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A \(7.00-\mathrm{kg}\) object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 2.60 \(\mathrm{s}\) . Find the force constant of the spring.

Short Answer

Expert verified
The force constant of the spring is approximately 10.25 N/m.

Step by step solution

01

Understand the Problem

Identify that the object oscillates with simple harmonic motion (SHM) which is characterized by the equation for the period of a mass-spring system, given as \(T = 2\pi\sqrt{\frac{m}{k}}\), where \(T\) is the period, \(m\) is the mass, and \(k\) is the force constant of the spring.
02

Isolate the Force Constant Variable

Rearrange the equation to solve for the force constant \(k\). Starting with \(T = 2\pi\sqrt{\frac{m}{k}}\), square both sides to get rid of the square root, then isolate \(k\) on one side of the equation. The rearranged equation will be \(k = \frac{4\pi^2 m}{T^2}\).
03

Plug in the Known Values

Substitute the given mass \(m = 7.00\ \mathrm{kg}\) and the period \(T = 2.60\ \mathrm{s}\) into the equation \(k = \frac{4\pi^2 m}{T^2}\) to calculate the force constant.
04

Calculate the Force Constant of the Spring

Use the equation from the previous step and calculate the value of \(k\): \(k = \frac{4\pi^2 \times 7.00\ \mathrm{kg}}{(2.60\ \mathrm{s})^2}\), and then evaluate this expression to find the spring constant.

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Key Concepts

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

Simple Harmonic Motion
Understanding simple harmonic motion (SHM) is fundamental when studying the behavior of oscillating systems like springs and pendulums. In SHM, the restoring force is directly proportional to the displacement from the equilibrium position but in the opposite direction. A perfect example of SHM is a mass attached to a spring, which when displaced and released, vibrates back and forth about an equilibrium position.

The motion is characterized by its periodic nature, meaning the motion repeats itself at regular intervals, known as the period. One of the beautiful aspects of SHM is its predictability. The period remains constant, regardless of the amplitude, as long as the system isn't damped or driven, and it follows the equation for the period of a mass-spring system, given by \(T = 2\pi\sqrt{\frac{m}{k}}\), where \(T\) is the period, \(m\) is the mass, and \(k\) is the force constant of the spring.

This relationship can be visualized as a sine or cosine wave when plotting displacement over time, illustrating the rhythmic and cyclic nature of SHM. It's a foundational concept not only in physics but also in various fields such as engineering, seismology, and even music.
Mass-Spring System
A mass-spring system is a classic example of a simple harmonic oscillator found in physics. It consists of a spring with a known stiffness (or force constant) attached to a mass that can move freely. The force constant, denoted as \(k\), is a measure of the spring's resistance to being compressed or stretched. \(k\) is unique to each spring and determines the system's behavior when the mass is set into motion.

When the mass is displaced from its equilibrium position and then released, it oscillates around that equilibrium due to the spring's tendency to restore it to the original position. This restoring force is governed by Hooke's Law, which states that the force exerted by the spring is proportional to the displacement.

The equation \( F = -kx \) represents Hooke's Law, where \(F\) is the restoring force, \(k\) is the force constant, and \(x\) is the displacement from equilibrium. The negative sign indicates that the force is in the opposite direction of the displacement. In the context of a vertical mass-spring system, such as the one in the given exercise, gravity plays a role, but for small oscillations, this system can still be considered a good approximation of SHM.
Oscillation Period
The oscillation period, denoted as \(T\), is a key characteristic of any simple harmonic oscillator, describing the time it takes for the system to complete one full cycle of motion. This period is constant for a given system and is dependent on properties such as the mass involved in the motion and the force constant of the spring in a mass-spring system.

The formula to find the period is \(T = 2\pi\sqrt{\frac{m}{k}}\), showing that the period is directly proportional to the square root of the mass and inversely proportional to the square root of the force constant. What's intriguing about this relationship is that it doesn't matter how far the mass is pulled or pushed from the equilibrium position; the period remains the same as long as the oscillations are small and the system is not subject to external forces or resistances.

