A spring of force constant `600 Nm^(-1)` is mounted on a horizontal table. A mass of 1.5 kg is attached to the free end of the spring,pulled sideways to a distance of 2 cm and released . The speed of the mass when the spring is compressed by 1 cm is

Options

A. `0.175 ms^(-1)`
B. ` 0. 35 ms^(-1)`
C. ` 0.7 ms^(-1)`
D. ` 1.4 ms^(-1)`

Answer

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy in the system remains constant if only conservative forces are acting on it, such as the spring force in this case.

### Step-by-step Solution:

1. **Identify Given Values:**
- Force constant of the spring, \( k = 600 \, \text{N/m} \)
- Mass attached to the spring, \( m = 1.5 \, \text{kg} \)
- Initial extension (amplitude), \( A = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Compression of the spring when we want to find the speed, \( x = -1 \, \text{cm} = -0.01 \, \text{m} \) (negative because it is a compression)

2. **Calculate Initial Potential Energy (PE_initial):**
The potential energy stored in the spring when it is stretched to its maximum amplitude is given by:
\[ PE_{\text{initial}} = \frac{1}{2} k A^2 = \frac{1}{2} \times 600 \times (0.02)^2 \]
\[ PE_{\text{initial}} = \frac{1}{2} \times 600 \times 0.0004 = 0.12 \, \text{J} \]

3. **Calculate Final Potential Energy (PE_final):**
The potential energy when the spring is compressed by 1 cm is:
\[ PE_{\text{final}} = \frac{1}{2} k x^2 = \frac{1}{2} \times 600 \times (0.01)^2 \]
\[ PE_{\text{final}} = \frac{1}{2} \times 600 \times 0.0001 = 0.03 \, \text{J} \]

4. **Apply Conservation of Mechanical Energy:**
According to the conservation of energy:
\[ PE_{\text{initial}} + KE_{\text{initial}} = PE_{\text{final}} + KE_{\text{final}} \]
Since the mass is released from rest, \( KE_{\text{initial}} = 0 \):
\[ 0.12 + 0 = 0.03 + KE_{\text{final}} \]
Rearranging gives:
\[ KE_{\text{final}} = 0.12 - 0.03 = 0.09 \, \text{J} \]

5. **Calculate Final Kinetic Energy (KE_final):**
The kinetic energy is given by:
\[ KE_{\text{final}} = \frac{1}{2} m v^2 \]
Substituting the known values:
\[ 0.09 = \frac{1}{2} \times 1.5 \times v^2 \]
\[ 0.09 = 0.75 v^2 \]
Solving for \( v^2 \):
\[ v^2 = \frac{0.09}{0.75} = 0.12 \]
Taking the square root gives:
\[ v = \sqrt{0.12} \approx 0.346 \, \text{m/s} \]

6. **Final Result:**
The speed of the mass when the spring is compressed by 1 cm is approximately:
\[ v \approx 0.35 \, \text{m/s} \]