A body of mass `0.10kg` is attached to vertical massless spring with force constant `4.0 xx 10^(3)N//m`. The body is displaced `10.0cm` from its equilibrium position and released. How much time elapses as the body moves from a point `8.0 cm` on one side of the equilibrium position to a point `6.0cm` on the same side of the equilibrium position ?

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the angular frequency (ω)
The angular frequency of the oscillation is given by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
Where:
- \( k = 4.0 \times 10^3 \, \text{N/m} \) (spring constant)
- \( m = 0.10 \, \text{kg} \) (mass)

Substituting the values:
\[ \omega = \sqrt{\frac{4.0 \times 10^3}{0.10}} = \sqrt{40000} = 200 \, \text{rad/s} \]

### Step 2: Write the equation of motion for SHM
The general equation for simple harmonic motion (SHM) can be expressed as:
\[ x(t) = A \sin(\omega t) \]
Where:
- \( A = 10 \, \text{cm} = 0.10 \, \text{m} \) (amplitude)
- \( \omega = 200 \, \text{rad/s} \)

Thus, the equation becomes:
\[ x(t) = 0.10 \sin(200 t) \]

### Step 3: Find the time \( t_1 \) when the displacement is \( 8 \, \text{cm} \)
Set \( x(t_1) = 0.08 \, \text{m} \):
\[ 0.08 = 0.10 \sin(200 t_1) \]
\[ \sin(200 t_1) = \frac{0.08}{0.10} = 0.8 \]
Now, taking the inverse sine:
\[ 200 t_1 = \arcsin(0.8) \]
Calculating \( \arcsin(0.8) \):
\[ 200 t_1 \approx 0.927 \, \text{radians} \]
Thus,
\[ t_1 = \frac{0.927}{200} \approx 0.004635 \, \text{s} = 4.635 \, \text{ms} \]

### Step 4: Find the time \( t_2 \) when the displacement is \( 6 \, \text{cm} \)
Set \( x(t_2) = 0.06 \, \text{m} \):
\[ 0.06 = 0.10 \sin(200 t_2) \]
\[ \sin(200 t_2) = \frac{0.06}{0.10} = 0.6 \]
Now, taking the inverse sine:
\[ 200 t_2 = \arcsin(0.6) \]
Calculating \( \arcsin(0.6) \):
\[ 200 t_2 \approx 0.6435 \, \text{radians} \]
Thus,
\[ t_2 = \frac{0.6435}{200} \approx 0.0032175 \, \text{s} = 3.218 \, \text{ms} \]

### Step 5: Calculate the time elapsed \( \Delta t \)
The time elapsed as the body moves from \( 8 \, \text{cm} \) to \( 6 \, \text{cm} \) is given by:
\[ \Delta t = t_2 - t_1 \]
Substituting the values:
\[ \Delta t = 3.218 \, \text{ms} - 4.635 \, \text{ms} \approx -1.417 \, \text{ms} \]
Since we are considering the absolute time elapsed:
\[ \Delta t \approx 1.417 \, \text{ms} \]

### Final Answer
The time elapsed as the body moves from \( 8.0 \, \text{cm} \) to \( 6.0 \, \text{cm} \) is approximately \( 1.42 \, \text{ms} \).

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