A body executing S.H.M.along a straight line has a velocity of `3 ms^(-1)` when it is at a distance of 4m from its mean position and`4ms^(-1)` when it is at a distance of 3m from its mean position.Its angular frequency and amplitude are

Options

A. `2 rad s^(-1) & 5m`
B. `1 rad s^(-1) & 10 m`
C. ` 2 rad s^(-1)& 10 m`
D. ` 1 rad s^(-1) & 5m`

Answer

To solve the problem step by step, we will use the equations of motion for Simple Harmonic Motion (SHM).

### Step 1: Write down the given data
We have the following information from the problem:
- Velocity \( v_1 = 3 \, \text{m/s} \) at a distance \( x_1 = 4 \, \text{m} \)
- Velocity \( v_2 = 4 \, \text{m/s} \) at a distance \( x_2 = 3 \, \text{m} \)

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

### Step 3: Write equations for both conditions
For the first condition:
\[ v_1 = \omega \sqrt{A^2 - x_1^2} \]
Substituting the values:
\[ 3 = \omega \sqrt{A^2 - 4^2} \quad \text{(Equation 1)} \]
This simplifies to:
\[ 3 = \omega \sqrt{A^2 - 16} \]

For the second condition:
\[ v_2 = \omega \sqrt{A^2 - x_2^2} \]
Substituting the values:
\[ 4 = \omega \sqrt{A^2 - 3^2} \quad \text{(Equation 2)} \]
This simplifies to:
\[ 4 = \omega \sqrt{A^2 - 9} \]

### Step 4: Divide the two equations
Now, we will divide Equation 1 by Equation 2:
\[ \frac{v_1}{v_2} = \frac{\sqrt{A^2 - 16}}{\sqrt{A^2 - 9}} \]
Substituting the values:
\[ \frac{3}{4} = \frac{\sqrt{A^2 - 16}}{\sqrt{A^2 - 9}} \]

### Step 5: Square both sides
Squaring both sides gives:
\[ \left(\frac{3}{4}\right)^2 = \frac{A^2 - 16}{A^2 - 9} \]
This simplifies to:
\[ \frac{9}{16} = \frac{A^2 - 16}{A^2 - 9} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 9(A^2 - 9) = 16(A^2 - 16) \]
Expanding both sides:
\[ 9A^2 - 81 = 16A^2 - 256 \]
Rearranging gives:
\[ 7A^2 = 175 \]
Thus:
\[ A^2 = 25 \quad \Rightarrow \quad A = 5 \, \text{m} \]

### Step 7: Calculate angular frequency \( \omega \)
Now, we can use either Equation 1 or Equation 2 to find \( \omega \). Using Equation 1:
\[ 3 = \omega \sqrt{5^2 - 4^2} \]
This simplifies to:
\[ 3 = \omega \sqrt{25 - 16} = \omega \sqrt{9} \]
Thus:
\[ 3 = 3\omega \quad \Rightarrow \quad \omega = 1 \, \text{rad/s} \]

### Final Answer
The amplitude \( A \) is \( 5 \, \text{m} \) and the angular frequency \( \omega \) is \( 1 \, \text{rad/s} \).

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