In a horizontal spring-block system force constant of spring is k = 16N/m, mass of the block is 1 kg. Maximum kinetic energy of the block is 8J. Then

Options

A. amplitude of osciallation is 1m
B. at half the amplitude, potential energy stored in the spring is 2J.
C. at half the amplitude kinetic energy is 6J
D. angular frequency of oscillation is 16 rad/s

Answer

To solve the problem step by step, we will analyze the given information and apply the principles of simple harmonic motion (SHM).

### Step 1: Understand the given data
- Spring constant (k) = 16 N/m
- Mass of the block (m) = 1 kg
- Maximum kinetic energy (KE_max) = 8 J

### Step 2: Find the amplitude of oscillation
The maximum kinetic energy in a spring-block system occurs at the mean position and is given by the formula:
\[ KE_{\text{max}} = \frac{1}{2} k A^2 \]
Where \( A \) is the amplitude. Rearranging the formula to find \( A \):
\[ A^2 = \frac{2 KE_{\text{max}}}{k} \]
Substituting the known values:
\[ A^2 = \frac{2 \times 8 \, \text{J}}{16 \, \text{N/m}} = \frac{16}{16} = 1 \]
Taking the square root gives:
\[ A = 1 \, \text{m} \]

### Step 3: Calculate potential energy at half the amplitude
At half the amplitude (\( \frac{A}{2} = \frac{1}{2} \, \text{m} \)), the potential energy (PE) stored in the spring is given by:
\[ PE = \frac{1}{2} k x^2 \]
Where \( x = \frac{A}{2} \). Substituting the values:
\[ PE = \frac{1}{2} \times 16 \, \text{N/m} \times \left(\frac{1}{2} \, \text{m}\right)^2 = \frac{1}{2} \times 16 \times \frac{1}{4} = 2 \, \text{J} \]

### Step 4: Calculate kinetic energy at half the amplitude
Using the conservation of mechanical energy:
\[ E_{\text{total}} = KE + PE \]
At half the amplitude, we know:
\[ E_{\text{total}} = 8 \, \text{J} \quad \text{and} \quad PE = 2 \, \text{J} \]
Thus, the kinetic energy (KE) at half the amplitude is:
\[ KE = E_{\text{total}} - PE = 8 \, \text{J} - 2 \, \text{J} = 6 \, \text{J} \]

### Step 5: Calculate the angular frequency of oscillation
The angular frequency (\( \omega \)) can be calculated using the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
Substituting the known values:
\[ \omega = \sqrt{\frac{16 \, \text{N/m}}{1 \, \text{kg}}} = \sqrt{16} = 4 \, \text{rad/s} \]

### Summary of Results
1. Amplitude \( A = 1 \, \text{m} \) (Correct)
2. Potential energy at half the amplitude = 2 J (Correct)
3. Kinetic energy at half the amplitude = 6 J (Correct)
4. Angular frequency \( \omega = 4 \, \text{rad/s} \) (Incorrect if stated as 16 rad/s)

### Final Answer
The correct options are:
1. Amplitude is 1 m (Correct)
2. Potential energy at half the amplitude is 2 J (Correct)
3. Kinetic energy at half the amplitude is 6 J (Correct)
4. Angular frequency is 4 rad/s (Incorrect)