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A 200 -g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the system is \(2.00 \mathrm{J},\) find (a) the force constant of the spring and (b) the amplitude of the motion.

Short Answer

Expert verified
The force constant of the spring is 126.7 N/m and the amplitude of motion is 0.4 m.

Step by step solution

01

Identify Given Variables

First, list the variables given in the problem. Here, the mass \(m = 200g = 0.2kg\), the period \(T = 0.250s\), and the total energy \(E = 2.00J\).
02

Compute Spring Constant

The formula for the period \(T\) of spring motion is given by \(T = 2\pi\sqrt{\frac{m}{k}}\), where \(k\) is the spring constant. Therefore, \(k\) can be calculated using \(k = \frac{4\pi^{2}m}{T^{2}}\). Substituting the given values, we compute \(k\).
03

Compute Amplitude of the Motion

The formula for the total energy of simple harmonic motion is \(E = \frac{1}{2}kA^{2}\), where \(A\) is the amplitude of the motion. So, \(A = \sqrt{\frac{2E}{k}}\). Substitute the computed \(k\) and given \(E\) into this formula to find \(A\).

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Key Concepts

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

Spring Constant
In simple harmonic motion, the spring constant is a crucial factor. It determines how stiff a spring is. The stiffer the spring, the larger the spring constant. Mathematically, the spring constant is denoted as \( k \). The spring constant can be calculated using different physical attributes of the motion.
For a system executing simple harmonic motion, the relationship between the period \( T \), mass \( m \), and the spring constant \( k \) is given by the formula: \[ T = 2\pi\sqrt{\frac{m}{k}} \] This formula can be rearranged to solve for the spring constant as: \[ k = \frac{4\pi^{2}m}{T^{2}} \] In this equation:
  • \( m \) is the mass attached to the spring.
  • \( T \) is the period of oscillation, which is the time it takes to complete one cycle of motion.
Using this formula in the example exercise, where \( m = 0.2 \text{ kg} \) and \( T = 0.250 \text{ s} \), you would find the spring constant \( k \). This allows you to determine how resistant the spring is to being compressed or stretched.
Amplitude
Amplitude in simple harmonic motion is the maximum displacement from the equilibrium position. It is one of the key properties of oscillatory motion. Amplitude is crucial because it relates directly to the energy of the system.
To find the amplitude \( A \), you can use the total energy \( E \) formula for a spring system. The total energy is expressed as: \[ E = \frac{1}{2}kA^{2} \] Rearranging the formula to solve for the amplitude gives: \[ A = \sqrt{\frac{2E}{k}} \] Here:
  • \( E \) is the total energy of the system (in this exercise, \( 2.00 \text{ J} \)).
  • \( k \) is the spring constant found using the earlier mentioned formula.
By substituting the value of \( k \) obtained in the exercise and the given energy \( E \), you calculate the amplitude \( A \). This tells you how far the spring can be stretched or compressed from its rest length during motion.
Total Energy
The concept of total energy in simple harmonic motion ties together both potential and kinetic energy present within the system. Total energy \( E \) remains constant throughout as long as there is no external force acting on the system. This energy can be understood as a mix of the spring's potential energy when compressed or stretched, and the mass's kinetic energy as it moves.
The formula that expresses total energy in a simple harmonic motion involving a spring is given as: \[ E = \frac{1}{2} k A^2 \] This equation highlights:
  • \( k \) is the spring constant, which shows the stiffness of the spring.
  • \( A \) is the amplitude, the maximum extent of movement from equilibrium.
Both the spring constant and amplitude affect the total energy. As you can see from the exercise, knowing the total energy and either the spring constant or amplitude allows you to calculate the other. The exercise shows us how energy is conserved and represented in a mathematical way, ensuring the amplitude and potential/kinetic transformation work seamlessly together. This preservation of energy is a core principle of mechanics in physics.

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Most popular questions from this chapter

A \(1.00-\mathrm{kg}\) object is attached to a horizontal spring. The spring is initially stretched by \(0.100 \mathrm{m},\) and the object is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.500 s later. What is the maximum speed of the object?

A light, cubical container of volume \(a^{3}\) is initially filled with a liquid of mass density \(\rho .\) The cube is initially supported by a light string to form a simple pendulum of length \(L_{i},\) measured from the center of mass of the filled container, where \(L_{i} \gg a\). The liquid is allowed to flow from the bottom of the container at a constant rate (dM/dt). At any time \(t,\) the level of the fluid in the container is \(h\) and the length of the pendulum is \(L\) (measured relative to the instantaneous center of mass). (a) Sketch the apparatus and label the dimensions \(a, h, L_{i},\) and \(L\).(b) Find the time rate of change of the period as a function of time \(t .(\mathrm{c})\) Find the period as a function of time.

A simple harmonic oscillator takes \(12.0 \mathrm{s}\) to undergo five complete vibrations. Find (a) the period of its motion, (b) the frequency in hertz, and (c) the angular frequency in radians per second.

A small object is attached to the end of a string to form a simple pendulum. The period of its harmonic motion is measured for small angular displacements and three lengths, each time clocking the motion with a stopwatch for 50 oscillations. For lengths of \(1.000 \mathrm{m}, 0.750 \mathrm{m}\) and \(0.500 \mathrm{m},\) total times of \(99.8 \mathrm{s}, 86.6 \mathrm{s},\) and \(71.1 \mathrm{s}\) are measured for 50 oscillations. (a) Determine the period of motion for each length. (b) Determine the mean value of \(g\) obtained from these three independent measurements, and compare it with the accepted value. (c) Plot \(T^{2}\) versus \(L,\) and obtain a value for \(g\) from the slope of your best-fit straight-line graph. Compare this value with that obtained in part (b).

(a) A hanging spring stretches by \(35.0 \mathrm{cm}\) when an object of mass \(450 \mathrm{g}\) is hung on it at rest. In this situation, we define its position as \(x=0 .\) The object is pulled down an additional \(18.0 \mathrm{cm}\) and released from rest to oscillate without friction. What is its position \(x\) at a time 84.4 s later? (b) What If? A hanging spring stretches by \(35.5 \mathrm{cm}\) when an object of mass \(440 \mathrm{g}\) is hung on it at rest. We define this new position as \(x=0 .\) This object is also pulled down an additional \(18.0 \mathrm{cm}\) and released from rest to oscillate without friction. Find its position 84.4 s later. (c) Why are the answers to (a) and (b) different by such a large percentage when the data are so similar? Does this circumstance reveal a fundamental difficulty in calculating the future? (d) Find the distance traveled by the vibrating object in part (a). (e) Find the distance traveled by the object in part (b).

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