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 simple harmonic motion (SHM) for the given object, we can use the relationship between kinetic energy, mass, angular frequency, amplitude, and position.

### Step-by-Step Solution:

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

2. **Calculate Angular Frequency (ω):**
The angular frequency (ω) is related to the frequency (f) by the formula:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \left(\frac{2}{\pi}\right) = 4 \text{ rad/s} \]

3. **Use the Kinetic Energy Formula:**
The formula for kinetic energy (KE) in terms of position (x) in SHM is given by:
\[ KE = \frac{1}{2} m \left(\omega^2 A^2 - x^2\right) \]
Where A is the amplitude. Substituting the known values:
\[ 24 = \frac{1}{2} \cdot 1 \cdot \left(4^2 A^2 - 1^2\right) \]

4. **Simplify the Equation:**
\[ 24 = \frac{1}{2} \left(16 A^2 - 1\right) \]
Multiply both sides by 2:
\[ 48 = 16 A^2 - 1 \]
Rearranging gives:
\[ 16 A^2 = 49 \]

5. **Solve for Amplitude (A):**
\[ A^2 = \frac{49}{16} \]
Taking the square root:
\[ A = \sqrt{\frac{49}{16}} = \frac{7}{4} = 2 \text{ m} \]

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

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