/*! 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 78 A uniform beam is suspended hori... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 \(\mathrm{kg}\) , and a \(175-\mathrm{kg}\) sack of gravel sits on the middle of it. The beam is oscillating in SHM, with an amplitude of 40.0 \(\mathrm{cm}\) and a frequency of 0.600 cycles \(/ \mathrm{s}\). (a) The sack of gravel falls off the beam when the beam has its maximum upward displacement. What are the frequency and amplitude of the subsequent SHM of the beam? (b) If the gravel instead falls off when the beam has its maximum speed, what are the frequency and amplitude of the subsequent SHM of the beam?

Short Answer

Expert verified
(a) Frequency: 0.800 cycles/s, Amplitude: 40.0 cm. (b) Frequency: 0.600 cycles/s, Amplitude: 53.3 cm.

Step by step solution

01

Understand given parameters

The mass of the beam is 225 kg, and the sack of gravel is 175 kg. The total mass when the sack is on the beam is 400 kg. The amplitude of the initial oscillation is 40.0 cm (0.4 m). The frequency is 0.600 cycles/s. Once the sack falls, the remaining mass is only 225 kg.
02

Using the properties of SHM

For SHM (Simple Harmonic Motion), the frequency of oscillation is given by the formula:\( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)where \( k \) is the spring constant and \( m \) is the mass. The amplitude and energy conservation will be used in different parts.
03

Apply conditions for part (a)

In part (a), the sack falls off at maximum upward displacement. The potential energy is at its maximum and kinetic energy is zero. Since the sack falls, now beam oscillates with reduced mass m = 225 kg. The frequency becomes:\( f' = \frac{1}{2\pi} \sqrt{\frac{k}{m'}} \)Since no kinetic energy loss occurs, amplitude stays same as both potential energies will be alike before and after sack removal.
04

Frequency calculation for part (a)

Given: original mass \( m = 400 \) kg, reduced mass \( m' = 225 \) kg. Using \( f' = f \cdot \sqrt{\frac{m}{m'}} \) Substitute the values: \( f' = 0.600 \cdot \sqrt{\frac{400}{225}} \approx 0.800 \) cycles/s.
05

Apply conditions for part (b)

In part (b), the sack falls at maximum speed which is at equilibrium; maximum kinetic, zero potential energy. Total energy is conserved, with no energy loss from sack removal, affecting amplitude. New amplitude \( A' \) can be derived from conservation of energy.
06

Calculate amplitude for part (b)

Equate energies:\( \frac{1}{2}kA^2 = \frac{1}{2}k'A'^2 \)Since \( k' \) and \( k \) unchanged:\( A' = A \cdot \sqrt{\frac{m}{m'}} \)\( A' = 0.4 \cdot \sqrt{\frac{400}{225}} \approx 0.533 \) m (53.3 cm).
07

Frequency remains same for part (b)

As the sack falls when the kinetic energy is maximum (at equilibrium position), the frequency doesn't change and we calculated the amplitude adjustment in step (6). Thus, the frequency remains 0.600 cycles/s for such energy scenarios.

