The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path.

Answer

To find the speed of a body of mass 2 kg performing Simple Harmonic Motion (SHM) while crossing the center of its path, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Total Energy in SHM**:
The total mechanical energy (E) in SHM is given by the sum of kinetic energy (KE) and potential energy (PE). At the center of the path (mean position), the potential energy is zero, and all the energy is kinetic.

\[ E = KE + PE \]
At the center:
\[ E = KE = \frac{1}{2}mv^2 \]

2. **Given Values**:
- Total energy, \( E = 40 \, \text{J} \)
- Mass, \( m = 2 \, \text{kg} \)

3. **Set Up the Equation**:
Since all the total energy is kinetic energy at the center:
\[ 40 = \frac{1}{2}mv^2 \]

4. **Substitute the Mass**:
Substitute the mass into the equation:
\[ 40 = \frac{1}{2} \times 2 \times v^2 \]

5. **Simplify the Equation**:
\[ 40 = 1 \times v^2 \]
\[ v^2 = 40 \]

6. **Solve for Speed**:
Taking the square root of both sides:
\[ v = \sqrt{40} = 6.324 \, \text{m/s} \]

### Final Answer:
The speed of the body while crossing the center of the path is approximately \( 6.324 \, \text{m/s} \).

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