An impulsive force gives an initial velocity of `-1.0 ms^(-1)` to the mass in the unstretched spring position. What is the amplitude of motion ? Give x as a function of time for the oscillating mass. Given m=3 kg, `k= 1200 Nm^(-1)`

Explanation

To solve the problem, we need to determine the amplitude of motion and express the position \( x \) as a function of time for a mass-spring system. Given the mass \( m = 3 \, \text{kg} \) and spring constant \( k = 1200 \, \text{N/m} \), we start with the initial conditions provided.

### Step 1: Calculate the Initial Kinetic Energy
The mass is given an initial velocity of \( v = -1.0 \, \text{m/s} \). The kinetic energy \( KE \) at this point can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 3 \, \text{kg} \times (-1.0 \, \text{m/s})^2 = \frac{1}{2} \times 3 \times 1 = 1.5 \, \text{J} \]

### Step 2: Relate Kinetic Energy to Potential Energy
At the maximum compression (which is the amplitude \( A \)), all the kinetic energy will be converted into potential energy \( PE \) stored in the spring. The potential energy in a spring is given by:
\[ PE = \frac{1}{2} k A^2 \]
Setting the kinetic energy equal to the potential energy at maximum compression:
\[ 1.5 \, \text{J} = \frac{1}{2} \times 1200 \, \text{N/m} \times A^2 \]

### Step 3: Solve for Amplitude \( A \)
Rearranging the equation to solve for \( A^2 \):
\[ 1.5 = 600 A^2 \implies A^2 = \frac{1.5}{600} = \frac{1}{400} \]
Taking the square root:
\[ A = \sqrt{\frac{1}{400}} = \frac{1}{20} \, \text{m} = 0.05 \, \text{m} = 5 \, \text{cm} \]

### Step 4: Calculate Angular Frequency \( \omega \)
The angular frequency \( \omega \) is calculated using the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
Substituting the values:
\[ \omega = \sqrt{\frac{1200 \, \text{N/m}}{3 \, \text{kg}}} = \sqrt{400} = 20 \, \text{rad/s} \]

### Step 5: Write the Equation of Motion
The equation of motion for simple harmonic motion (SHM) can be expressed as:
\[ x(t) = A \sin(\omega t) \]
Substituting the values of \( A \) and \( \omega \):
\[ x(t) = \frac{1}{20} \sin(20t) \]

### Final Answers
1. The amplitude of motion is \( A = 5 \, \text{cm} \).
2. The position as a function of time is:
\[ x(t) = \frac{1}{20} \sin(20t) \]