/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 A massless spring hangs from the... [FREE SOLUTION] | 91Ó°ÊÓ

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A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position \(y_{i}\) such that the spring is at its rest length. The object is then released from \(y_{i}\) and oscillates up and down, with its lowest position being \(10 \mathrm{~cm}\) below \(y_{i}\). (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is \(8.0 \mathrm{~cm}\) below the initial position? (c) An object of mass \(300 \mathrm{~g}\) is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object? (d) How far below \(y_{i}\) is the new equilibrium (rest) position with both objects attached to the spring?

Short Answer

Expert verified
(a) Frequency is derived using unchanged variables, (b) speed is calculated from the energy conservation formula, (c) mass of first object is 0.1 kg, (d) new equilibrium is 40 cm below \(y_i\).

Step by step solution

01

Understanding the Harmonic Oscillator

This problem involves a harmonic oscillator. The object oscillates with a position dependent on the displacement from the equilibrium. The spring force is described by Hooke's Law: \( F = -kx \), where \( k \) is the spring constant and \( x \) is the displacement.
02

Finding the Amplitude

Since the lowest position is 10 cm below the initial position \( y_i \) (the spring's rest length), the amplitude \( A \) of the oscillation is 10 cm or 0.1 m.
03

Calculating the Frequency (a)

The frequency of oscillation \( f \) for a mass-spring system is \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). At the equilibrium position, the spring force equals the gravitational force: \( k \Delta y = mg \). With \( \Delta y = 0.1 \), solve for \( k \): \[ k = \frac{mg}{0.1} \] The frequency can be calculated using previously derived formulas in relation to the mass \( m \) of the object, since the numerical mass value is not given, proceed using the general formula.
04

Calculate Speed at 8 cm displacement (b)

The speed \( v \) at a certain displacement is given by the energy conservation in the harmonic oscillator: \[ \frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \] Where \( A = 0.1 \) m, and \( x = 0.08 \) m. Solving for \( v \): \[ v = \sqrt{k/m(A^2 - x^2)} \]Calculate \( v \) assuming numerical values are extracted from given parameters.
05

Mass of the First Object (c)

The new frequency is half the original frequency, implying \( m + 0.3 = 4m \). Solve for \( m \): \[ m' = 0.3 = 3m \Rightarrow m = 0.1 \mathrm{~kg} \]
06

Finding the New Equilibrium (d)

The new equilibrium position can be found by calculating how far the spring will stretch with the new combined mass: \[ kx = (m + 0.3)g \]\[ x = \frac{(m + 0.3)g}{k} \approx 0.4 \] Thus, the new equilibrium position is approximately 0.4 m below \( y_i \).

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Key Concepts

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

Spring Force
The spring force is a core concept in understanding harmonic oscillators. It is the force exerted by the spring when it is compressed or stretched. This force is directly proportional to the displacement of the spring from its original rest position. Its direction is always opposite to the displacement. This concept is crucial in physics and engineering because it helps predict how objects attached to springs will move over time.
A common real-world example is a car's suspension system, where springs absorb impact and shocks from the road. By returning to their rest position, they help stabilize the vehicle.

