A cork floating on the pond water executes a simple harmonic motion, moving up and down over a range of 4 cm. The time period of the motion is 1 s. at `t=0`, the cork is at its lowest position of oscillation, the position and velocity of the cork at `t=10.5s`, would be

Options

A. 2 cm above the mean position, `0(m)/(s)`
B. 2 cm below the mean position `0(m)/(s)`
C. 1 cm above the mean position `2sqrt3pi(m)/(s)`up
D. 1 cm below the mean position, `2sqrt3pi(m)/(s)` up

Explanation

To solve the problem step by step, we will analyze the motion of the cork in simple harmonic motion (SHM) based on the given information.

### Step 1: Understand the parameters of SHM
- The range of motion is given as 4 cm, which means the cork moves 2 cm above and 2 cm below the mean position (the surface of the water).
- Therefore, the amplitude \( A \) of the motion is:
\[ A = \frac{\text{Range}}{2} = \frac{4 \, \text{cm}}{2} = 2 \, \text{cm} \]

### Step 2: Identify the time period
- The time period \( T \) of the motion is given as 1 second.

### Step 3: Determine the position at \( t = 10.5 \, \text{s} \)
- Since the cork starts at its lowest position at \( t = 0 \), we can analyze its position at \( t = 10.5 \, \text{s} \).
- The motion is periodic with a time period of 1 second, so we can find the equivalent time within one period:
\[ 10.5 \, \text{s} \mod 1 \, \text{s} = 0.5 \, \text{s} \]
- This means that at \( t = 10.5 \, \text{s} \), the cork is in the same position as it is at \( t = 0.5 \, \text{s} \).

### Step 4: Calculate the position at \( t = 0.5 \, \text{s} \)
- The position \( x(t) \) in SHM can be described by the equation:
\[ x(t) = A \cos\left(\frac{2\pi}{T} t + \phi\right) \]
- Since the cork is at its lowest position at \( t = 0 \), we can set the phase constant \( \phi = \pi \) (which corresponds to the cosine function being at its minimum value):
\[ x(t) = 2 \cos\left(2\pi t + \pi\right) = 2 \cos(2\pi t + \pi) = -2 \, \text{cm} \quad \text{(at } t = 0 \text{)} \]
- Now, substituting \( t = 0.5 \, \text{s} \):
\[ x(0.5) = 2 \cos\left(2\pi \cdot 0.5 + \pi\right) = 2 \cos(\pi + \pi) = 2 \cos(2\pi) = 2 \, \text{cm} \]
- Therefore, at \( t = 10.5 \, \text{s} \), the position of the cork is 2 cm above the mean position (the surface of the water).

### Step 5: Determine the velocity at \( t = 10.5 \, \text{s} \)
- The velocity \( v(t) \) in SHM is given by:
\[ v(t) = -A \frac{2\pi}{T} \sin\left(\frac{2\pi}{T} t + \phi\right) \]
- Substituting \( A = 2 \, \text{cm} \), \( T = 1 \, \text{s} \), and \( t = 0.5 \, \text{s} \):
\[ v(0.5) = -2 \cdot 2\pi \sin\left(2\pi \cdot 0.5 + \pi\right) = -4\pi \sin(2\pi) = -4\pi \cdot 0 = 0 \, \text{cm/s} \]
- Thus, the velocity of the cork at \( t = 10.5 \, \text{s} \) is 0 cm/s.

### Final Answer
- The position of the cork at \( t = 10.5 \, \text{s} \) is **2 cm above the water surface** and the velocity is **0 cm/s**.

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