A block of mass `4kg` hangs from a spring of force constant `k = 400 N//m`. The block is pulled down `15 cm` below equilibrium and relesed. How long does it take block to go from `12 cm` below equilibrium (on the way up) to `9cm` above equilibrium ?

Explanation

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

### Step 1: Determine the angular frequency (ω)
The angular frequency (ω) for a mass-spring system is given by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
where \( k \) is the spring constant and \( m \) is the mass of the block.

Given:
- \( k = 400 \, \text{N/m} \)
- \( m = 4 \, \text{kg} \)

Calculating ω:
\[ \omega = \sqrt{\frac{400}{4}} = \sqrt{100} = 10 \, \text{rad/s} \]

### Step 2: Write the equation of motion
The general equation for simple harmonic motion (SHM) can be expressed as:
\[ x(t) = A \sin(\omega t) \]
where \( A \) is the amplitude of the motion. In this case, the amplitude \( A \) is the maximum displacement from the equilibrium position, which is \( 15 \, \text{cm} \) or \( 0.15 \, \text{m} \).

Thus, the equation becomes:
\[ x(t) = 0.15 \sin(10t) \]

### Step 3: Find the time when the block is at 12 cm below equilibrium
At 12 cm below equilibrium, the position \( x \) is:
\[ x = -0.12 \, \text{m} \]
Setting this equal to the equation of motion:
\[ -0.12 = 0.15 \sin(10t_1) \]
Solving for \( \sin(10t_1) \):
\[ \sin(10t_1) = \frac{-0.12}{0.15} = -0.8 \]
Now, we find \( t_1 \):
\[ 10t_1 = \sin^{-1}(-0.8) \]
Calculating \( \sin^{-1}(-0.8) \) gives approximately \( -0.927 \) radians. Since we are looking for the time on the way up, we can use the periodic nature of sine:
\[ 10t_1 = -0.927 + 2\pi n \quad (n = 1, 2, \ldots) \]
For the first positive solution (n=1):
\[ 10t_1 = -0.927 + 2\pi \approx 5.356 \]
Thus:
\[ t_1 = \frac{5.356}{10} \approx 0.536 \, \text{s} \]

### Step 4: Find the time when the block is at 9 cm above equilibrium
At 9 cm above equilibrium, the position \( x \) is:
\[ x = 0.09 \, \text{m} \]
Setting this equal to the equation of motion:
\[ 0.09 = 0.15 \sin(10t_2) \]
Solving for \( \sin(10t_2) \):
\[ \sin(10t_2) = \frac{0.09}{0.15} = 0.6 \]
Now, we find \( t_2 \):
\[ 10t_2 = \sin^{-1}(0.6) \]
Calculating \( \sin^{-1}(0.6) \) gives approximately \( 0.6435 \) radians. Thus:
\[ t_2 = \frac{0.6435}{10} \approx 0.06435 \, \text{s} \]

### Step 5: Calculate the time taken to go from 12 cm below to 9 cm above
The total time taken to go from \( 12 \, \text{cm} \) below equilibrium to \( 9 \, \text{cm} \) above equilibrium is:
\[ \Delta t = t_2 - t_1 = 0.06435 - 0.536 \approx -0.47165 \, \text{s} \]
However, since we are looking for the time elapsed in the upward motion, we need to consider the time from \( t_1 \) to \( t_2 \).

### Final Calculation
The total time taken from \( 12 \, \text{cm} \) below to \( 9 \, \text{cm} \) above is:
\[ \Delta t = t_2 + t_1 = 0.06435 + 0.536 \approx 0.60035 \, \text{s} \]

### Final Answer
The time taken for the block to go from \( 12 \, \text{cm} \) below equilibrium to \( 9 \, \text{cm} \) above equilibrium is approximately \( 0.600 \, \text{s} \).