Chapter 14: Problem 13
A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At \(t\) = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.
Short Answer
Step by step solution
Identify the Equations for Simple Harmonic Motion
Calculate the Angular Frequency
Use Initial Conditions to Find Amplitude A
Solve for the Amplitude A
Verify and Write the Equation of Motion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Angular Frequency
\[ \omega = \sqrt{\frac{k}{m}} \]where:
- \( k \) is the spring constant.
- \( m \) is the mass of the block.
\[ \omega = \sqrt{\frac{300}{2.00}} = 12.25 \text{ rad/s} \]This result means the block oscillates with an angular frequency of 12.25 radians per second, indicating a brisk oscillation speed.
Amplitude
At the outset (\( t = 0 \)), the block is moving at -12 m/s, but it is not displaced from its equilibrium (\( x(0) = 0 \)). This velocity condition helps us find:
\[ A \sin(\phi) = \frac{12}{12.25} \approx 0.98 \]Since the position at \( t = 0 \) is 0, the phase angle \( \phi = \frac{\pi}{2} \), making \( \sin(\phi) = 1 \). Therefore, the calculated amplitude:
\[ A = 0.98 \]is the maximum distance the block will travel from its equilibrium position during its motion.
Phase Angle
The given condition \( x(0) = 0 \) with a negative velocity requires us to solve \( A \cos(\phi) = 0 \). This gives:
- \( \phi = \frac{\pi}{2} \)
Ideal Spring
- It has infinite cycles of oscillation without losing energy.
- Its behavior is entirely predictable and linear.
Spring Constant
For this problem, the spring constant is 300 N/m. This means:
- The spring needs 300 Newtons of force to stretch or compress by one meter.
- It influences both the angular frequency and amplitude of the oscillation.