For a damped oscillation m=500 gm ,k=100 N/m,b=75 gm/sec. What is the ratio of amplitue of damped oscillation to the intial amplitude at the end of 20 cycles
`["given" 1/(e^(0.675)=0.5,1/(e^0.51)=0.6)]`

Options

A. 0.6
B. 0.51
C. 0.4
D. none

Answer

To solve the problem of finding the ratio of the amplitude of damped oscillation to the initial amplitude at the end of 20 cycles, we can follow these steps:

### Step 1: Convert Given Values
Convert the mass and damping coefficient to SI units.
- Mass \( m = 500 \, \text{g} = 0.5 \, \text{kg} \)
- Damping coefficient \( b = 75 \, \text{g/s} = 0.075 \, \text{kg/s} \)
- Spring constant \( k = 100 \, \text{N/m} \)

### Step 2: Calculate the Time Period
The time period \( T \) of the oscillation is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting the values:
\[ T = 2\pi \sqrt{\frac{0.5}{100}} = 2\pi \sqrt{0.005} = 2\pi \cdot 0.0707 \approx 0.444 \, \text{s} \]

### Step 3: Calculate Total Time for 20 Cycles
The total time \( t \) for 20 cycles is:
\[ t = 20 \cdot T = 20 \cdot 0.444 \approx 8.88 \, \text{s} \]

### Step 4: Calculate the Damping Ratio
The ratio of the amplitude of damped oscillation \( y \) to the initial amplitude \( y_0 \) is given by:
\[ \frac{y}{y_0} = e^{-\frac{b}{2m} t} \]
Substituting the values:
\[ \frac{y}{y_0} = e^{-\frac{0.075}{2 \cdot 0.5} \cdot 8.88} \]
Calculating \( \frac{b}{2m} \):
\[ \frac{b}{2m} = \frac{0.075}{1} = 0.075 \]
Now substituting into the equation:
\[ \frac{y}{y_0} = e^{-0.075 \cdot 8.88} \approx e^{-0.666} \]

### Step 5: Use the Given Values
Using the provided approximation:
\[ e^{-0.675} \approx 0.5 \quad \text{and} \quad e^{-0.51} \approx 0.6 \]
Since \( -0.666 \) is close to \( -0.675 \), we can approximate:
\[ \frac{y}{y_0} \approx 0.5 \]

### Final Answer
Thus, the ratio of the amplitude of damped oscillation to the initial amplitude at the end of 20 cycles is approximately:
\[ \frac{y}{y_0} \approx 0.51 \]

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