A particle executes `SHM` with an amplitude of `10cm` and frequency `2 Hz`. At `t = 0`, the particle is at a point where potential energy and kinetic energy are same. The equation for its displacement is

Options

A. `x = 0.1 sin (4pi t+(pi)/(4))m`
B. `x = 0.1 (sin 4pi t)m`
C. `x = 0.1 cos(4pi t+(pi)/(3))`
D. `x = 0.1(sin4pit-(pi)/(3))m`

Explanation

To find the equation for the displacement of a particle executing Simple Harmonic Motion (SHM) with given parameters, we can follow these steps:

### Step 1: Identify the parameters
- Amplitude (A) = 10 cm = 0.1 m
- Frequency (f) = 2 Hz

### Step 2: Calculate angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2\pi \times 2 = 4\pi \, \text{rad/s} \]

### Step 3: Write the general equation for SHM
The general equation for the displacement (x) of a particle in SHM can be written as:
\[ x(t) = A \sin(\omega t + \phi) \]
Substituting the values we have:
\[ x(t) = 0.1 \sin(4\pi t + \phi) \]

### Step 4: Analyze the condition at \(t = 0\)
At \(t = 0\), we are given that the potential energy (PE) and kinetic energy (KE) are equal.

The formulas for kinetic energy and potential energy in SHM are:
- Kinetic Energy (KE) = \(\frac{1}{2} m \omega^2 (A^2 - x^2)\)
- Potential Energy (PE) = \(\frac{1}{2} m \omega^2 x^2\)

Setting KE = PE at \(t = 0\):
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2 \]

### Step 5: Simplify the equation
Canceling \(\frac{1}{2} m \omega^2\) from both sides gives:
\[ A^2 - x^2 = x^2 \]
\[ A^2 = 2x^2 \]
\[ x^2 = \frac{A^2}{2} \]
Substituting for A:
\[ x^2 = \frac{(0.1)^2}{2} = \frac{0.01}{2} = 0.005 \quad \Rightarrow \quad x = \sqrt{0.005} = 0.0707 \, \text{m} \]

### Step 6: Find the phase angle (φ)
At \(t = 0\):
\[ x(0) = A \sin(\phi) \Rightarrow 0.0707 = 0.1 \sin(\phi) \]
\[ \sin(\phi) = \frac{0.0707}{0.1} = 0.707 \]
This implies:
\[ \phi = \frac{\pi}{4} \, \text{(or 45 degrees)} \]

### Step 7: Write the final equation
Substituting the value of φ back into the displacement equation:
\[ x(t) = 0.1 \sin(4\pi t + \frac{\pi}{4}) \]

### Final Answer
The equation for the displacement of the particle is:
\[ x(t) = 0.1 \sin(4\pi t + \frac{\pi}{4}) \]