To solve the problem of calculating the period of a damped oscillator, we will follow these steps:
### Step 1: Convert the mass to SI units
The mass \( m \) is given as 200 g. To convert this to kilograms (the SI unit for mass), we use the conversion factor:
\[
m = 200 \, \text{g} = 200 \times 10^{-3} \, \text{kg} = 0.2 \, \text{kg}
\]
### Step 2: Identify the spring constant
The spring constant \( k \) is given as:
\[
k = 90 \, \text{N/m}
\]
### Step 3: Convert the damping constant to SI units
The damping constant \( b \) is given as 40 g/s. To convert this to kg/s, we use the conversion factor:
\[
b = 40 \, \text{g/s} = 40 \times 10^{-3} \, \text{kg/s} = 0.04 \, \text{kg/s}
\]
### Step 4: Calculate the damped angular frequency \( \omega_d \)
The formula for the damped angular frequency \( \omega_d \) is given by:
\[
\omega_d = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2}
\]
First, we calculate \( \frac{k}{m} \):
\[
\frac{k}{m} = \frac{90 \, \text{N/m}}{0.2 \, \text{kg}} = 450 \, \text{s}^{-2}
\]
Next, we calculate \( \frac{b}{2m} \):
\[
\frac{b}{2m} = \frac{0.04 \, \text{kg/s}}{2 \times 0.2 \, \text{kg}} = \frac{0.04}{0.4} = 0.1 \, \text{s}^{-1}
\]
Now, we square this value:
\[
\left(\frac{b}{2m}\right)^2 = (0.1)^2 = 0.01 \, \text{s}^{-2}
\]
Now we can find \( \omega_d \):
\[
\omega_d = \sqrt{450 - 0.01} = \sqrt{449.99} \approx 21.21 \, \text{rad/s}
\]
### Step 5: Calculate the period of oscillation \( T \)
The period \( T \) of the damped oscillator is given by:
\[
T = \frac{2\pi}{\omega_d}
\]
Substituting the value of \( \omega_d \):
\[
T = \frac{2\pi}{21.21} \approx \frac{6.2832}{21.21} \approx 0.296 \, \text{s}
\]
### Step 6: Round the answer
Rounding \( 0.296 \, \text{s} \) gives us approximately:
\[
T \approx 0.3 \, \text{s}
\]
### Final Answer
The period of oscillation for the damped oscillator is approximately \( 0.3 \, \text{s} \).
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