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\(\bullet\) If the frequency of a mass-spring system is \(1.50 \mathrm{~Hz}\) and the mass on the spring is \(5.00 \mathrm{~kg}\), what is the spring constant?

Short Answer

Expert verified
The spring constant is approximately 443.5 N/m.

Step by step solution

01

Understand the Relationship

The frequency of a mass-spring system is related to the spring constant (\(k\)) and the mass (\(m\)) through the formula \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\). This tells us how the cyclic frequency depends on these variables.
02

Rearrange the Formula

We need to find the spring constant \(k\). Reordering the formula gives us \(k = (2\pi f)^2 m\). This allows us to express \(k\) in terms of known quantities.
03

Insert Known Values

Substitute the given frequency \(f = 1.50 \, \mathrm{Hz}\) and mass \(m = 5.00 \, \mathrm{kg}\) into the modified formula. Thus, we calculate \(k = (2\pi \times 1.50)^2 \times 5.00\).
04

Calculate the Spring Constant

First, calculate \(2\pi \times 1.50\), which is approximately \(9.42\). Then, square this result: \(9.42^2 \approx 88.7\). Finally, multiply by the mass: \[k \approx 88.7 \times 5.00 = 443.5 \, \mathrm{N/m}\].

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Key Concepts

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

Mass-Spring System
A mass-spring system is a fundamental concept in physics that involves a mass attached to a spring. When the mass is displaced, the spring applies a force to bring it back to its equilibrium position. This system can oscillate back and forth, creating a motion known as simple harmonic motion.

In these systems, the movement of the mass is determined by the spring's properties and the mass itself. The elasticity of the spring is characterized by the spring constant, which measures how stiff the spring is. Meanwhile, the mass affects the system's inertia. The interaction between the mass and the spring leads to periodic oscillations.

Understanding a mass-spring system is crucial for solving problems involving oscillations and vibrations in various applications, from mechanical engineering to everyday objects like car suspensions. The simple relationship between mass, spring constant, and oscillation frequency lays the foundation for more complex dynamics in physics.
Cyclic Frequency
Cyclic frequency, often just called frequency, is a measure of how often a repeating event, such as an oscillation, occurs per unit time. In physics, it's usually measured in hertz (Hz), which represents cycles per second.

For a mass-spring system, the cyclic frequency (\(f\)) is crucial in understanding how fast the system oscillates. The formula for frequency in a mass-spring system is given by:\[f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\]where \(k\) is the spring constant and \(m\) is the mass.

This relationship implies that:
  • Higher spring constants lead to higher frequencies, meaning the system oscillates more rapidly.
  • Increasing the mass reduces the frequency, slowing the system's oscillation.
Therefore, by analyzing the frequency, we can infer certain qualities about the mass and the spring in the system used, allowing us to solve for unknowns like the spring constant as shown in the problem.
Spring Constant Formula
The spring constant formula is an essential tool in determining the stiffness of a spring in a mass-spring system. It defines the relationship between the spring's force, displacement, and the spring constant itself, denoted by \(k\). The general formula that relates these variables is Hooke's Law: \[F = k \cdot x\]where \(F\) is the force applied to the spring, \(x\) is the displacement from the equilibrium position, and \(k\) is the spring constant.

In the context of the cyclic frequency, we use a rearranged version to find \(k\) in terms of known quantities like frequency \(f\) and mass \(m\): \[k = (2\pi f)^2 m\]This variation is particularly useful when we know the frequency and mass but need to determine the spring constant. Knowing how to manipulate this formula is essential for accurately describing the dynamics of a mass-spring system.
Mass in Physics
In physics, mass is a fundamental property of matter that measures an object's resistance to acceleration when a force is applied. It also determines the gravitational interaction with other bodies. Mass plays a key role in the behavior of a mass-spring system.

In the context of the given exercise, the mass of the object attached to the spring directly affects the system's oscillation frequency. As shown by the formula:\[f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\]we see that mass is inversely proportional to the cyclic frequency. Thus, a heavier object attached to the same spring will oscillate more slowly compared to a lighter one.

Understanding mass in physics helps us predict how changes to an object's mass will affect the oscillatory motion in a spring system. This knowledge is crucial for designing systems in engineering and technology where precise control of motion is required.

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Most popular questions from this chapter

\(\bullet\) A wave on a rope that measures \(10 \mathrm{~m}\) long takes \(2.0 \mathrm{~s}\) to travel the whole rope. If the wavelength of the wave is \(2.5 \mathrm{~m},\) what is the frequency of oscillation of any piece of the rope?

\(\bullet\bullet\) (a) At what position is the magnitude of the force on a mass in a mass-spring system minimum: \((1) x=0\), (2) \(x=-A,\) or \((3) x=+A ?\) Why? \((b)\) If \(m=0.500 \mathrm{~kg}\) \(k=150 \mathrm{~N} / \mathrm{m},\) and \(A=0.150 \mathrm{~m},\) what are the magnitude of the force on the mass and the acceleration of the mass at \(x=0,0.050 \mathrm{~m},\) and \(0.150 \mathrm{~m} ?\)

\(\bullet\bullet\) A \(0.350-\mathrm{kg}\) block moving vertically upward collides with a light vertical spring and compresses it \(4.50 \mathrm{~cm}\) before coming to rest. If the spring constant is \(50.0 \mathrm{~N} / \mathrm{m}\) what was the initial speed of the block? (Ignore energy losses to sound and other factors during the collision.)

\(\bullet\bullet\) A 0.25 -kg object is suspended on a light spring of spring constant \(49 \mathrm{~N} / \mathrm{m}\) and the system is allowed to come to rest at its equilibrium position. The object is then pulled down \(0.10 \mathrm{~m}\) from the equilibrium position and released. What is the speed of the object when it goes through the equilibrium position?

You are setting up two standing string waves. You have a length of uniform piano wire that is 3.0 m long and has a mass of 0.150 kg. You cut this into two lengths, one of 1.0 m and the other of 2.0 m, and place each length under tension. What should be the ratio of tensions (expressed as short to long) so that their fundamental frequencies are the same?

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