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A particle with a mass of \(0.500 \mathrm{kg}\) is attached to a spring with a force constant of \(50.0 \mathrm{N} / \mathrm{m} .\) At time \(t=0\) the particle has its maximum speed of \(20.0 \mathrm{m} / \mathrm{s}\) and is moving to the left. (a) Determine the particle's equation of motion, specifying its position as a function of time.(b) Where in the motion is the potential energy three times the kinetic energy? (c) Find the length of a simple pendulum with the same period. (d) Find the minimum time interval required for the particle to move from \(x=0\) to \(x=1.00 \mathrm{m}\).

Short Answer

Expert verified
a) The equation of motion is \(x(t) = 20.0cos(10t)\). b) The potential energy is three times the kinetic energy when the particle is located at positions \(x = -1.8m\) or \(x = 1.8m\). c) The length of a simple pendulum with the same period is \(0.254m\). d) The time required for the particle to move from \(x=0\) to \(x=1.00m\) is \(0.157s\).

Step by step solution

01

Derive the equation of motion

We know a mass \(m\) attached to a spring oscillates with a frequency given by \(f=\sqrt{\frac{k}{m}}\), where \(k = 50.0 N/m\) is the spring constant, and \(m = 0.50 kg\) is the mass of the particle. Because the motion is also sinusoidal, it can be described by the equation of motion \(x(t) = A\cos(\omega t + \phi)\), where \(A\) is the amplitude and \(\phi\) is the phase constant. This particle is said to have maximum speed \(v = 20.0 m/s\) at \(t = 0\), which happens at equilibrium position (\(x = 0\)). Therefore, \(\phi = 0\). Given \(\omega = \sqrt{\frac{k}{m}}\), one can find the amplitude \(A = \frac{v}{\omega}\).
02

Determine where potential energy equals thrice the kinetic energy

At any given point during the motion of the spring, total energy would be conserved and equals to the sum of the kinetic energy and potential energy. Kinetic energy is given by \(KE = 0.5 * m*v^2\), and potential energy by \(PE = 0.5 * k*x^2\). The question needs us to find the values of \(x\) when \(PE = 3*KE\). By substituting known values and solving this equation, we obtain the possible values for \(x\).
03

Find the period of the simple pendulum

The period of the motion in the mass-spring system is given by \(T = 2*\pi*\sqrt{\frac{m}{k}}\). For a simple pendulum with length \(l\) and under the small angle approximation, its period is given by \(T = 2*\pi*\sqrt{\frac{l}{g}}\), where \(g = 9.8 m/s^2\) is acceleration due to gravity. Equating the two equations and solve for \(l\).
04

Calculate the time required to move from \(x=0\) to \(x=1.00 m\)

From the equation of motion obtained in step 1, we know the object reaches its amplitude \(A\) in a quarter of a period. So the time to move from \(x=0\) to \(x=1.00m\) (a quarter of the amplitude) would be \(T/4\).

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Key Concepts

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

Equation of Motion
In harmonic oscillation, the equation of motion describes a particle's position as a function of time. For a particle attached to a spring, this equation is typically represented as \(x(t) = A \cos(\omega t + \phi)\).
Here, \(A\) stands for the amplitude, which is the maximum distance from equilibrium, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant. The angular frequency \(\omega\) is determined by the spring constant \(k\) and the mass \(m\) using the formula \(\omega = \sqrt{\frac{k}{m}}\).
At maximum speed, the phase constant \(\phi\) can be zero if the maximum speed occurs at the equilibrium position. This sets the basis for understanding motion in harmonic systems.
Potential Energy
Potential energy in the context of springs is a form of energy stored due to the position of the particle. It is expressed as \(PE = \frac{1}{2} k x^2\), where \(x\) is the displacement from the equilibrium position and \(k\) is the spring constant.
  • It peaks when the particle is at its furthest point from equilibrium.
  • For harmonic oscillators, potential energy is part of the constant total mechanical energy, balanced by kinetic energy.
Understanding potential energy is crucial for analyzing where the energy is at any part of the oscillation cycle, such as determining when it is three times the kinetic energy.
Kinetic Energy
Kinetic energy is the energy of motion for a particle with mass \(m\) and velocity \(v\). The formula for calculating kinetic energy is \(KE = \frac{1}{2} m v^2\).
In a harmonic oscillator, the kinetic energy
  • is maximum when the particle is at the equilibrium position, meaning the potential energy is zero.
  • Switches back to potential energy as the particle moves away from this point.
In the given scenario, finding when the kinetic energy balances with the potential energy is key to understanding the energy distribution during oscillation.
Simple Pendulum
A simple pendulum is a classic example of harmonic motion. It is a weight suspended from a pivot, swinging under gravity's influence.
  • The pendulum's period, the time it takes for one complete swing, is \(T = 2\pi\sqrt{\frac{l}{g}}\).
  • The length \(l\) of the pendulum can be equated with the period of a spring-mass system for comparative analysis.
  • Both the pendulum and spring-mass systems exhibit remarkable similarities in their motion, making them excellent subjects of study for periodic motion.
This concept helps bridge the understanding of the motion characteristics shared between different oscillating systems.
Spring Constant
The spring constant \(k\) is a measure of a spring's stiffness and plays a fundamental role in determining the behavior of oscillating masses. The higher the spring constant, the stiffer the spring.
  • It appears in both the formulas for potential energy \(PE = \frac{1}{2}kx^2\) and angular frequency \(\omega = \sqrt{\frac{k}{m}}\).
  • Determining \(k\) helps in calculating the natural frequency of oscillation of the mass attached to the spring.
Understanding the spring constant is crucial for predicting how quickly and vigorously a spring system will respond to force, thus forming the backbone of harmonic oscillation analysis.

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Most popular questions from this chapter

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression $$x=(5.00 \mathrm{cm}) \cos (2 t+\pi / 6)$$. where \(x\) is in centimeters and \(t\) is in seconds. At \(t=0\) find (a) the position of the piston, (b) its velocity, and (c) its acceleration. (d) Find the period and amplitude of the motion.

The angular position of a pendulum is represented by the equation \(\theta=(0.320 \mathrm{rad}) \cos \omega t,\) where \(\theta\) is in radians and \(\omega=4.43 \mathrm{rad} / \mathrm{s} .\) Determine the period and length of the pendulum.

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).

One end of a light spring with force constant \(100 \mathrm{N} / \mathrm{m}\) is attached to a vertical wall. A light string is tied to the other end of the horizontal spring. The string changes from horizontal to vertical as it passes over a solid pulley of diameter \(4.00 \mathrm{cm} .\) The pulley is free to turn on a fixed smooth axle. The vertical section of the string supports a \(200-\mathrm{g}\) object. The string does not slip at its contact with the pulley. Find the frequency of oscillation of the object if the mass of the pulley is (a) negligible, (b) \(250 \mathrm{g}\), and (c) \(750 \mathrm{g}\)

A block-spring system oscillates with an amplitude of \(3.50 \mathrm{cm} .\) If the spring constant is \(250 \mathrm{N} / \mathrm{m}\) and the mass of the block is \(0.500 \mathrm{kg},\) determine (a) the mechanical energy of the system, (b) the maximum speed of the block, and (c) the maximum acceleration.

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