Problem 50 - College Physics

Question

\(\bullet\) 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.

Step-by-Step Solution

Answer

The system lost approximately 0.297 J to damping forces such as air resistance.

Step 1: Convert units

Convert the spring constant from \( \mathrm{N/cm} \) to \( \mathrm{N/m} \). We have \( 2.5\, \mathrm{N/cm} = 250\, \mathrm{N/m} \).

Step 2: Calculate initial energy

The initial energy can be calculated using the formula for potential energy in a spring: \[ E_i = \frac{1}{2} k x^2 \] where \( k = 250 \) N/m and \( x = 0.06 \) m (6.0 cm converted to meters). Thus, \[ E_i = \frac{1}{2} \times 250 \times (0.06)^2 = 0.45 \mathrm{\, J} \].

Step 3: Calculate final energy

The final energy can be calculated using the displacement after eight cycles: \[ E_f = \frac{1}{2} k x_f^2 \] where \( x_f = 0.035 \) m (3.5 cm converted to meters). Thus, \[ E_f = \frac{1}{2} \times 250 \times (0.035)^2 = 0.153125 \mathrm{\, J} \].

Step 4: Calculate energy lost

Subtract the final energy from the initial energy to find the energy lost: \[ E_{\text{lost}} = E_i - E_f = 0.45 - 0.153125 = 0.296875 \mathrm{\, J} \].

Step 5: Explain energy loss

The energy was likely lost to damping forces such as air resistance or internal friction within the system, which slowly dissipates the energy as thermal energy or sound.

Key Concepts

Energy Loss

In a damped harmonic oscillator, energy loss refers to the decrease in the system's mechanical energy over time. This usually occurs due to damping forces such as air resistance or internal friction. In the context of our spring-mass system, the energy loss can be understood by recognizing that not all the energy remains in the form of mechanical oscillations. Instead, a portion of it is transformed into other forms of energy like heat or sound, reducing the amplitude of oscillation.

Over the course of eight cycles, our system loses energy, evidenced by the decrease in the maximum displacement of the spring from 6.0 cm to 3.5 cm. Using the equations for potential energy, we find that the system initially has an energy of 0.45 J and ends with 0.153125 J. This results in an energy loss of 0.296875 J.

This loss highlights the practical concept that no real system is entirely lossless, and damping plays a significant role in the gradual decrease of energy in oscillating systems.

Spring Constant

The spring constant, symbolized as \( k \), is a fundamental property of a spring, representing its stiffness. It is defined as the force required to produce a unit displacement in the spring. In our exercise, the spring constant is given as 2.5 N/cm, which we convert to 250 N/m to match standard SI units.

Understanding the spring constant is key when studying the behavior of harmonic oscillators. The magnitude of \( k \) determines how "strong" the spring is; a higher \( k \) implies a stiffer spring that requires more force to stretch or compress.

• A large spring constant means a quick return to equilibrium after a disturbance.
• A small spring constant results in slower, more gradual oscillations.

Knowing the spring constant allows us to calculate the potential energy stored in the spring and predict how the system will behave when set into motion.

Potential Energy

Potential energy in the context of springs refers to the stored energy resulting from the spring's position. This type of energy can be calculated using the formula: \[ E = \frac{1}{2} k x^2 \] where \( k \) is the spring constant and \( x \) is the displacement from equilibrium position.

In our given system, the initial potential energy when the spring is stretched 6.0 cm is 0.45 J. This energy owes its magnitude to both the displacement and the characteristic stiffness of the spring. As the system oscillates, this potential energy is continually converted to kinetic energy and back, defining the oscillatory motion.

• It represents the energy "stored" at the furthest points of displacement during oscillation.
• It decreases as the maximum displacement reduces due to energy dissipation.

Analyzing potential energy gives insight into how much energy the system starts with and helps track energy loss over time.

Oscillation

Oscillation describes the repetitive variation, typically in time, of some measure about a central value. For our system, it refers to the back-and-forth motion of the mass attached to the spring.

Oscillations can be linear or angular and are characterized by properties such as amplitude, frequency, and period. In this scenario,

• Amplitude refers to the maximum displacement from equilibrium, initially 6.0 cm, reducing to 3.5 cm due to damping effects.
• Frequency is how often the oscillations occur per unit time, and is influenced by the mass and spring constant.
• The period is the duration of one complete cycle.

Understanding oscillation helps elucidate how energy is transferred in the system. The reduction in amplitude over multiple cycles highlights the system's energy dissipation. This gradual energy loss is typical in damped oscillators and contributes to the understanding of real-world dynamic systems.