give answer in 2 step with explanation at the end of each step and final answer at the end:6.10 In this problem you will derive the equations of motion for the three - spring system. The only physics

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:6.10 In this problem you will derive the equations of motion for the three - spring system. The only physics you need is F = ma and the fact that a spring exerts a force F = -kx where k is the spring constant and x is the displacement. (a) The position x1 = x2 = 0 represents the equilibrium position. Now imagine that Ball 1 is at this position precisely, but Ball 2 is slightly to the right of this position, as shown in Figure 6.1. Which springs now push on Ball 1, and in which directions? Which springs now push on Ball 2, and in which directions? (b) Now imagine that Ball 1 is displaced to the right by a distance x1, and Ball 2 is displaced to the right by x2. This time the answers will be quantitative, and they will include the three spring constants k1, k2, and k3. We begin with Ball 1. i. The force of Spring 1 on Ball 1 depends only on the position x1 (the other position is irrelevant). Write a formula for this force. ii. How much is Spring 2 stretched? Your answer should be a function of x1 and x 2. For example, if both balls are displaced the same amount to the right then Spring 2 isn't

VIPSolutions · Accepted Answer

# Three-Spring System: Equations of Motion

## (a) Force Directions When Only Ball 2 Is Displaced

**Step 1:**  
**Which springs push on Ball 1, and in which directions?**

- **Spring 1:** Only attached to Ball 1 and the wall; with $x_1 = $, this spring is at equilibrium (no force).
- **Spring 2:** Connects Ball 1 and Ball 2. With $x_1 = $ and $x_2 > $, Spring 2 is stretched to the right.
    - This means Spring 2 pulls Ball 1 to the right (positive direction).
- **Spring 3:** Only attached to Ball 2 and the right wall, so it does not act on Ball 1.

**Explanation:**  
Ball 1 feels a force only from Spring 2, pulling to the right.

---

**Step 2:**  
**Which springs push on Ball 2, and in which directions?**

- **Spring 1:** Does not act on Ball 2.
- **Spring 2:** Connects Ball 1 and Ball 2. With $x_2 > $, Spring 2 pulls Ball 2 to the left (back toward Ball 1).
- **Spring 3:** Ball 2 is right of equilibrium, so Spring 3 is stretched and pulls Ball 2 to the left (negative direction).

**Explanation:**  
Ball 2 feels forces from both Spring 2 and Spring 3, both pulling it to the left.

---

## (b) Forces and Equation for Ball 1

### (i) Force of Spring 1 on Ball 1

**Step 1:**  
The force from Spring 1 (attached to wall and Ball 1) is:
$$
F_{\text{Spring 1 on 1}} = -k_1 x_1
$$

**Explanation:**  
This is Hooke's Law; force is proportional and opposite to displacement from equilibrium.

---

### (ii) Stretch of Spring 2

**Step 2:**  
Spring 2 connects Ball 1 ($x_1$) and Ball 2 ($x_2$). The amount Spring 2 is stretched from its equilibrium length is:
$$
\Delta x_2 = x_2 - x_1
$$

**Explanation:**  
If both balls are displaced equally, $\Delta x_2 = $, as required. Positive $\Delta x_2$ means Spring 2 is stretched.

---

### (iii) Force of Spring 2 on Ball 1

**Step 3:**  
Spring 2 pulls Ball 1 toward Ball 2, so the force on Ball 1 is:
$$
F_{\text{Spring 2 on 1}} = k_2 (x_2 - x_1)
$$

**Explanation:**  
The force is positive if $x_2 > x_1$ (stretched right), negative if compressed.

---

### (iv) Differential Equation for $x_1(t)$

**Step 4:**  
Net force on Ball 1:
$$
F_{\text{net,1}} = F_{\text{Spring 1 on 1}} + F_{\text{Spring 2 on 1}}
= -k_1 x_1 + k_2 (x_2 - x_1)
$$

Newton’s Second Law ($F = m_1 \ddot{x}_1$):
$$
m_1 \ddot{x}_1 = -k_1 x_1 + k_2 (x_2 - x_1)
$$

**Explanation:**  
Both spring forces included, with correct signs for restoring forces.

---

## (c) Equation for Ball 2

**Step 1:**  
Forces on Ball 2:
- From Spring 2: $F_{2,\text{from S2}} = -k_2 (x_2 - x_1)$ (opposite to force on Ball 1)
- From Spring 3: $F_{2,\text{from S3}} = -k_3 x_2$ (attached to wall)

Net force:
$$
F_{\text{net,2}} = -k_2 (x_2 - x_1) - k_3 x_2
$$

Newton’s Second Law ($F = m_2 \ddot{x}_2$):
$$
m_2 \ddot{x}_2 = -k_2 (x_2 - x_1) - k_3 x_2
$$

**Explanation:**  
Both forces are restoring, acting to return Ball 2 to equilibrium or equalize positions.

---

## **Final Answers**

### (b) Equation for Ball 1:
$$
m_1 \ddot{x}_1 = -k_1 x_1 + k_2 (x_2 - x_1)
$$

### (c) Equation for Ball 2:
$$
m_2 \ddot{x}_2 = -k_2 (x_2 - x_1) - k_3 x_2
$$

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