The velocities of a body executing `SHM` at displacement 'a' and 'b' are 'b' and 'a' respectively. The amplitude of `SHM` is

Options

A. `a+b`
B. `a-b`
C. `sqrt(a^(2)+b^(2))`
D. `sqrt(a^(2)-b^(2))`

Explanation

To find the amplitude of a body executing Simple Harmonic Motion (SHM) given its velocities at two different displacements, we can follow these steps:

### Step 1: Write down the equations for velocity in SHM
The velocity \( v \) of a body in SHM at a displacement \( y \) is given by the formula:
\[ v = \omega \sqrt{A^2 - y^2} \]
where:
- \( A \) is the amplitude,
- \( y \) is the displacement,
- \( \omega \) is the angular frequency.

### Step 2: Set up the equations for the given conditions
We have two conditions:
1. At displacement \( a \), the velocity \( v_1 = b \):
\[ b = \omega \sqrt{A^2 - a^2} \tag{1} \]

2. At displacement \( b \), the velocity \( v_2 = a \):
\[ a = \omega \sqrt{A^2 - b^2} \tag{2} \]

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

From equation (2):
\[ a^2 = \omega^2 (A^2 - b^2) \tag{4} \]

### Step 4: Rearrange both equations to express \( A^2 \)
From equation (3):
\[ A^2 = \frac{b^2}{\omega^2} + a^2 \tag{5} \]

From equation (4):
\[ A^2 = \frac{a^2}{\omega^2} + b^2 \tag{6} \]

### Step 5: Set equations (5) and (6) equal to each other
Since both expressions equal \( A^2 \), we can set them equal:
\[ \frac{b^2}{\omega^2} + a^2 = \frac{a^2}{\omega^2} + b^2 \]

### Step 6: Rearranging the equation
Multiply through by \( \omega^2 \) to eliminate the denominator:
\[ b^2 + a^2 \omega^2 = a^2 + b^2 \omega^2 \]

Rearranging gives:
\[ b^2 (1 - \omega^2) = a^2 (1 - \omega^2) \]

### Step 7: Solve for \( A^2 \)
This implies:
\[ A^2 = a^2 + b^2 \]

### Step 8: Take the square root to find the amplitude
Thus, the amplitude \( A \) is:
\[ A = \sqrt{a^2 + b^2} \]

### Final Answer
The amplitude of the SHM is:
\[ A = \sqrt{a^2 + b^2} \]

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