Statement-1 : In S.H.M., the motion is ‘to and fro’ and
periodic.
Statement-2 : Velocity of the particle
(v)`=omegasqrt(k^(2)-x^(2))`(where x is the displacement and k is amplitude)

Options

A. If both the assertion and reason are true and reason is correct explanation of the assertion .
B. If both the assertion and reason are true but reason is not a correct explanation of the assertion .
C. If the assertion is true but the reason is false.
D. If both assertion and reason are false.

Explanation

To solve the problem, we need to analyze both statements regarding Simple Harmonic Motion (S.H.M.) and determine their correctness.

### Step-by-Step Solution:

1. **Understanding Statement 1**:
- Statement 1 claims that in S.H.M., the motion is "to and fro" and periodic.
- This is true. In S.H.M., a particle oscillates around an equilibrium position, moving back and forth (to and fro) and repeating this motion at regular intervals (periodic).

2. **Understanding Statement 2**:
- Statement 2 provides a formula for the velocity of a particle in S.H.M.:
\[ v = \omega \sqrt{k^2 - x^2} \]
where \( x \) is the displacement and \( k \) is the amplitude.
- We need to derive this expression to verify its correctness.

3. **Deriving the Velocity Formula**:
- The equation of motion for S.H.M. can be expressed as:
\[ y = k \sin(\omega t) \]
- Differentiate this with respect to time \( t \) to find the velocity \( v \):
\[ v = \frac{dy}{dt} = k \omega \cos(\omega t) \]
- Now, square the velocity:
\[ v^2 = (k \omega \cos(\omega t))^2 = k^2 \omega^2 \cos^2(\omega t) \]
- The displacement \( y \) can also be squared:
\[ y^2 = (k \sin(\omega t))^2 = k^2 \sin^2(\omega t) \]
- Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we can relate \( \sin^2(\omega t) \) and \( \cos^2(\omega t) \):
\[ \cos^2(\omega t) = 1 - \sin^2(\omega t) \]
- Thus, we can express \( v^2 \) in terms of \( y \):
\[ v^2 + \omega^2 y^2 = k^2 \omega^2 \]
- Rearranging gives:
\[ v^2 = k^2 \omega^2 - \omega^2 y^2 \]
- Substituting \( y \) with \( x \) (the displacement):
\[ v^2 = \omega^2 (k^2 - x^2) \]
- Taking the square root provides the velocity formula:
\[ v = \omega \sqrt{k^2 - x^2} \]

4. **Conclusion**:
- Both statements are correct. Statement 1 accurately describes the nature of S.H.M., and Statement 2 correctly provides the formula for the velocity of a particle in S.H.M.
- Therefore, the answer is that both statements are true, and Statement 2 explains Statement 1.