A particle is executing S.H.M. with amplitude A and has maximum velocity `v_(0)`. Its speed at displacement `(3A)/(4)` will be

Options

A. `(sqrt7)/(4) V_(0)`
B. `(v_(0))/(sqrt2)`
C. `v_(0)`
D. `(sqrt3)/(2) v_(0)`

Answer

To find the speed of a particle executing Simple Harmonic Motion (S.H.M.) at a displacement of \( \frac{3A}{4} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between velocity and displacement in S.H.M.:**
The velocity \( V \) of a particle in S.H.M. can be expressed in terms of its amplitude \( A \) and its displacement \( x \) as:
\[ V = \omega \sqrt{A^2 - x^2} \]
where \( \omega \) is the angular frequency.

2. **Substitute the given displacement:**
We need to find the speed when the displacement \( x = \frac{3A}{4} \):
\[ V = \omega \sqrt{A^2 - \left(\frac{3A}{4}\right)^2} \]

3. **Calculate \( \left(\frac{3A}{4}\right)^2 \):**
\[ \left(\frac{3A}{4}\right)^2 = \frac{9A^2}{16} \]

4. **Substitute this back into the velocity equation:**
\[ V = \omega \sqrt{A^2 - \frac{9A^2}{16}} \]

5. **Simplify the expression inside the square root:**
\[ A^2 - \frac{9A^2}{16} = \frac{16A^2}{16} - \frac{9A^2}{16} = \frac{7A^2}{16} \]

6. **Now substitute this back into the velocity equation:**
\[ V = \omega \sqrt{\frac{7A^2}{16}} = \omega \cdot \frac{A\sqrt{7}}{4} \]

7. **Relate \( \omega \) to the maximum velocity \( V_0 \):**
The maximum velocity \( V_{max} \) in S.H.M. is given by:
\[ V_{max} = \omega A \]
Given that \( V_{max} = V_0 \), we can express \( \omega \) as:
\[ \omega = \frac{V_0}{A} \]

8. **Substitute \( \omega \) back into the velocity equation:**
\[ V = \left(\frac{V_0}{A}\right) \cdot \frac{A\sqrt{7}}{4} = \frac{V_0 \sqrt{7}}{4} \]

9. **Final Result:**
Therefore, the speed of the particle at a displacement of \( \frac{3A}{4} \) is:
\[ V = \frac{V_0 \sqrt{7}}{4} \]