91Ó°ÊÓ

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

Step by step solution

01

Determine total mechanical energy

We first determine the total mechanical energy, which is equal to the potential energy at the maximum displacement (which is $0.100 \mathrm{m}$ from the equilibrium position). This is given by \(\frac{1}{2} kx^2 = mgh\), where \(k\) is the spring constant, \(x\) is the displacement, \(m\) is the mass of the partridge, and \(h\) is the maximum height. We can find the spring constant using the known period of oscillation \(T = 2\pi \sqrt{\frac{m}{k}}\). Rearranging for \(k\), we get \(k = \frac{4\pi^2m}{T^2}\), subbing this into the equation for total mechanical energy we will get a value for speed.

02

Determine acceleration

Acceleration in simple harmonic motion is given by \(a = -\omega^2x\), where \(\omega\) is the angular frequency and \(x\) is displacement from the equilibrium position. Angular frequency \(\omega\) is equal to \( \frac{2\pi}{T}\). Here, \(x = 0.050 \mathrm{m}\), the displacement above the equilibrium position.

03

Determine time for specific displacement

The time required for the partridge to move from $0.050 \mathrm{m}$ below its equilibrium position to $0.050 \mathrm{m}$ above it is half of the total period of oscillation, which is \(T/2 = 4.20 \mathrm{s}/2\).

04

Determine spring shortness

The displacement of the spring when the partridge is removed is equal to the equilibrium position of the partridge. This can be determined using Hooke’s Law: \(F = -kx\). Here, \(F = mg\) and \(x\) is the spring shortening we’re trying to determine.

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Key Concepts

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

Spring Constant

In simple harmonic motion, the spring constant, denoted by \( k \), plays a crucial role. It quantifies the stiffness of a spring, indicating how much force is needed to change its length by a certain amount. The larger the spring constant, the stiffer the spring is.

• The spring force follows Hooke's Law: \( F = -kx \), where \( F \) is the force applied, \( x \) is the displacement from the equilibrium position, and \( k \) is the spring constant.
• This constant is determined using data from simple harmonic motion. For instance, in the original exercise, given the mass \( m \) of the partridge and the period \( T \) of oscillation, we could find \( k \) using the formula: \[ k = \frac{4\pi^2m}{T^2} \]

Understanding \( k \) helps in knowing how a system will oscillate when a mass is attached to the spring. A higher \( k \) means quicker oscillations because the spring is stiffer.

Period of Oscillation

The period of oscillation, represented by \( T \), is the time taken for one complete cycle of movement in an oscillatory system. For a mass-spring system, knowing the period is essential to understanding the system's dynamics.

• The formula linking the period with mass and spring constant is: \[ T = 2\pi \sqrt{\frac{m}{k}} \]
• The period conditions how quickly a system returns to its starting point. A larger period means the system oscillates more slowly.
• In the exercise, the calculated period of \( 4.20 \) seconds helped us derive other parameters such as the spring constant and the mechanical energy.

Recognizing the period allows one to predict how and when certain events in the oscillatory cycle will occur, crucial for timing calculations in real-world applications.

Mechanical Energy

Mechanical energy in the context of simple harmonic motion is the sum of potential energy stored in the spring and the kinetic energy of the oscillating mass.

• The potential energy at maximum displacement (where velocity is zero) is given by: \[ PE = \frac{1}{2}kx^2 \]
• Kinetic energy is maximum at the equilibrium point where potential energy is zero: \[ KE = \frac{1}{2}mv^2 \]
• Total mechanical energy remains constant during the motion, i.e., it oscillates between kinetic and potential forms as energy is conserved in perfect circumstances: \[ E = \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \]

In solving problems, it’s crucial to know that energy transformation within the system doesn’t change the total energy, providing a method to find velocities or other unknowns with ease.

Angular Frequency

Angular frequency, denoted by \( \omega \), is a measure of how quickly an object oscillates in radians per second in circular or harmonic motion. It is connected to both the frequency of oscillation and the period.

• Angular frequency is determined through the formula: \[ \omega = \frac{2\pi}{T} \]
• It connects with acceleration in simple harmonic motion through the expression: \[ a = -\omega^2x \]
• Angular frequency provides insight into how quickly the system repeats its cycles.

Understanding \( \omega \) helps in comprehending the motion's rapidity and forms a bridge between linear and rotational perspectives. In the exercise, knowing \( \omega \) allowed the calculation of parameters such as acceleration and contributed to a deeper understanding of the partridge's motion. By mastering these concepts, students can better grasp the dynamics of oscillatory systems in physics.