A particle performs SHM with an amplitude of 5 m. If at x = 4 m, the magnitude of velocity and accelertion are equal, then the time period of SHM (in seconds) is

Options

A. `(2pi)/(3)`
B. `(4pi)/(3)`
C. `(8pi)/(3)`
D. `(6pi)/(3)`

Answer

To solve the problem, we need to find the time period of a particle performing Simple Harmonic Motion (SHM) given that the amplitude is 5 m and at x = 4 m, the magnitudes of velocity and acceleration are equal.

### Step-by-Step Solution:

1. **Understand the parameters of SHM**:
- Amplitude (A) = 5 m
- Position (x) = 4 m

2. **Formulas for velocity and acceleration in SHM**:
- The velocity \( v \) at position \( x \) is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
- The acceleration \( a \) at position \( x \) is given by:
\[ a = -\omega^2 x \]
- Since we are interested in magnitudes, we can ignore the negative sign in the acceleration formula.

3. **Set the magnitudes of velocity and acceleration equal**:
\[ |\omega \sqrt{A^2 - x^2}| = |\omega^2 x| \]
Since \( \omega \) is common in both expressions, we can cancel it out (assuming \( \omega \neq 0 \)):
\[ \sqrt{A^2 - x^2} = \omega x \]

4. **Substitute the values of A and x**:
- Substitute \( A = 5 \) m and \( x = 4 \) m into the equation:
\[ \sqrt{5^2 - 4^2} = \omega \cdot 4 \]
- Calculate \( 5^2 - 4^2 \):
\[ \sqrt{25 - 16} = \omega \cdot 4 \]
\[ \sqrt{9} = \omega \cdot 4 \]
\[ 3 = \omega \cdot 4 \]

5. **Solve for \( \omega \)**:
\[ \omega = \frac{3}{4} \]

6. **Relate \( \omega \) to the time period \( T \)**:
- The relationship between angular frequency \( \omega \) and time period \( T \) is given by:
\[ \omega = \frac{2\pi}{T} \]
- Rearranging gives:
\[ T = \frac{2\pi}{\omega} \]

7. **Substitute \( \omega \) into the time period formula**:
\[ T = \frac{2\pi}{\frac{3}{4}} = \frac{2\pi \cdot 4}{3} = \frac{8\pi}{3} \]

8. **Final answer**:
The time period \( T \) of the SHM is:
\[ T = \frac{8\pi}{3} \text{ seconds} \]