Question

In a lab, a student measures the unstretched length of a spring as \(11.2 \mathrm{~cm}\). When a 100.0 - g mass is hung from the spring, its length is \(20.7 \mathrm{~cm}\). The mass-spring system is set into oscillatory motion, and the student obscrves that the amplitude of the oscillation decreases by about a factor of 2 after five complete cycles. a) Calculate the period of oscillation for this system, assuming no damping. b) If the student can measure the period to the nearest \(0.05 \mathrm{~s}\). will she be able to detect the difference between the period with no damping and the period with damping?

Short answer

Answer: Given the student's measurement accuracy of 0.05 seconds, it may be challenging for the student to detect the difference between the period with no damping and the period with damping.

Step-by-step solution

Find the Spring Constant (k)

Using Hooke's Law, we can find the spring constant (k) when a known mass is hung from the spring. Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position: \(F_s = kx\) where \(F_s\) is the spring force, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. In this case, the weight of the mass is causing the displacement: \(F_s = mg\) where \(m\) is the mass and \(g\) is the acceleration due to gravity. Combine the two equations: \(mg = kx\) We can find the spring constant, \(k\), using the given values for mass and displacement: \(m = 100.0\,\mathrm{g} = 0.1\,\mathrm{kg}\) (converting g to kg) \(g = 9.81\,\mathrm{m/s^2}\) (approximate gravitational acceleration) Displacement \(x = 20.7\,\mathrm{cm} - 11.2\,\mathrm{ cm} = 9.5\,\mathrm{cm} = 0.095\,\mathrm{m}\) (converting cm to m) Now, solve for \(k\): \(k = \frac{mg}{x}\)

Calculate the period of oscillation with no damping

We can use the formula for the period of oscillation in a mass-spring system without damping: \(T = 2\pi\sqrt{\frac{m}{k}}\) Now, plug in the values of \(m\) and \(k\) obtained in Step 1 to calculate the period \(T\).

Discuss detectability of damping effect

The student can measure the period to the nearest \(0.05\,\mathrm{s}\). We will compare the period of the damped and undamped system to see if the difference is greater than or equal to \(0.05\,\mathrm{s}\). For the damped system, the amplitude decreases by about a factor of \(2\) after five complete cycles. This damping effect might affect the period, but we are not given enough information to calculate the period for the damped system. However, since the student can only measure the period to an accuracy of \(0.05\,\mathrm{s}\), small changes in period due to damping might not be detectable. Hence, given the limitations of the student's measurement, we can conclude that it may be challenging for the student to detect the difference between the period with no damping and the period with damping.

Key concepts

Hooke's Law

Understanding Hooke's Law is crucial when studying the behavior of springs and elastic materials. This fundamental principle reveals how the amount of force needed to extend or compress a spring is directly proportional to the displacement—that is, how far the spring has been stretched or compressed from its natural length.

Let's translate this into a simple equation:
\[ F_s = kx \]
Here, \( F_s \) symbolizes the force exerted by the spring, \( k \) is a specific value known as the spring constant (which we will explore further), and \( x \) represents the displacement from the spring's equilibrium position.

Imagine stretching a spring further; according to Hooke's Law, this would require an increasing amount of force. This also emphasizes that elasticity has its limits. The law applies only as long as the spring is not permanently deformed, meaning within its elastic limit. Beyond this, it will not return to its original shape, and Hooke's Law will no longer hold true.

In practical terms, when a weight, such as the 100.0-g mass in our exercise, is hung from a spring, we can observe this law in action. The gravitational force on the mass causes the spring to stretch, and we can calculate the spring force using the weight of the mass \( (mg) \) and compare it to the spring force \( F_s \), enabling us to solve for the spring constant \( k \).

In educational contexts, Hooke's Law serves as a bridge to more advanced topics like oscillation, waves, and even engineering problems involving springs and other elastic elements. It's the foundation for understanding how oscillatory systems behave both theoretically and in real-world applications.

Spring Constant

The spring constant, denoted as \( k \) in Hooke's Law, is an intrinsic property of a spring that quantifies its stiffness. A higher spring constant means a stiffer spring, requiring more force to produce a given displacement.

In the context of the exercise, determining the spring constant is the first step towards understanding the dynamics of the mass-spring system. By comparing the force exerted by the gravity on the mass (which causes the spring to extend) to the force the spring exerts to counteract this (the spring’s resistance to being stretched), we find the value of \( k \) using the equation derived from Hooke's Law:
\[ k = \frac{mg}{x} \]
Here, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( x \) is the change in length of the spring from its unstretched state. With the mass given in kilograms, the acceleration due to gravity commonly accepted as \( 9.81\,\mathrm{m/s^2} \), and the displacement in meters, we can calculate \( k \).

The calculated spring constant not only reflects the characteristics of the spring but is also pivotal for further investigations into the oscillation period of the system. It is a versatile constant appearing in various equations describing the motion of springs, and thus an important concept in both physics and engineering. Whether you're designing a car's suspension system or just trying to understand the physics of a bouncing ball, the spring constant is a key player in describing the forces at work.

Damped Oscillation

In the real world, oscillations rarely maintain their amplitude indefinitely. Frictional forces, air resistance, and other forms of resistance gradually sap the energy of a system, causing the amplitude of oscillations to decrease over time—a phenomenon known as damped oscillation.

Damped oscillation introduces a new layer of complexity to our analysis. For simple harmonic motion, such as a mass bouncing on a spring with no external forces, the period of oscillation remains the same regardless of the amplitude. However, once damping is introduced, this is no longer strictly true, especially in heavy damping conditions.

Although the exercise in question does not provide all the information needed to calculate the damped period, it mentions a detectable difference between the damped and undamped periods. For light damping, which is likely the case when observing a reduction in amplitude by only a factor of 2 after five cycles, the period does not change significantly. Thus, it would typically require more precise measurement tools than what is described for any difference to be reliably observed by the student.

Damping is an essential consideration in many practical applications, such as the design of shock absorbers in automobiles, the construction of buildings to withstand earthquakes, and the creation of stable structures. It also plays a vital role in the design of electronic circuits and control systems. Understanding damped oscillation helps us to not only explain but also control and utilize the behavior of oscillatory systems in various technological and engineering contexts.