To solve the problem step by step, we will analyze the motion of the mass described by the equation \( x = 0.20 \sin(3.0 t) \, \text{m} \).
### Step 1: Identify the parameters
From the equation of motion:
- Amplitude \( a = 0.20 \, \text{m} \)
- Angular frequency \( \omega = 3.0 \, \text{rad/s} \)
### Step 2: Convert the displacement to meters
We need to find the velocity and acceleration when the object is \( 5 \, \text{cm} \) from its equilibrium position.
- Convert \( 5 \, \text{cm} \) to meters:
\[
x = 5 \, \text{cm} = 0.05 \, \text{m}
\]
### Step 3: Calculate the velocity
The formula for velocity \( v \) in simple harmonic motion (SHM) is given by:
\[
v = \pm \omega \sqrt{a^2 - x^2}
\]
Substituting the values:
- \( a = 0.20 \, \text{m} \)
- \( x = 0.05 \, \text{m} \)
- \( \omega = 3.0 \, \text{rad/s} \)
Now, calculate:
\[
v = \pm 3.0 \sqrt{(0.20)^2 - (0.05)^2}
\]
Calculating the squares:
\[
(0.20)^2 = 0.04 \quad \text{and} \quad (0.05)^2 = 0.0025
\]
Thus,
\[
0.04 - 0.0025 = 0.0375
\]
Now take the square root:
\[
\sqrt{0.0375} \approx 0.1936
\]
Now substitute back into the velocity formula:
\[
v = \pm 3.0 \times 0.1936 \approx \pm 0.5808 \, \text{m/s}
\]
So, the velocity when the object is \( 5 \, \text{cm} \) from the equilibrium position is:
\[
v \approx \pm 0.58 \, \text{m/s}
\]
### Step 4: Calculate the acceleration
The formula for acceleration \( a \) in SHM is given by:
\[
a = -\omega^2 x
\]
Substituting the values:
\[
a = - (3.0)^2 \times 0.05
\]
Calculating:
\[
a = -9.0 \times 0.05 = -0.45 \, \text{m/s}^2
\]
### Summary for Part A
- Velocity when \( x = 5 \, \text{cm} \): \( v \approx \pm 0.58 \, \text{m/s} \)
- Acceleration when \( x = 5 \, \text{cm} \): \( a = -0.45 \, \text{m/s}^2 \)
### Step 5: Repeat for \( x = 0 \)
Now, we will find the velocity and acceleration when \( x = 0 \).
#### Velocity at \( x = 0 \)
Using the same velocity formula:
\[
v = \pm \omega \sqrt{a^2 - x^2}
\]
Since \( x = 0 \):
\[
v = \pm \omega a = \pm 3.0 \times 0.20 = \pm 0.60 \, \text{m/s}
\]
#### Acceleration at \( x = 0 \)
Using the acceleration formula:
\[
a = -\omega^2 x
\]
Since \( x = 0 \):
\[
a = -9.0 \times 0 = 0 \, \text{m/s}^2
\]
### Summary for Part B
- Velocity when \( x = 0 \): \( v \approx \pm 0.60 \, \text{m/s} \)
- Acceleration when \( x = 0 \): \( a = 0 \, \text{m/s}^2 \)
### Final Answers
1. When \( x = 5 \, \text{cm} \):
- Velocity: \( \pm 0.58 \, \text{m/s} \)
- Acceleration: \( -0.45 \, \text{m/s}^2 \)
2. When \( x = 0 \):
- Velocity: \( \pm 0.60 \, \text{m/s} \)
- Acceleration: \( 0 \, \text{m/s}^2 \)
