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Chapter 14: Problem 76

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 kg, and a 175-kg sack of gravel sits on the middle of it. The beam is oscillating in SHM with an amplitude of 40.0 cm and a frequency of 0.600 cycle/s. (a) The sack 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 sack instead falls off when the beam has its maximum speed, repeat part (a).

Short Answer

Expert verified
(a) New frequency decreases, amplitude stays same; (b) New frequency decreases, amplitude calculated from energy conservation.

Step by step solution

01

Understanding the System and Given Variables

The beam of mass 225 kg is oscillating in simple harmonic motion (SHM) with a sack of gravel (mass 175 kg) initially on it. The system oscillates with an amplitude of 40.0 cm and a frequency of 0.600 cycle/s. We need to analyze two scenarios where the sack falls off at different points in the oscillation cycle to find the new frequency and amplitude after the sack is removed.
02

Calculating Effective Mass

Initially, the total mass of the system is the sum of the beam and the sack: \( m_1 = 225 \text{ kg} + 175 \text{ kg} = 400 \text{ kg} \). When the sack falls off, the new mass becomes \( m_2 = 225 \text{ kg} \).
03

Converting Amplitude and Frequency

Convert amplitude to meters: \( A = 40.0 \text{ cm} = 0.40 \text{ m} \). Convert frequency from cycles per second to radians per second: \( \omega_1 = 2\pi \times 0.600 \text{ s}^{-1} \).
04

Scenario (a): Sack Falls at Maximum Displacement

At maximum displacement, velocity is zero, so energy is purely potential. Initial potential energy: \(E_1 = \frac{1}{2} k A^2\) remains constant since sack falls when beam isn't moving. After sack falls: \(E_2 = \frac{1}{2} k A_2^2\), solve for new amplitude \(A_2\). Since mass changes, the same spring constant \(k\) applies, yielding \( \omega_2 = \sqrt{\frac{k}{m_2}} = \omega_1 \sqrt{\frac{m_1}{m_2}}\). Find \( \omega_2\) and \( A_2\).
05

Scenario (b): Sack Falls at Maximum Speed

At maximum speed, energy is kinetic. Initial kinetic energy: \(E_1 = \frac{1}{2} m_1 v^2\), where \(v = \omega_1 A\). Solve for new frequency given sack falls at maximum speed using energy conservation: initial kinetic energy equals final combination of potential and kinetic with \(m_2\). New frequency: \(\omega_2\) and explore amplitude repairing initial distribution through same factors.

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Key Concepts

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

Uniform Beam
A uniform beam is a crucial element in the study of oscillations and harmonic motion. In our given scenario, a uniform beam is suspended horizontally by two identical vertical springs. This setup allows the beam to maintain an even distribution of mass and balance. The beam itself has a mass of 225 kg. When you add a 175-kg sack of gravel to its center and let the system oscillate, it continues to display even weight distribution which is ideal for studying simple harmonic motion. Let's summarize some key points:
  • The beam is "uniform", indicating it's evenly distributed in mass.
  • Springs are attached at both ends, ensuring steady support and oscillation.
  • The uniform nature allows us to predict and calculate oscillations effectively.
Understanding uniform beams helps us appreciate how mass distribution impacts oscillation behavior.
Oscillations
Oscillations take place when an object moves back and forth around an equilibrium or central position. In simple harmonic motion (SHM), which is a type of oscillation, this movement is periodic and typically follows a sine or cosine wave pattern. In our case, the uniform beam oscillates vertically, influenced by the forces from the two springs. Reflecting on some characteristics:
  • The initial oscillation period is defined by the combined mass on the beam and the characteristics of the springs.
  • When the sack of gravel falls off, the oscillation dynamics are altered, requiring recalculation.
  • The energy transformation between kinetic and potential forms during oscillations is a key focus here.
By considering both scenarios of the sack falling, we can better understand how oscillation behaviors adapt to changes in mass.
Frequency
Frequency is one of the core attributes of oscillation, determining how many cycles an object completes per unit of time. In the scenario at hand, the system initially operates with a frequency of 0.600 cycles per second. Considerations when analyzing frequency include:
  • Initially, the frequency factors in the total mass (beam plus sack).
  • After the sack's removal, the system's mass decreases, increasing the frequency's value.
  • New frequency calculations take into account the energy conservation principles, where frequency is directly related to the angular speed.
Accurately determining frequency assists in predicting how the oscillation phases shift post any sudden mass change.
Amplitude
Amplitude in oscillations refers to the utmost extent a point within the system is displaced from its equilibrium position. For our oscillating beam, the initial amplitude is given as 40.0 cm. Key points about amplitude you should note:
  • When the sack drops at maximum displacement, only potential energy is conserved, leading to a direct calculation of new amplitude.
  • When the sack drops at maximum speed, kinetic energy is conserved, affecting amplitude differently.
  • The change in amplitude reflects adjustment in the system's energy distribution after a mass is removed.
Understanding amplitude helps clarify how the system responds visually and physically to alterations in mass or conditions. This ensures accurate tracking and modeling of the oscillation behavior.

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Most popular questions from this chapter

A proud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.180 m. (a) Find the force constant of the spring. The fish is now pulled down 5.00 cm and released. (b) What is the period of oscillation of the fish? (c) What is the maximum speed it will reach?

A 0.500-kg mass on a spring has velocity as a function of time given by \({v_x}(t) = -\)(3.60 cm/s) sin[ (4.71 rad/s)\(t - \pi\)/2) ]. What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

A 0.0200-kg bolt moves with SHM that has an amplitude of 0.240 m and a period of 1.500 s. The displacement of the bolt is \(+\)0.240 m when \(t =\) 0. Compute (a) the displacement of the bolt when \(t =\) 0.500 s; (b) the magnitude and direction of the force acting on the bolt when \(t =\) 0.500 s; (c) the minimum time required for the bolt to move from its initial position to the point where \(x = -\)0.180 m; (d) the speed of the bolt when \(x = -\)0.180 m.

When a 0.750-kg mass oscillates on an ideal spring, the frequency is 1.75 Hz. What will the frequency be if 0.220 kg are (a) added to the original mass and (b) subtracted from the original mass? Try to solve this problem \(without\) finding the force constant of the spring.

A 5.00-kg partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.20 s. (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is 0.050 m above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?
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