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Chapter 14: Problem 77

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?

Short Answer

Expert verified
(a) 0.149 m/s. (b) -0.111 m/s虏. (c) 1.05 s. (d) 0.438 m.

Step by step solution

01

Determine the spring constant

To find the spring constant \(k\), we use the formula for the period \(T\) of a mass-spring system: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where \(m = 5.00\, \text{kg}\) and \(T = 4.20\, \text{s}\). Solving for \(k\), we get: \[ k = \frac{4\pi^2m}{T^2} \] Substituting the given values: \[ k = \frac{4\cdot(3.1416)^2 \cdot 5.00}{(4.20)^2} = 11.17 \, \text{N/m} \]
02

Calculate speed at equilibrium position

The maximum speed occurs at the equilibrium position. We use the formula \( v_{\text{max}} = \omega A \), where \( \omega = \sqrt{\frac{k}{m}} \) and \(A = 0.100\, \text{m}\). First, find \(\omega\): \[ \omega = \sqrt{\frac{11.17}{5.00}} = 1.49 \, \text{s}^{-1} \] Then calculate the speed: \[ v_{\text{max}} = 1.49 \, \text{s}^{-1} \times 0.100\, \text{m} = 0.149\, \text{m/s} \]
03

Find acceleration 0.050 m above equilibrium

The acceleration \(a\) in a mass-spring system is given by \( a = -\omega^2 x \), where \( x = 0.050\, \text{m} \) and \(\omega = 1.49 \, \text{s}^{-1} \). Substitute the values: \[ a = -(1.49)^2 \times 0.050 = -0.111 \, \text{m/s}^2 \] The negative sign indicates direction.
04

Calculate time to move through equilibrium region

The object moves from \(-0.050\, \text{m}\) to \(0.050\, \text{m}\). The amplitude \(A\) is \(0.100\, \text{m}\). This motion covers a quarter of a period. The time \(t_{1/4}\) to move this distance is \[ t_{1/4} = \frac{T}{4} = \frac{4.20}{4} = 1.05\, \text{s} \]
05

Calculate spring shortening after stop and removal

When the mass is removed, the spring shortens by its equilibrium extension. Using Hooke's Law, \(F = kx = mg\), solving for \(x\): \[ x = \frac{mg}{k} = \frac{5.00 \times 9.81}{11.17} = 4.38 \times 10^{-1}\, \text{m} \] Converts to: \( 0.438\, \text{m} \), so the spring shortens 0.438 m.

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Key Concepts

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

Spring Constant
The spring constant, often denoted by the symbol \(k\), is a fundamental property of any spring, representing its stiffness. In essence, it defines how much force is needed to stretch or compress the spring by a certain amount. To find the spring constant, we use the formula for the period of a mass-spring system:
  • \(T = 2\pi \sqrt{\frac{m}{k}}\)
Rearranging this formula helps us solve for \(k\):
  • \(k = \frac{4\pi^2m}{T^2}\)
Understanding the spring constant is crucial because it directly impacts the oscillation frequency of any harmonic oscillator like the mass-spring system being studied. A higher spring constant means a stiffer spring, requiring more force to stretch or compress.
Harmonic Oscillator
A harmonic oscillator is a system in which an object experiences a force that is directly proportional to its displacement from its equilibrium position, but in the opposite direction. Essentially, if you pull or push the object away from equilibrium, it will tend to return to that position. This returning force also attempts to restore the object to its original state, causing oscillations.
  • Period (\(T\)): the time it takes for one complete cycle of motion.
  • Frequency (\(f\)): how often the cycles occur per second, depends on the spring constant \(k\) and mass \(m\).
  • Angular Frequency (\(\omega\)): \(\omega = \sqrt{\frac{k}{m}}\).
In the case of our partridge suspended from a spring, this behavior is a prime example of simple harmonic motion as it vibrates through its equilibrium position.
Hooke's Law
Hooke's Law is a principle that defines the behavior of springs. It states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed. Mathematically, it's expressed as:
  • \(F = kx\)
Where:
  • \(F\) is the force exerted by the spring.
  • \(k\) is the spring constant.
  • \(x\) is the displacement of the spring from its rest position.
In the exercise, when the partridge is removed from the spring, Hooke's Law helps us calculate how much the spring shortens. This calculation is crucial for understanding how energy and force are stored and released in the system.
Mass-Spring System
A mass-spring system is a classic example of a simple harmonic oscillator, consisting of a mass attached to a spring that can move vertically or horizontally. This type of system is ideal for studying motion because of its predictable and repetitive nature.
  • The mass affects the period and frequency of the oscillations.
  • The spring constant determines the stiffness and the system's response to displacement.
In this system: - The equilibrium position is the point where the net force is zero, and the spring is not stretched or compressed. - As the mass moves, it experiences restoring forces due to the spring, leading to oscillatory motion. Such a system is crucial for understanding broader concepts in physics, such as energy conservation, as potential energy is converted to kinetic energy during the damping of the oscillations. When you understand a mass-spring system, you gain insights into a fundamental part of mechanical systems in physics.

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Most popular questions from this chapter

A 1.35-kg object is attached to a horizontal spring of force constant 2.5 N/cm. The object is started oscillating by pulling it 6.0 cm from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only 3.5 cm. (a) How much energy has this system lost to damping during these eight cycles? (b) Where did the "lost" energy go? Explain physically how the system could have lost energy.

Two uniform solid spheres, each with mass \(M =\) 0.800 kg and radius \(R =\) 0.0800 m, are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant \(k =\) 160 N/m has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

A 2.00-kg bucket containing 10.0 kg of water is hanging from a vertical ideal spring of force constant 450 N/m and oscillating up and down with an amplitude of 3.00 cm. Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 g/s. When the bucket is half full, find (a) the period of oscillation and (b) the rate at which the period is changing with respect to time. Is the period getting longer or shorter? (c) What is the shortest period this system can have?

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 m from its equilibrium position and released with zero initial speed, then after 0.800 s its displacement is found to be 0.120 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.

A 2.00-kg frictionless block attached to an ideal spring with force constant 315 N/m is undergoing simple harmonic motion. When the block has displacement \(+\)0.200 m, it is moving in the negative \(x\)-direction with a speed of 4.00 m/s. Find (a) the amplitude of the motion; (b) the block's maximum acceleration; and (c) the maximum force the spring exerts on the block.
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