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This problem involves a mass-spring system, so we can use the simple harmonic motion model to solve for \( u(t) \), the position of the mass over time.

### Given Data:
- Mass \( m = 102 \, \text{g} = 0.102 \, \text{kg} \)
- Spring stretch due to weight \( \Delta x = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Initial velocity \( v(0) = -12 \, \text{cm/s} = -0.12 \, \text{m/s} \)
- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)

### Step 1: Find the Spring Constant \( k \)
The force due to gravity is balanced by the spring's restoring force at equilibrium:
\[
mg = k \Delta x
\]
Solving for \( k \):
\[
k = \frac{mg}{\Delta x} = \frac{(0.102 \, \text{kg})(9.8 \, \text{m/s}^2)}{0.1 \, \text{m}} = 9.996 \, \text{N/m}
\]
We'll approximate \( k \approx 10 \, \text{N/m} \).

### Step 2: Write the Differential Equation
The equation of motion for a mass-spring system without damping is:
\[
m \frac{d^2u}{dt^2} + k u = 0
\]
Substituting \( m = 0.102 \, \text{kg} \) and \( k = 10 \, \text{N/m} \):
\[
0.102 \frac{d^2u}{dt^2} + 10 u = 0
\]
Dividing through by \( 0.102 \):
\[
\frac{d^2u}{dt^2} + \frac{10}{0.102} u = 0
\]
\[
\frac{d^2u}{dt^2} + 98.04 u = 0
\]

This is a simple harmonic oscillator with angular frequency:
\[
\omega = \sqrt{98.04} \approx 9.9 \, \text{rad/s}
\]

### Step 3: General Solution
The general solution for simple harmonic motion is:
\[
u(t) = A \cos(\omega t) + B \sin(\omega t)
\]
where \( A \) and \( B \) are constants determined by initial conditions.

### Step 4: Apply Initial Conditions
1. **Initial Position**: At \( t = 0 \), the mass is at equilibrium, so \( u(0) = 0 \).
   \[
   u(0) = A \cos(0) + B \sin(0) = A = 0
   \]
   Therefore, \( A = 0 \).

2. **Initial Velocity**: The initial downward velocity is \( v(0) = -0.12 \, \text{m/s} \).
   \[
   v(t) = \frac{du}{dt} = -A \omega \sin(\omega t) + B \omega \cos(\omega t)
   \]
   At \( t = 0 \):
   \[
   v(0) = B \omega = -0.12
   \]
   Solving for \( B \):
   \[
   B = \frac{-0.12}{\omega} = \frac{-0.12}{9.9} \approx -0.0121
   \]

Thus, the solution for \( u(t) \) is:
\[
u(t) = -0.0121 \sin(9.9 t)
\]

### Part (a): Final Answer for \( u(t) \)
\[
u(t) = -0.0121 \sin(9.9 t) \, \text{m}
\]

### Part (b): Time to Return to Equilibrium
The mass returns to its equilibrium position when \( u(t) = 0 \). Since \( u(t) = -0.0121 \sin(9.9 t) \), the first time this happens after \( t = 0 \) is when:
\[
\sin(9.9 t) = 0
\]
The smallest positive value for \( t \) is when \( 9.9 t = \pi \):
\[
t = \frac{\pi}{9.9} \approx 0.317 \, \text{s}
\]

### Final Answer for Part (b)
\[
t \approx 0.317 \, \text{s}
\]

Would you like further details on any part of this solution, or do you have any questions?

Here are five additional questions that build on this topic:
1. How would damping affect the motion of the mass?
2. What would happen to the period if the mass were doubled?
3. How would the system behave if there were an initial displacement but no initial velocity?
4. How does the spring constant \( k \) influence the frequency of oscillation?
5. What is the effect of gravity on this system in the vertical direction?

**Tip:** When working with oscillatory systems, keep an eye on units to ensure consistent results, especially when calculating spring constants and forces.

A mass of 102 g stretches a spring 10 cm. The mass is set in motion from its equilibrium position with a downward velocity of 12 cm/s and no damping is applied. a. Determine the position u of the mass at any time t. Use 9.8 m/s^2 as the acceleration due to gravity. b. When does the mass first return to its equilibrium position?

Explore how to determine the position of a mass in simple harmonic motion and calculate the time it returns to equilibrium. This problem uses principles of harmonic oscillation, differential equations, and Hooke's Law, ideal for advanced high school or introductory college physics students.

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