A particle is moving with shm in a-straight line. When the distance of the particle from the equilibrium position has the values 2 cm and 4 cm, the corresponding values of velocities are `5 cms^(-1) and 3 cms^(-1)` respectively. Find the length of the path and the period.

Explanation

To solve the problem, we need to find the amplitude and the time period of a particle moving in simple harmonic motion (SHM) given two positions and their corresponding velocities.

### Step 1: Write down the given data
We have:
- Displacement \( x_1 = 2 \, \text{cm} \) with velocity \( v_1 = 5 \, \text{cm/s} \)
- Displacement \( x_2 = 4 \, \text{cm} \) with velocity \( v_2 = 3 \, \text{cm/s} \)

### Step 2: Use the SHM velocity formula
The velocity in SHM can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

### Step 3: Set up equations for both positions
For the first position:
\[ v_1 = \omega \sqrt{A^2 - x_1^2} \]
Substituting the values:
\[ 5 = \omega \sqrt{A^2 - 2^2} \quad \text{(1)} \]

For the second position:
\[ v_2 = \omega \sqrt{A^2 - x_2^2} \]
Substituting the values:
\[ 3 = \omega \sqrt{A^2 - 4^2} \quad \text{(2)} \]

### Step 4: Square both equations to eliminate the square root
From equation (1):
\[ 25 = \omega^2 (A^2 - 4) \quad \text{(3)} \]

From equation (2):
\[ 9 = \omega^2 (A^2 - 16) \quad \text{(4)} \]

### Step 5: Solve for \( \omega^2 \) from both equations
From equation (3):
\[ \omega^2 = \frac{25}{A^2 - 4} \quad \text{(5)} \]

From equation (4):
\[ \omega^2 = \frac{9}{A^2 - 16} \quad \text{(6)} \]

### Step 6: Set equations (5) and (6) equal to each other
\[ \frac{25}{A^2 - 4} = \frac{9}{A^2 - 16} \]

### Step 7: Cross-multiply and simplify
\[ 25(A^2 - 16) = 9(A^2 - 4) \]
Expanding both sides:
\[ 25A^2 - 400 = 9A^2 - 36 \]
Rearranging gives:
\[ 25A^2 - 9A^2 = 400 - 36 \]
\[ 16A^2 = 364 \]
\[ A^2 = \frac{364}{16} = 22.75 \]

### Step 8: Calculate the amplitude \( A \)
\[ A = \sqrt{22.75} \approx 4.77 \, \text{cm} \]

### Step 9: Substitute \( A^2 \) back to find \( \omega^2 \)
Using equation (5):
\[ \omega^2 = \frac{25}{22.75 - 4} = \frac{25}{18.75} \approx 1.33 \]

### Step 10: Calculate \( \omega \)
\[ \omega \approx \sqrt{1.33} \approx 1.15 \, \text{rad/s} \]

### Step 11: Find the time period \( T \)
The time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} \approx \frac{2\pi}{1.15} \approx 5.47 \, \text{s} \]

### Final Answers
- Amplitude \( A \approx 4.77 \, \text{cm} \)
- Time period \( T \approx 5.47 \, \text{s} \)