A particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passes through a certain point P(APltBP) at successive intervals of 0.5s and 1.5 s with a speed of 3m/s.

Options

A. the maximum speed of particle is `3sqrt(2)`m/s
B. the maximum speed of particle is `sqrt(2)m/s`.
C. the ratio (AP/BP) is `(sqrt(2)-1)/(sqrt(2)+1)`
D. the ratio AP/BP is `1/sqrt(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the motion of the particle
The particle is executing Simple Harmonic Motion (SHM) between points A and B, where the velocity is zero. The particle passes through point P at intervals of 0.5 seconds and 1.5 seconds with a speed of 3 m/s.

### Step 2: Determine the time period of the motion
The total time taken to go from P to A and back to P is 0.5 seconds, and the time taken to go from P to B and back to P is 1.5 seconds. Therefore, the total time period (T) of the motion can be calculated as:
\[ T = 0.5 \, \text{s} + 1.5 \, \text{s} = 2 \, \text{s} \]

### Step 3: Calculate the angular frequency (ω)
The angular frequency (ω) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{2} = \pi \, \text{rad/s} \]

### Step 4: Relate the speed at point P to the amplitude (A)
At point P, the speed (v) is given as 3 m/s. The velocity in SHM can be expressed as:
\[ v = A \omega \cos(\omega t + \phi) \]
At t = 0 (when the particle is at point P), we can simplify this to:
\[ 3 = A \cdot \pi \cdot \cos(\phi) \]

### Step 5: Determine the time taken to go from P to A
Since the particle takes 0.5 seconds to go from P to A and back, it takes 0.25 seconds to go from P to A.

### Step 6: Find the position at point A
At point A, the velocity is zero. Therefore, we can use the fact that:
\[ 0 = A \omega \cos(\omega t + \phi) \]
At t = 0.25 seconds, we have:
\[ \cos(\pi \cdot 0.25 + \phi) = 0 \]
This implies:
\[ \pi \cdot 0.25 + \phi = \frac{\pi}{2} \]
Solving for φ gives:
\[ \phi = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Step 7: Substitute φ back to find A
Now substituting φ back into the equation for speed at point P:
\[ 3 = A \cdot \pi \cdot \cos\left(\frac{\pi}{4}\right) \]
Since \(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\):
\[ 3 = A \cdot \pi \cdot \frac{1}{\sqrt{2}} \]
Solving for A gives:
\[ A = \frac{3\sqrt{2}}{\pi} \]

### Step 8: Calculate the maximum speed
The maximum speed (V_max) in SHM is given by:
\[ V_{\text{max}} = A \omega \]
Substituting the values of A and ω:
\[ V_{\text{max}} = \left(\frac{3\sqrt{2}}{\pi}\right) \cdot \pi = 3\sqrt{2} \, \text{m/s} \]

### Step 9: Calculate the ratio of distances AP and BP
Using the definitions:
- \(AP = A - PO\)
- \(BP = OB + OP\)

Where \(PO = \frac{A}{\sqrt{2}}\) (the position at P):
\[ AP = A - \frac{A}{\sqrt{2}} = A\left(1 - \frac{1}{\sqrt{2}}\right) \]
\[ BP = A + \frac{A}{\sqrt{2}} = A\left(1 + \frac{1}{\sqrt{2}}\right) \]
Thus, the ratio \( \frac{AP}{BP} \) becomes:
\[ \frac{AP}{BP} = \frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} \]

### Final Answers
1. Maximum speed of the particle: \(3\sqrt{2} \, \text{m/s}\)
2. Ratio \(AP:BP = \frac{\sqrt{2}-1}{\sqrt{2}+1}\)