VIPSolutions- TE 2/18 Compltod mm 01:25:18 — . 2 RE [Mieston A fi Ce . BRR 1) chine hs as of 50 ky anc 5 supp on deshors an sins. The sping sess 45 i. The 401909 1: 30% fe cl vake Cacee decrement. gi 137 Ri 191 NN Boe | RN ki REL TEE HON 128 Bh | RE 6 hone sm SR i Cero Previous Guessen i RR AA RRR 1 RN = LA ARR Er IRA ai
Question:
- TE
2/18 Compltod mm 01:25:18 — .
2 RE
[Mieston A
fi Ce
. BRR
1) chine hs as of 50 ky anc 5 supp on deshors an sins. The sping sess 45 i. The 401909 1: 30% fe cl vake Cacee
decrement. gi
137 Ri
191 NN
Boe
| RN
ki REL
TEE
HON
128 Bh
| RE
6 hone sm SR
i Cero
Previous Guessen i
RR
AA
RRR
1 RN
= LA
ARR
Er IRA
ai
This is AI generated Response.
Asked by: Veera manikanta
Created at: 2025-08-06 22:53:57
Asked by: Veera manikanta
Created at: 2025-08-06 22:53:57
Final Answer:
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Let's break down the given problem:
**Given:**
- Mass, \( m = 150 \) kg
- Spring stiffness, \( k = 45 \) kN/m = \( 45000 \) N/m
- Damping is 30% of the critical value (\( \xi = 0.3 \))
We are to calculate **logarithmic decrement** \( \delta \).
---
### Step 1: Formula for Logarithmic Decrement
\[
\delta = \frac{2\pi \xi}{\sqrt{1-\xi^2}}
\]
Where \( \xi \) is the damping ratio.
Given: \( \xi = 0.3 \)
---
### Step 2: Substitute the values
\[
\delta = \frac{2\pi \times 0.3}{\sqrt{1 - (0.3)^2}}
\]
\[
= \frac{2\pi \times 0.3}{\sqrt{1 - 0.09}}
\]
\[
= \frac{2\pi \times 0.3}{\sqrt{0.91}}
\]
\[
= \frac{1.884}{0.9539}
\]
\[
\approx 1.97
\]
---
### **Final Answer**
The value of logarithmic decrement is **1.97**.
So, the correct option is:
- **1.97**
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