An object of mass 1kg executes simple harmonic oscillations along the x-axis, with a frequency of `(2)/(pi)Hz`. At the position x = 1 m, the object has a kinetic energy of 24 J. The amplitude of the oscillation is

Options

A. `(3)/(2)m`
B. 2m
C. 4m
D. 8m

Answer

To find the amplitude of the oscillation for an object executing simple harmonic motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Mass, \( m = 1 \, \text{kg} \)
- Frequency, \( f = \frac{2}{\pi} \, \text{Hz} \)
- Position, \( x = 1 \, \text{m} \)
- Kinetic Energy, \( KE = 24 \, \text{J} \)

2. **Calculate Angular Frequency (\( \omega \))**:
\[ \omega = 2\pi f = 2\pi \left(\frac{2}{\pi}\right) = 4 \, \text{rad/s} \]

3. **Use Kinetic Energy Formula**:
The kinetic energy in SHM is given by:
\[ KE = \frac{1}{2} m v^2 \]
We can express the velocity \( v \) in terms of amplitude \( A \) and position \( x \):
\[ v = \omega \sqrt{A^2 - x^2} \]

4. **Substituting for Velocity**:
Substituting \( v \) into the kinetic energy formula:
\[ KE = \frac{1}{2} m (\omega \sqrt{A^2 - x^2})^2 \]
Simplifying this gives:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

5. **Plugging in Known Values**:
Now substituting the known values:
\[ 24 = \frac{1}{2} \cdot 1 \cdot 4^2 (A^2 - 1^2) \]
Simplifying further:
\[ 24 = \frac{1}{2} \cdot 16 (A^2 - 1) \]
\[ 24 = 8 (A^2 - 1) \]

6. **Solving for \( A^2 \)**:
Dividing both sides by 8:
\[ 3 = A^2 - 1 \]
Adding 1 to both sides:
\[ A^2 = 4 \]

7. **Finding Amplitude \( A \)**:
Taking the square root:
\[ A = \sqrt{4} = 2 \, \text{m} \]

### Final Answer:
The amplitude of the oscillation is \( A = 2 \, \text{m} \).

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