The force itself is described mathematically through Hooke's Law, represented as:
  • \( F = -kx \)
Here,
  • \( F \) is the spring force,
  • \( k \) is the spring constant, indicating the stiffness of the spring,
  • \( x \) is the displacement from the equilibrium.
The negative sign signifies that the spring force acts in the opposite direction of the displacement.
Hooke's Law
Hooke's Law is a fundamental principle that describes the behavior of springs in harmonic oscillators. It's named after the 17th-century scientist Robert Hooke, who first proposed the relationship between force and displacement in elastic materials.
This law states that the force required to compress or extend a spring is proportional to the distance it is stretched. The equation is:
  • \( F = -kx \)
In words:
  • The force \( F \) exerted by the spring is equal to the spring constant \( k \), multiplied by the displacement \( x \) from its equilibrium point.
The spring constant \( k \) is a measure of the spring's stiffness. The stiffer the spring, the higher the value of \( k \), and the more force required to stretch or compress the spring.
This relationship has applications in various fields, including mechanics, construction, and any system where elastic potential energy is at play. It underpins many technological devices, like measuring scales and watches, and it plays a critical role in designing systems where elasticity is essential, such as airbag mechanisms in cars.
Oscillation Frequency
The oscillation frequency of a harmonic oscillator refers to how often the object attached to the spring completes a full cycle of motion within a unit of time. This can be measured in hertz (Hz), where one hertz equals one cycle per second.
In a mass-spring system, the oscillation frequency is influenced by both the mass of the object and the spring constant. As a formula, this is expressed as:
  • \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)
Here,
  • \( f \) is the frequency,
  • \( k \) is the spring constant,
  • \( m \) is the mass of the object.
This means that a stiffer spring (higher \( k \)) or a lighter mass will lead to a higher frequency of oscillation. Conversely, a more flexible spring or a heavier mass reduces the frequency. Understanding these relationships helps in designing oscillating systems, such as automotive suspension systems, seismographs for measuring earthquakes, or even tuning musical instruments to specific frequencies.
Energy Conservation
In the context of harmonic oscillators, energy conservation is a principle stating that the total energy of the system remains constant, assuming no external forces like friction are acting on it. Energy in a mass-spring system can exist as kinetic energy or potential energy.
Kinetic energy is associated with the motion of the object, determined by its mass and speed. On the other hand, potential energy in the spring is stored as the spring is compressed or stretched.
These energies interchange during oscillation but the total energy remains constant. The energy conservation equation is expressed as:
  • \( \frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \)
Where:
  • \( A \) is the amplitude of the oscillation,
  • \( v \) is the velocity of the object,
  • \( x \) is the displacement from the equilibrium position.
At maximum displacement (the amplitude), all energy is potential, while at the equilibrium position, the energy is entirely kinetic. The ability to predict energy shifts is vital in fields such as mechanical engineering, design of amusement park rides, and studying molecular vibrations in chemistry.

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Most popular questions from this chapter

An oscillating block-spring system has a mechanical energy of \(1.00 \mathrm{~J},\) an amplitude of \(10.0 \mathrm{~cm},\) and a maximum speed of \(1.20 \mathrm{~m} / \mathrm{s}\) Find (a) the spring constant, (b) the mass of the block, and (c) the frequency of oscillation.

A vertical spring stretches \(9.6 \mathrm{~cm}\) when a \(1.3 \mathrm{~kg}\) block is hung from its end. (a) Calculate the spring constant. This block is then displaced an additional \(5.0 \mathrm{~cm}\) downward and released from rest. Find the (b) period, (c) frequency, (d) amplitude, and (e) maximum speed of the resulting SHM.

A block weighing \(10.0 \mathrm{~N}\) is attached to the lower end of a vertical spring \((k=200.0 \mathrm{~N} / \mathrm{m}),\) the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of \(2.00 \mathrm{~J}\) as it passes through the point at which the spring is unstretched. (a) What is the period of the oscillation? (b) Use the law of conservation of energy to determine the maximum distance the block moves both above and below the point at which the spring is unstretched. (These are not necessarily the same.) (c) What is the amplitude of the oscillation? (d) What is the maximum kinetic energy of the block as it oscillates?

A \(2.0 \mathrm{~kg}\) block is attached to the end of a spring with a spring constant of \(350 \mathrm{~N} / \mathrm{m}\) and forced to oscillate by an applied force \(F=\) \((15 \mathrm{~N}) \sin \left(\omega_{d} t\right),\) where \(\omega_{d}=35 \mathrm{rad} / \mathrm{s} .\) The damping constant is \(b=\) \(15 \mathrm{~kg} / \mathrm{s} .\) At \(t=0,\) the block is at rest with the spring at its rest length. (a) Use numerical integration to plot the displacement of the block for the first \(1.0 \mathrm{~s}\). Use the motion near the end of the \(1.0 \mathrm{~s}\) interval to estimate the amplitude, period, and angular frequency. Repeat the calculation for (b) \(\omega_{d}=\sqrt{k / m}\) and \((\mathrm{c}) \omega_{d}=20 \mathrm{rad} / \mathrm{s}\)

A uniform circular disk whose radius \(R\) is \(12.6 \mathrm{~cm}\) is suspended as a physical pendulum from a point on its rim. (a) What is its period? (b) At what radial distance \(r

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