Passage I) In simple harmonic motion force acting on a particle is given as `F=-4x`, total mechanical energy of the particle is 10 J and amplitude of oscillations is 2m, At time t=0 acceleration of the particle is `-16m/s^(2)`. Mass of the particle is 0.5 kg.
Displacement time equation equation of the particle is

Options

A. `x=2sin2t`
B. `x= 2sin4t`
C. `x=2cos2t`
D. None of these

Answer

To find the displacement-time equation of the particle in simple harmonic motion, we can follow these steps:

### Step 1: Identify the spring constant (k)
The force acting on the particle is given by:
\[ F = -4x \]
This can be compared to the standard form of Hooke's law:
\[ F = -kx \]
From this, we can identify that:
\[ k = 4 \, \text{N/m} \]

### Step 2: Find the angular frequency (ω)
The angular frequency (ω) is related to the spring constant (k) and the mass (m) of the particle by the formula:
\[ k = m\omega^2 \]
Given that the mass \( m = 0.5 \, \text{kg} \), we can rearrange the equation to solve for ω:
\[ \omega^2 = \frac{k}{m} = \frac{4}{0.5} = 8 \]
Thus,
\[ \omega = \sqrt{8} = 2\sqrt{2} \, \text{rad/s} \]

### Step 3: Use the acceleration to find the displacement at \( t = 0 \)
The acceleration (a) at any position x in simple harmonic motion is given by:
\[ a = -\omega^2 x \]
At \( t = 0 \), the acceleration is given as:
\[ a = -16 \, \text{m/s}^2 \]
Substituting the value of ω:
\[ -16 = -8x \]
Solving for x gives:
\[ x = \frac{16}{8} = 2 \, \text{m} \]

### Step 4: Write the displacement-time equation
The general form of the displacement-time equation for simple harmonic motion is:
\[ x(t) = A \sin(\omega t + \phi) \]
Where:
- \( A \) is the amplitude (given as 2 m)
- \( \omega \) is the angular frequency (calculated as \( 2\sqrt{2} \))
- \( \phi \) is the phase constant

Since at \( t = 0 \), \( x(0) = 2 \, \text{m} \), we can substitute:
\[ 2 = 2 \sin(\phi) \]
This implies:
\[ \sin(\phi) = 1 \]
Thus, \( \phi = \frac{\pi}{2} \) (since sine is maximum at \( \frac{\pi}{2} \)).

### Final Displacement-Time Equation
Substituting the values into the equation gives:
\[ x(t) = 2 \sin(2\sqrt{2} t + \frac{\pi}{2}) \]
Using the identity \( \sin\left(\theta + \frac{\pi}{2}\right) = \cos(\theta) \):
\[ x(t) = 2 \cos(2\sqrt{2} t) \]

### Conclusion
The displacement-time equation of the particle is:
\[ x(t) = 2 \cos(2\sqrt{2} t) \]

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