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 step by step, we will use the properties of Simple Harmonic Motion (SHM).

### Step 1: Identify Given Values
- Amplitude \( A = 5 \, \text{m} \)
- Displacement \( x = 4 \, \text{m} \)

### Step 2: Write the Formulas for Velocity and Acceleration in SHM
The formulas for velocity \( v \) and acceleration \( a \) in SHM are:
- Velocity:
\[ v = \omega \sqrt{A^2 - x^2} \]
- Acceleration:
\[ a = -\omega^2 x \]

### Step 3: Set the Magnitude of Velocity Equal to the Magnitude of Acceleration
According to the problem, at \( x = 4 \, \text{m} \), the magnitudes of velocity and acceleration are equal:
\[ |v| = |a| \]

### Step 4: Substitute the Formulas into the Equation
Substituting the expressions for \( v \) and \( a \):
\[ \omega \sqrt{A^2 - x^2} = \omega^2 x \]

### Step 5: Simplify the Equation
Since \( \omega \) is common in both sides and is non-zero, we can divide both sides by \( \omega \):
\[ \sqrt{A^2 - x^2} = \omega x \]

### Step 6: Substitute the Known Values
Substituting \( A = 5 \, \text{m} \) and \( x = 4 \, \text{m} \):
\[ \sqrt{5^2 - 4^2} = \omega \cdot 4 \]
\[ \sqrt{25 - 16} = \omega \cdot 4 \]
\[ \sqrt{9} = \omega \cdot 4 \]
\[ 3 = \omega \cdot 4 \]

### Step 7: Solve for Angular Frequency \( \omega \)
\[ \omega = \frac{3}{4} \, \text{rad/s} \]

### Step 8: Calculate the Time Period \( T \)
The time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{\frac{3}{4}} = \frac{2\pi \cdot 4}{3} = \frac{8\pi}{3} \, \text{s} \]

### Final Answer
The time period of SHM is:
\[ T = \frac{8\pi}{3} \, \text{s} \]

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