—_— 01:29:24 =
| Ouse: :
BI Abeer mace ras mos 1500901 Pring The spring stlness 15 45 kim The damping is 30% of the crtcal ake. Cakculate he ye I.
Ee -
Cs 3
1
- a

Question

—_— 01:29:24 =
| Ouse: :
BI Abeer mace ras mos 1500901 Pring The spring stlness 15 45 kim The damping is 30% of the crtcal ake. Cakculate he ye I.
Ee -
Cs 3
1
- a

VIPSolutions · Accepted Answer

Let's break down the problem:

**Given:**
- Mass, $ m = 150 $ kg
- Spring stiffness, $ k = 45\, \text{kN/m} = 45000\, \text{N/m} $
- Damping is 30% of the critical value ($ \xi = 0.3 $)
- Need to calculate the **logarithmic decrement ($ \delta $)**

### Step 1: Natural Frequency ($ \omega_n $)
$$
\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{45000}{150}} = \sqrt{300} \approx 17.32\, \text{rad/s}
$$

### Step 2: Logarithmic Decrement ($ \delta $)
$$
\delta = 2\pi \xi / \sqrt{1-\xi^2}
$$
where $ \xi $ is the damping ratio.

Given $ \xi = 0.3 $:

$$
\sqrt{1-\xi^2} = \sqrt{1 - 0.3^2} = \sqrt{0.91} \approx 0.9539
$$

$$
\delta = \frac{2\pi \times 0.3}{0.9539} \approx \frac{1.884}{0.9539} \approx 1.975
$$

### Step 3: Match with options
The closest answer is **1.97**, which matches option 1.

---

**Final Answer:**  
$$
\boxed{1.97}
$$

✨ VIPSolutions

—_— 01:29:24 = | Ouse: : BI Abeer mace ras mos 1500901 Pring The spring stlness 15 45 kim The damping is 30% of the crtcal ake. Cakculate he ye I. Ee - Cs 3 1 - a

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