Problem 83 - University Physics with Modern Physics

Question

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 \(\mathrm{m}\) 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

Answer

(a) 0.150 m/s, (b) -0.112 m/s², (c) 1.05 s, (d) 27.6 cm.

Step 1: Understanding Simple Harmonic Motion

The partridge attached to a spring exhibits simple harmonic motion. Given the period \( T = 4.20 \text{ s} \) and amplitude \( A = 0.100 \text{ m} \). We will use these properties to find velocity, acceleration, and time in various cases.

Step 2: Find the Spring Constant

We use the formula for the period of a spring-mass system: \(T = 2\pi \sqrt{\frac{m}{k}}\)where \( m = 5.00 \text{ kg} \). Solving for \( k \), we have:\(k = \frac{4\pi^2m}{T^2}\)Substitute the given values:\[k = \frac{4\pi^2 \times 5.00}{4.20^2} \approx 1.78 \, \text{N/m}\]

Step 3: Calculate Maximum Speed

The maximum speed occurs at the equilibrium position:\(v_{max} = A\omega\)where \( \omega = \frac{2\pi}{T} = \frac{2\pi}{4.20} \approx 1.496 \text{ rad/s} \), and \( A = 0.100 \text{ m} \). So,\(v_{max} \approx 0.100 \times 1.496 \approx 0.150 \text{ m/s}\)

Step 4: Calculate Acceleration at a Point

At \( x = -0.050 \text{ m} \), the displacement is \( -0.050 \text{ m} \). Use the formula:\(a = -\omega^2 x\)Substitute \( x = 0.050 \text{ m} \):\(a = -(1.496)^2 \times 0.050 \approx -0.112 \text{ m/s}^2\)

Step 5: Determine Time to Move Between Two Points

To find the time taken to move from \( x = -0.050 \text{ m} \) to \( x = 0.050 \text{ m} \), consider one quarter of the period because the motion is symmetrical. Thus, the time taken is \( \frac{T}{4} \approx \frac{4.20}{4} = 1.05 \text{ s} \).

Step 6: Determine the Stretched Length of the Spring

The spring force equals the weight of the partridge at equilibrium:\(k \Delta x = mg\)Solve for \( \Delta x \):\(\Delta x = \frac{mg}{k} = \frac{5.00 \times 9.81}{1.78} \approx 27.6 \text{ cm}\)

Key Concepts

Spring Constant

The spring constant, often symbolized as \( k \), is a crucial parameter in Hooke's Law of elasticity. It measures the stiffness of a spring. In simple harmonic motion, such as that of a mass-spring system, the spring constant is vital.
The formula for the period \( T \) of a mass \( m \) attached to a spring is given by:\[T = 2\pi \sqrt{\frac{m}{k}}\]where:

• \( T \) is the period of oscillation.
• \( m \) is the mass attached to the spring.
• \( k \) is the spring constant.

Solving for \( k \), you rearrange the formula:\[k = \frac{4\pi^2m}{T^2}\]Using this expression, you determine how stiff or flexible the spring attached to the partridge is. The spring constant influences both the frequency and the energy stored in the spring during oscillation.

Maximum Speed

The maximum speed of an object in simple harmonic motion is a crucial aspect. This speed occurs when the object passes through the equilibrium position.
The formula to calculate the maximum speed \( v_{max} \) is:\[v_{max} = A\omega\]where:

• \( A \) is the amplitude of motion, the maximum displacement from the equilibrium position.
• \( \omega \) is the angular frequency, given by \( \omega = \frac{2\pi}{T} \).

This speed is of particular interest because, at this point, all of the system's potential energy is converted into kinetic energy, showing the energy dynamics of the system. Calculating \( v_{max} \) helps to understand how rapidly the mass moves through its path.

Acceleration in Simple Harmonic Motion

Acceleration in simple harmonic motion is all about how quickly the velocity of the object changes. It is directly related to the position of the object.
The formula for acceleration \( a \) is:\[a = -\omega^2 x\]where:

• \( \omega \) is the angular frequency.
• \( x \) is the displacement from the equilibrium position.

The negative sign indicates that the acceleration is always directed towards the equilibrium position—this is a characteristic feature of restorative forces in simple harmonic motion. Knowing the acceleration at a specific point can provide insights into how forces act within the oscillating system.

Period of Oscillation

The period of oscillation \( T \) refers to the time taken for one complete cycle of motion. It is a fundamental aspect of the motion's dynamics.
In the context of simple harmonic motion, the period is determined by the mass of the object and the spring constant using:\[T = 2\pi \sqrt{\frac{m}{k}}\]Periodic motion is predictable and repetitive, making it easier to analyze. Knowing the period allows you to calculate how long it takes for an object to return to the same point in its motion. Understanding the changes in period with different masses or spring constants can offer deeper insights into the mechanical properties of oscillating systems.