A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm . Its maximum velocity is 100 cm / sec . Its velocity will be 50 cm / sec at a distance

Options

A. 5 cm
B. `5 sqrt2` cm
C. `5 sqrt3` cm
D. `10 sqrt2` cm

Answer

To solve the problem step by step, we will follow the principles of Simple Harmonic Motion (SHM).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass (m) = 5 gm (which is not needed for the calculation of distance)
- Amplitude (A) = 10 cm
- Maximum Velocity (Vmax) = 100 cm/s
- Velocity (V) = 50 cm/s

2. **Use the Relationship Between Maximum Velocity and Angular Frequency:**
The maximum velocity in SHM is given by the formula:
\[ V_{max} = A \cdot \omega \]
Where:
- \( \omega \) is the angular frequency.
Rearranging gives:
\[ \omega = \frac{V_{max}}{A} \]
Substituting the known values:
\[ \omega = \frac{100 \, \text{cm/s}}{10 \, \text{cm}} = 10 \, \text{rad/s} \]

3. **Use the Velocity Formula in SHM:**
The velocity at a distance \( y \) from the mean position is given by:
\[ V = \omega \sqrt{A^2 - y^2} \]
Substituting the known values:
\[ 50 = 10 \sqrt{10^2 - y^2} \]

4. **Simplify the Equation:**
Divide both sides by 10:
\[ 5 = \sqrt{100 - y^2} \]

5. **Square Both Sides:**
\[ 25 = 100 - y^2 \]

6. **Rearrange to Solve for \( y^2 \):**
\[ y^2 = 100 - 25 = 75 \]

7. **Take the Square Root:**
\[ y = \sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3} \, \text{cm} \]

### Final Answer:
The distance \( y \) at which the velocity is 50 cm/s is:
\[ y = 5\sqrt{3} \, \text{cm} \]