A particle executing SHM completes 1200 oscillations per minute and passes through the mean position with a velocity of `31.4ms^(-1)`. Determine the maximum displacement of the particle from the mean position also obtain the displacement equation of the aprticle if its displacemet be zero at the instant t=0.

Answer

To solve the problem step-by-step, we will follow these steps:

### Step 1: Convert Oscillations per Minute to Frequency
The particle completes 1200 oscillations per minute. To find the frequency in hertz (oscillations per second), we divide the number of oscillations by the number of seconds in a minute.

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

### Step 2: Use the Maximum Velocity Formula
The maximum velocity \( V_{max} \) at the mean position is given as \( 31.4 \, \text{m/s} \). The relationship between maximum velocity, angular frequency \( \omega \), and amplitude \( A \) is given by the formula:

\[ V_{max} = \omega A \]

### Step 3: Calculate Angular Frequency
Angular frequency \( \omega \) can be calculated using the frequency:

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

### Step 4: Rearrange the Maximum Velocity Formula to Find Amplitude
Now we can rearrange the formula to solve for amplitude \( A \):

\[ A = \frac{V_{max}}{\omega} \]

Substituting the known values:

\[ A = \frac{31.4 \, \text{m/s}}{40 \pi \, \text{rad/s}} \]

### Step 5: Calculate the Amplitude
Now we can calculate the amplitude:

\[ A = \frac{31.4}{40 \times 3.14} = \frac{31.4}{125.6} \approx 0.25 \, \text{m} \]

### Step 6: Write the Displacement Equation
The displacement \( y \) in simple harmonic motion can be expressed as:

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

Given that the displacement is zero at \( t = 0 \), the phase angle \( \phi \) is zero. Thus, the equation simplifies to:

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

Substituting the values of \( A \) and \( \omega \):

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

### Final Answers
- The maximum displacement (amplitude) of the particle from the mean position is \( 0.25 \, \text{m} \).
- The displacement equation of the particle is \( y(t) = 0.25 \sin(40 \pi t) \).

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