An undamped 2.11 kg horizontal spring oscillator has a spring constant of 37.9 N/m. While oscillating, it is found to have a speed of 2.33 m/s as it passes through its equilibrium position. What is its amplitude A of oscillation? A = 0.38874296 m Incorrect What is the oscillator's total mechanical energy Eot as it passes through a position that is 0.711 of the amplitude away from the equilibrium position? E1ot J II
Simple harmonic motion
Simple harmonic motion is a type of periodic motion in which an object undergoes oscillatory motion. The restoring force exerted by the object exhibiting SHM is proportional to the displacement from the equilibrium position. The force is directed towards the mean position. We see many examples of SHM around us, common ones are the motion of a pendulum, spring and vibration of strings in musical instruments, and so on.
Simple Pendulum
A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.
Oscillation
In Physics, oscillation means a repetitive motion that happens in a variation with respect to time. There is usually a central value, where the object would be at rest. Additionally, there are two or more positions between which the repetitive motion takes place. In mathematics, oscillations can also be described as vibrations. The most common examples of oscillation that is seen in daily lives include the alternating current (AC) or the motion of a moving pendulum.
![## Oscillator Dynamics and Energy Calculation
### Problem Statement
An undamped 2.11 kg horizontal spring oscillator has a spring constant of 37.9 N/m. While oscillating, it is found to have a speed of 2.33 m/s as it passes through its equilibrium position.
1. **Amplitude of Oscillation:**
- **Given:**
- Mass (\( m \)) = 2.11 kg
- Spring constant (\( k \)) = 37.9 N/m
- Speed at equilibrium position (\( v \)) = 2.33 m/s
- **Objective:** Calculate the amplitude (\( A \)) of oscillation.
- **Incorrect Attempt:** \( A = 0.38874296 \) m
2. **Total Mechanical Energy:**
- **Position:** 0.711 of the amplitude away from equilibrium.
- **Objective:** Calculate the total mechanical energy (\( E_{\text{tot}} \)) at this position.
- **Incorrect Attempt:** The answer for \( E_{\text{tot}} \) was not provided.
### Explanation
1. **Amplitude Calculation:**
- The energy conservation in a spring oscillator:
\[
\frac{1}{2}mv^2 = \frac{1}{2}kA^2
\]
- Solve for amplitude (\( A \)):
\[
A = \sqrt{\frac{mv^2}{k}}
\]
2. **Total Mechanical Energy Calculation:**
- Total mechanical energy remains constant and is given by:
\[
E_{\text{tot}} = \frac{1}{2}kA^2
\]
- At position \( x = 0.711A \), decompose energy into kinetic and potential parts:
\[
E_{\text{tot}} = \frac{1}{2}k(x^2) + \frac{1}{2}mv^2
\]
**Note:** The specific numeric inputs need to be recalculated for both the amplitude and total energy to verify correctness.
### Graph/Diagram Explanation
There are no graphs or diagrams included in this problem. The focus is on solving equations for amplitude and total energy in a spring oscillator scenario.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4fcfcba-e514-40cd-a23f-a99b46943f83%2F6cce9487-9e0e-4e95-a32d-1689f7f20bdf%2Fijwjfl4_processed.png&w=3840&q=75)
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