A particle simple harmonic motion completes `1200` oscillations per minute and passes through the mean position with a velocity `3.14 ms^(-1)`. Determine displacement of the particle from its mean position. Also obtain the displacement equation of the particle if its displacement be zero at the instant `t = 0`.

Explanation

To solve the problem step by step, we will break it down into parts.

### Step 1: Calculate the Frequency
The frequency \( f \) of the particle is given by the number of oscillations per minute. We need to convert this to oscillations per second (Hertz).

\[ f = \frac{1200 \text{ oscillations}}{60 \text{ seconds}} = 20 \text{ Hz} \]

**Hint:** Remember that 1 minute = 60 seconds. To convert oscillations per minute to oscillations per second, divide by 60.

### Step 2: Calculate the Angular Frequency
The angular frequency \( \omega \) is related to the linear frequency \( f \) by the formula:

\[ \omega = 2 \pi f \]

Substituting the value of \( f \):

\[ \omega = 2 \pi \times 20 = 40 \pi \text{ rad/s} \]

**Hint:** The angular frequency gives you the rate of oscillation in radians per second. Use \( 2\pi \) for one complete cycle.

### Step 3: Relate Velocity to Displacement
The velocity \( v \) at the mean position can be expressed in terms of the amplitude \( r \) (maximum displacement) and angular frequency \( \omega \):

\[ v = r \omega \]

We are given that \( v = 3.14 \text{ m/s} \). Rearranging the formula to find \( r \):

\[ r = \frac{v}{\omega} \]

Substituting the known values:

\[ r = \frac{3.14}{40 \pi} \]

Calculating \( r \):

\[ r \approx \frac{3.14}{125.66} \approx 0.025 \text{ m} \quad (\text{approximately } 0.25 \text{ m}) \]

**Hint:** When calculating \( r \), ensure that you are using the correct values for \( v \) and \( \omega \).

### Step 4: Write the Displacement Equation
The displacement \( y \) of a particle in simple harmonic motion can be expressed as:

\[ y(t) = A \sin(\omega t) \]

Where \( A \) is the amplitude (maximum displacement). Since the displacement is zero at \( t = 0 \), we can write:

\[ y(t) = 0.25 \sin(40 \pi t) \]

**Hint:** The sine function starts at zero, which is why we can use it directly for the displacement equation.

### Final Answer
1. The maximum displacement (amplitude) of the particle from its mean position is approximately \( 0.25 \text{ m} \).
2. The displacement equation of the particle is:

\[ y(t) = 0.25 \sin(40 \pi t) \]