For students to fully grasp the concept, it's essential to note that this consistent time frame of oscillation makes it possible to predict the system's motion at any given time. This is incredibly useful in engineering applications where timing is crucial, like in watches and clocks, where a stable oscillation period ensures accurate timekeeping.

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Most popular questions from this chapter

A \(1.00-\mathrm{kg}\) glider attached to a spring with a force constant of 25.0 \(\mathrm{N} / \mathrm{m}\) oscillates on a horizontal, frictionless air track. At \(t=0\) the glider is released from rest at \(x=-3.00 \mathrm{cm} .\) (That is, the spring is compressed by \(3.00 \mathrm{cm} . )\) Find (a) the period of its motion, (b) the maximum values of its speed and acceleration, and (c) the position, velocity, and acceleration as functions of time.

consider a bob on a light stiff rod, forming a simple pendulum of length \(L=1.20 \mathrm{m} .\) It is displaced from the vertical by an angle \(\theta_{\max }\) and then released. Predict the subsequent angular positions if \(\theta_{\max }\) is small or if it is large. Proceed as follows: Set up and carry out a numerical method to integrate the equation of motion for the simple pendulum: $$ \frac{d^{2} \theta}{d t^{2}}=-\frac{g}{L} \sin \theta $$ Take the initial conditions to be \(\theta=\theta_{\max }\) and \(d \theta / d t=0\) at \(t=0 .\) On one trial choose \(\theta_{\max }=5.00^{\circ},\) and on another trial take \(\theta_{\max }=100^{\circ} .\) In each case find the position \(\theta\) as a function of time. Using the same values of \(\theta_{\max },\) compare your results for \(\theta\) with those obtained from \(\theta(t)=\) \(\theta_{\max } \cos \omega t .\) How does the period for the large value of \(\theta_{\text { max }}\) compare with that for the small value of \(\theta_{\text { max }} ?\) Note: Using the Euler method to solve this differential equation, you may find that the amplitude tends to increase with time. The fourth-order Runge-Kutta method would be a better choice to solve the differential equation. However, if you choose \(\Delta t\) small enough, the solution using Euler's method can still be good.

A block of mass \(M\) is connected to a spring of mass \(m\) and oscillates in simple harmonic motion on a horizontal, frictionless track (Fig. P15.66). The force constant of the spring is \(k\) and the equilibrium length is \(\ell .\) Assume that all portions of the spring oscillate in phase and that the velocity of a segment \(d x\) is proportional to the distance \(x\) from the fixed end; that is, \(v_{x}=(x / \ell) v\) . Also, note that the mass of a segment of the spring is \(d m=(m / \ell) d x\) . Find \((a)\) the kinetic energy of the system when the block has a speed \(v\) and \((b)\) the period of oscillation.

The mass of the deuterium molecule \(\left(\mathrm{D}_{2}\right)\) is twice that of the hydrogen molecule \(\left(\mathrm{H}_{2}\right) .\) If the vibrational frequency of \(\mathrm{H}_{2}\) is \(1.30 \times 10^{14} \mathrm{Hz}\) , what is the vibrational frequency of \(\mathrm{D}_{2}\) ? Assume that the "spring constant" of attracting forces is the same for the two molecules.

A particle moving along the \(x\) axis in simple harmonic motion starts from its equilibrium position, the origin, at \(t=0\) and moves to the right. The amplitude of its motion is \(2.00 \mathrm{cm},\) and the frequency is 1.50 \(\mathrm{Hz}\) . (a) Show that the position of the particle is given by $$ x=(2.00 \mathrm{cm}) \sin (3.00 \pi t) $$ Determine (b) the maximum speed and the earliest time \((t>0)\) at which the particle has this speed, (c) the maximum acceleration and the earliest time \((t>0)\) at which the particle has this acceleration, and \((d)\) the total distance traveled between \(t=0\) and \(t=1.00 \mathrm{s}\) .

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