A particle starts from rest at the extreme position and makes SHM with an amplitude A and angular frequency `omega`. Its displacement at time 't' is

Options

A. `A sin omega t`
B. `A cos omega t`
C. `A sin 2 pi omega t`
D. `A cos 2 pi omega t`

Answer

To find the displacement of a particle undergoing simple harmonic motion (SHM) at time 't', we can follow these steps:

### Step 1: Understand the General Equation of SHM
The general equation for simple harmonic motion is given by:
\[ y(t) = A \sin(\omega t + \phi) \]
where:
- \( y(t) \) is the displacement at time \( t \),
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

### Step 2: Identify Initial Conditions
The problem states that the particle starts from rest at the extreme position. This means:
- At \( t = 0 \), the displacement \( y(0) = A \) (the maximum displacement).
- The velocity at this point is zero.

### Step 3: Determine the Phase Constant \( \phi \)
Since the particle is at the extreme position at \( t = 0 \), we can substitute \( t = 0 \) into the general equation:
\[ y(0) = A \sin(\omega \cdot 0 + \phi) = A \sin(\phi) \]
Given that \( y(0) = A \), we have:
\[ A \sin(\phi) = A \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \sin(\phi) = 1 \]
The sine function equals 1 at \( \phi = \frac{\pi}{2} \).

### Step 4: Substitute \( \phi \) Back into the Equation
Now that we have determined \( \phi \):
\[ \phi = \frac{\pi}{2} \]
We can substitute this value back into the general equation:
\[ y(t) = A \sin(\omega t + \frac{\pi}{2}) \]

### Step 5: Simplify the Equation
Using the trigonometric identity \( \sin\left(\theta + \frac{\pi}{2}\right) = \cos(\theta) \), we can rewrite the equation:
\[ y(t) = A \cos(\omega t) \]

### Final Answer
Thus, the displacement of the particle at time \( t \) is given by:
\[ y(t) = A \cos(\omega t) \]

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