A spring executes SHM with mass of 10 kg attached to it. The force constant of spring is 10 N/m. If at any instant its velocity is 40 cm/sec, the displacement will be (where amplitude is 0.5 m )

Options

A. 0.9m
B. 0.3m
C. 0.03 m
D. 0.9 m

Answer

To solve the problem, we will use the relationship between velocity, displacement, and angular frequency in simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass (m) = 10 kg (not directly needed for this calculation)
- Spring constant (k) = 10 N/m
- Velocity (v) = 40 cm/s = 0.4 m/s (conversion from cm to m)
- Amplitude (A) = 0.5 m

2. **Calculate the angular frequency (ω):**
- The formula for angular frequency in SHM is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]
- Substituting the values:
\[ \omega = \sqrt{\frac{10 \, \text{N/m}}{10 \, \text{kg}}} = \sqrt{1} = 1 \, \text{rad/s} \]

3. **Use the relationship between velocity, amplitude, and displacement:**
- The relationship is given by:
\[ v^2 = \omega^2 A^2 - \omega^2 x^2 \]
- Rearranging this gives:
\[ x^2 = A^2 - \frac{v^2}{\omega^2} \]

4. **Substituting the known values into the equation:**
- First, calculate \(A^2\):
\[ A^2 = (0.5 \, \text{m})^2 = 0.25 \, \text{m}^2 \]
- Now calculate \(v^2\):
\[ v^2 = (0.4 \, \text{m/s})^2 = 0.16 \, \text{m}^2/\text{s}^2 \]
- Substitute \(A\), \(v\), and \(\omega\) into the equation:
\[ x^2 = 0.25 - \frac{0.16}{1^2} = 0.25 - 0.16 = 0.09 \]

5. **Calculate the displacement (x):**
- Taking the square root gives:
\[ x = \sqrt{0.09} = 0.3 \, \text{m} \]

### Final Answer:
The displacement of the mass at that instant is **0.3 m**.

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