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Key Concepts

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

Oscillation Frequency
The oscillation frequency in Simple Harmonic Motion (SHM) describes how fast an object moves back and forth through its cycle. For a spring-mass system, it's vital for determining how quickly the oscillations occur.
The frequency is largely dependent on two factors:
  • The spring constant, denoted as \( k \)
  • The mass of the system, denoted as \( m \)
The mathematical expression linking frequency \( f \), spring constant \( k \), and mass \( m \) is:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]In this exercise, when the sack of gravel leaves the beam, the mass changes from 400 kg to 225 kg. This affects the frequency, as shown in the calculation:\[ f' = f \cdot \sqrt{\frac{400}{225}} \approx 0.800 \text{ cycles/s} \]The frequency increases since the system mass decreases. This highlights the inverse relationship between frequency and mass.
Spring Constant
The spring constant, \( k \), provides a measure of a spring's ability to resist deformation. It is a fixed value for a given spring and plays a crucial role in SHM.
This constant strongly influences both the frequency and the potential energy stored in the spring. The spring constant is a reflection of the spring's stiffness and is critical in determining how the system behaves:
  • A higher \( k \) value means a stiffer spring.
  • A lower \( k \) value means a more flexible spring.
In the exercise, as the gravel leaves the beam, the spring constant remains constant, but its effect on the system changes due to the reduced mass. As seen in the frequency and amplitude equations, the spring constant helps determine the new motion properties when mass changes, while ensuring the total energy conservation in a different mass scenario.
Energy Conservation
In SHM, energy conservation is a fundamental principle that predicts how the system evolves over time. Here, the total mechanical energy is a mix of kinetic and potential energy that stays constant if no external force acts upon it.
Consider:
  • At maximum displacement (either upward or downward), all the energy is potential.
  • At equilibrium, the energy is entirely kinetic.
The exercise illustrates the two cases:- **Case (a):** When the sack falls at maximum upward displacement, potential energy is at peak.The system conserves energy, thus keeping the amplitude constant at 40 cm. The kinetic component holds no value at this point.- **Case (b):** When the sack falls off at maximum speed, energy is again fully conserved. This time, energy is primarily kinetic as we are at the equilibrium point. The amplitude changes due to the new balance of kinetic energy causing dynamics in displacement capability, calculated by:\[ A' = A \cdot \sqrt{\frac{m}{m'}} \approx 0.533 \text{ m} \]So ultimately, whether potential or kinetic, energy rules the outcomes of SHM.
Amplitude Change
Amplitude in this context refers to how far the object moves from its equilibrium position during oscillation. It shows the maximum extent of the oscillation, and is directly linked to the energy in the system.
If no energy is lost, the amplitude reflects the total energy's potential axis:
  • In part (a) of the problem, since the system changes as the sack of gravel departs at maximum displacement, the amplitude stays constant as energy mainly transitions in potential energy form.
  • For part (b), a shift due to loss of mass at equilibrium results in a new amplitude.
This new amplitude accounts for changes in the potential for maximum displacement capability, calculated as approximately 53.3 cm. The derivation of this change relied on the conservation principles and the altered mass. The amplitude change factors significantly into determining the properties and dynamics of SHM post any mass alteration during the motion cycle.

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Most popular questions from this chapter

A \(1.50-\mathrm{kg},\) horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 \(\mathrm{N} / \mathrm{m}\) and a \(275-\mathrm{g}\) metal ball is in the tray. The spring is below the tray, so it can oscillate up-and-down. The tray is then pushed down 15.0 \(\mathrm{cm}\) below its equilibrium point (call this point \(A\) ) and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (Hint: This does not oceur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?

A tuning fork labeled 392 Hz has the tip of each of its two prongs vibrating with an amplitude of 0.600 \(\mathrm{mm}\) . (a) What is the maximum speed of the tip of a prong? (b) A housefly (Musca domestica) with mass 0.0270 \(\mathrm{g}\) is holding on to the tip of one of the prongs. As the prong vibrates, what is the fly's maximum kinetic energy? Assume that the fly's mass has a negligible effect on the frequency of oscillation.

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 \(\mathrm{cm}\) and a frequency of 0.850 \(\mathrm{Hz}\) . Find (a) the maximum magnitude of the acceleration and of the velocity; \((6)\) the acceleration and speed when the poin-poin's coordinate is \(x=+9.0 \mathrm{cm}\); (c) the time required to move from the equilibrium position directly to a point 12.0 \(\mathrm{cm}\) away. (d) Which of the quantities asproach used parts \((\mathrm{a}),(\mathrm{b}),\) and \((\mathrm{c})\) can be found using the energy approach used in Section \(13.3,\) and which cannot? Explain.

Two point masses \(m\) are held in place a distance \(d\) apart. Another point mass \(M\) is midway between them. \(M\) is then displaced a small distance \(x\) perpendicular to the line connecting the two fixed masses and released. (a) Show that the magnitude of the net gravitational force on \(M\) due to the fixed masses is given approximately by \(F_{\mathrm{na}}=\frac{16 \mathrm{GmMx}}{d^{3}}\) if \(x \ll d\) . What is the direction of this force? Is it a restoring force? (b) Show that the mass \(M\) will oscillate with an angular frequency of \((4 / d) \sqrt{G m / d}\) and period \(\pi d / 2 \sqrt{d / G m} .\) (c) What would the period be if \(m=100 \mathrm{kg}\) and \(d=25.0 \mathrm{cm} ?\) Does it seem that you could easily measure this period? What things prevent this experiment from easily being performed in an ordinary physics lab? (d) Will \(M\) oscillate if it is dis- placed from the center a small distance \(x\) toward either of the fixed masses? Why?

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced 0.120 \(\mathrm{m}\) from its equilibrium position and released with zero initial speed, then after 0.800 \(\mathrm{s}\) its displacement is found to be 0.120 \(\mathrm{m}\) on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

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