Act like a helpful tutor and exlain me :
A mass of
1
0
0
kg is suspended from a vertical coil spring of stiffness
2
0
k
N
m
.
The mass is pulled downwards a definite distance, and then released.

Question

Act like a helpful tutor and exlain me :
A mass of 
1
0
0
 ﻿kg is suspended from a vertical coil spring of stiffness 
2
0
k
N
m
.
 ﻿The mass is pulled downwards a definite distance, and then released. The vibration amplitude is reduced to 
0
.
0
1
0
 ﻿of the original in three complete oscillations. What is the damping ratio of the system?
a
.
 
0
.
5
9
b
.
 
0
.
1
8
c
.
 
0
.
9
1
d
.
 
0
.
2
4
e
.
 ﻿None of the above
Natural frequency of a given system is
a
.
 ﻿variable
b
.
 ﻿constant
c
.
 ﻿complex
d
.
 ﻿infinite
e
.
 ﻿None of the above
The most salient features of any vibration are
a
.
 ﻿Amplitude Find the amplitude of transient oscillation 
(
mm
)
a
.
 
,
5
0
.
9
0
b
.
 
1
8
.
9
7
c
.
 
6
3
.
2
1
d
.
 
7
1
.
5
5
e
.
 ﻿None of the above
Find the phase angle 
(
rad
)
 ﻿of steady state response
a
.
 
2
.
1
6
b
.
 
1
.
0
2
c
.
 
4
.
2
1
d
.
 
1
.
3
3
e
.
 ﻿None of the above

VIPSolutions · Accepted Answer

# Damped Oscillation of a Mass-Spring System

Let's break down your questions, step-by-step.

---

## 1. **Damping Ratio Calculation**

### **Given:**
- Mass, $ m = 100 $ kg
- Spring stiffness, $ k = 20\, \text{kN/m} = 20,000\, \text{N/m} $
- Amplitude after 3 oscillations is $ 0.010 $ (i.e., $ 1\% $) of original amplitude.

### **Step 1: Calculate Natural Frequency**
$$
\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{20,000}{100}} = \sqrt{200} \approx 14.14\, \text{rad/s}
$$

### **Step 2: Logarithmic Decrement**
The ratio of amplitudes after $ n $ cycles is related to damping by:
$$
\delta = \frac{1}{n} \ln\left(\frac{A_0}{A_n}\right)
$$
Given $ n = 3 $, $ A_3 = 0.01\,A_0 $:
$$
\delta = \frac{1}{3} \ln\left(\frac{A_0}{0.01A_0}\right) = \frac{1}{3} \ln(100) = \frac{1}{3} \times 4.605 = 1.535
$$

### **Step 3: Damping Ratio**
$$
\delta = 2\pi\zeta / \sqrt{1-\zeta^2}
$$
For small $\zeta$, $ \sqrt{1-\zeta^2} \approx 1 $, so:
$$
\delta \approx 2\pi\zeta
$$
$$
\zeta \approx \frac{\delta}{2\pi} = \frac{1.535}{6.283} \approx 0.244
$$

**Closest answer:**  
**d. 0.24**

---

## 2. **Natural Frequency of a System**

- The **natural frequency** of a given mass-spring system (with fixed mass and stiffness) is a **constant** value.

**Correct answer:**  
**b. constant**

---

## 3. **Salient Features of Vibration**

- The most salient (important) features of any vibration are:  
  - Amplitude
  - Frequency/natural frequency
  - Phase
  - Damping

**From your options:**  
**a. Amplitude** (but full answer would list all above)

---

## 4. **Amplitude of Transient Oscillation**

This requires more context (possibly a missing calculation or data), but usually, you use the initial conditions and exponential decay due to damping to find the amplitude.

---

## 5. **Phase Angle of Steady State Response**

Without a forced frequency or more detail, we cannot calculate exactly.  
But, if you provide the frequency ratio or excitation frequency, use:
$$
\tan\phi = \frac{2\zeta (\omega/\omega_n)}{1 - (\omega/\omega_n)^2}
$$
Choose the answer that best matches the calculation given the system's parameters.

---

# **Summary Table**

| Question                                 | Answer      |
|-------------------------------------------|-------------|
| Damping ratio                            | d. 0.24     |
| Natural frequency is                     | b. constant |
| Salient feature                          | a. Amplitude|
| Amplitude of transient (needs data)      | -           |
| Phase angle (needs frequency info)        | -           |

---

**Let me know if you need the missing calculations or if you have additional details!**

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