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 (SHM) at a displacement of \( \frac{3A}{4} \), we can follow these steps:

### Step 1: Understand the formula for velocity in SHM
The velocity \( v \) of a particle in SHM is given by the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude,
- \( x \) is the displacement from the mean position.

### Step 2: Relate maximum velocity to angular frequency and amplitude
The maximum velocity \( v_0 \) in SHM occurs when \( x = 0 \):
\[ v_{\text{max}} = \omega A \]
Given that \( v_{\text{max}} = v_0 \), we can write:
\[ v_0 = \omega A \]

### Step 3: Substitute the displacement into the velocity formula
Now, we need to find the velocity when the displacement \( x = \frac{3A}{4} \):
\[ v = \omega \sqrt{A^2 - \left(\frac{3A}{4}\right)^2} \]

### Step 4: Simplify the expression
Calculate \( \left(\frac{3A}{4}\right)^2 \):
\[ \left(\frac{3A}{4}\right)^2 = \frac{9A^2}{16} \]
Now substitute this back into the velocity formula:
\[ v = \omega \sqrt{A^2 - \frac{9A^2}{16}} \]
To combine the terms under the square root, find a common denominator:
\[ A^2 = \frac{16A^2}{16} \]
Thus,
\[ A^2 - \frac{9A^2}{16} = \frac{16A^2 - 9A^2}{16} = \frac{7A^2}{16} \]
Now substitute this back into the velocity equation:
\[ v = \omega \sqrt{\frac{7A^2}{16}} = \omega \frac{A \sqrt{7}}{4} \]

### Step 5: Substitute \( \omega A \) with \( v_0 \)
Since \( \omega A = v_0 \), we can replace \( \omega A \) in our equation:
\[ v = \frac{v_0 \sqrt{7}}{4} \]

### Final Answer
Thus, the speed of the particle at a displacement of \( \frac{3A}{4} \) is:
\[ v = \frac{v_0 \sqrt{7}}{4} \]