergoes simple harmonic motion 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 \(x=-0.015 \mathrm{~m},\) and (e) the total mechanical energy of the glider at any point in its motion.

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