A particle performing S.H.M. has velocities of 8 cm/s and 6 cm/s at displacements of 3 cm and 4 cm respectively. Calculate the amplitude and period of S.H.M.

Answer

To solve the problem, we will use the formula for the velocity of a particle in Simple Harmonic Motion (SHM):

\[ V = \omega \sqrt{A^2 - x^2} \]

where:
- \( V \) is the velocity,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude,
- \( x \) is the displacement.

Given:
1. \( V_1 = 8 \, \text{cm/s} \) at \( x_1 = 3 \, \text{cm} \)
2. \( V_2 = 6 \, \text{cm/s} \) at \( x_2 = 4 \, \text{cm} \)

### Step 1: Write the equations for both velocities

From the first condition:
\[ 8 = \omega \sqrt{A^2 - 3^2} \]
\[ 8 = \omega \sqrt{A^2 - 9} \quad \text{(1)} \]

From the second condition:
\[ 6 = \omega \sqrt{A^2 - 4^2} \]
\[ 6 = \omega \sqrt{A^2 - 16} \quad \text{(2)} \]

### Step 2: Solve for \( \omega \) from both equations

From equation (1):
\[ \omega = \frac{8}{\sqrt{A^2 - 9}} \]

From equation (2):
\[ \omega = \frac{6}{\sqrt{A^2 - 16}} \]

### Step 3: Set the two expressions for \( \omega \) equal to each other

\[ \frac{8}{\sqrt{A^2 - 9}} = \frac{6}{\sqrt{A^2 - 16}} \]

### Step 4: Cross-multiply to eliminate the fractions

\[ 8 \sqrt{A^2 - 16} = 6 \sqrt{A^2 - 9} \]

### Step 5: Square both sides to eliminate the square roots

\[ 64(A^2 - 16) = 36(A^2 - 9) \]

### Step 6: Expand both sides

\[ 64A^2 - 1024 = 36A^2 - 324 \]

### Step 7: Rearrange the equation to isolate terms involving \( A^2 \)

\[ 64A^2 - 36A^2 = 1024 - 324 \]
\[ 28A^2 = 700 \]

### Step 8: Solve for \( A^2 \)

\[ A^2 = \frac{700}{28} = 25 \]
\[ A = 5 \, \text{cm} \]

### Step 9: Find \( \omega \) using the value of \( A \)

Substituting \( A = 5 \) into either equation for \( \omega \). Using equation (1):
\[ \omega = \frac{8}{\sqrt{5^2 - 9}} = \frac{8}{\sqrt{25 - 9}} = \frac{8}{\sqrt{16}} = \frac{8}{4} = 2 \, \text{rad/s} \]

### Step 10: Calculate the period \( T \)

The period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} \]
\[ T = \frac{2\pi}{2} = \pi \, \text{s} \]

### Final Answer:
- Amplitude \( A = 5 \, \text{cm} \)
- Period \( T = \pi \, \text{s} \)