A particle executing a linear S.H.M. performs 1200 oscillations/minute. The velocity at the midpoint of its path is 3.142 m/s. What is its equation of its displacement, if at time t=0, it is in the extreme right position ?

Options

A. `y=0.025 "sin" (40 pi t)`
B. `y=0.05 "cos" (40 pi t)`
C. `y=0.025 "cos" (40 pi t)`
D. `y= 0.025 "cos" (20 pi t)`

Explanation

To find the equation of displacement for a particle executing simple harmonic motion (SHM), we can follow these steps:

### Step 1: Determine the frequency in Hertz
The particle performs 1200 oscillations per minute. To convert this to oscillations per second (Hertz), we divide by 60.

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

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

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) can be calculated using 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 is related to the frequency by the factor of \(2\pi\).

### Step 3: Use the velocity at the midpoint to find the amplitude (A)
The maximum velocity (V_max) in SHM is given by the formula:

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

We know the velocity at the midpoint is 3.142 m/s, so we can set this equal to V_max:

\[ 3.142 = (40\pi) A \]

Now, solve for A:

\[ A = \frac{3.142}{40\pi} \]

Calculating A:

\[ A \approx \frac{3.142}{125.66} \approx 0.025 \text{ m} \]

**Hint:** The maximum velocity occurs at the midpoint of the oscillation, and it is directly proportional to the amplitude and angular frequency.

### Step 4: Write the equation of displacement
The general equation for displacement in SHM is given by:

\[ x(t) = A \cos(\omega t + \phi) \]

Since the particle is at the extreme right position at \(t = 0\), we know that \(x(0) = A\). Therefore, \(\phi = 0\) (since cosine of 0 is 1).

Thus, the equation simplifies to:

\[ x(t) = A \cos(\omega t) \]

Substituting the values of A and ω:

\[ x(t) = 0.025 \cos(40\pi t) \]

### Final Answer:
The equation of displacement is:

\[ x(t) = 0.025 \cos(40\pi t) \]

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