Equations y = 2 A `cos^(2)omega`t and y = A`(sin omegat + sqrt(3) cosomegat)` represent the motion of two particles.

Options

A. Only one of these is S.H.M
B. Ratio of maximum speeds is 2 : 1
C. Ratio of maximum speeds is 1 : 1
D. Ratio of maximum accelerations is 1:4

Answer

To solve the problem, we need to analyze the two equations given for the motion of the particles and determine which one represents Simple Harmonic Motion (SHM) and calculate the ratios of their maximum speeds and accelerations.

### Step 1: Identify the equations
The equations given are:
1. \( y_1 = 2A \cos^2(\omega t) \)
2. \( y_2 = A(\sin(\omega t) + \sqrt{3} \cos(\omega t)) \)

### Step 2: Check if the first equation represents SHM
The general form of SHM is:
- \( y = A \sin(\omega t + \phi) \) or \( y = A \cos(\omega t + \phi) \)

The first equation \( y_1 = 2A \cos^2(\omega t) \) can be rewritten using the double angle identity:
\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]
Thus,
\[ y_1 = 2A \left(\frac{1 + \cos(2\omega t)}{2}\right) = A + A \cos(2\omega t) \]
This shows that \( y_1 \) is not in the standard form of SHM because it includes a constant term \( A \) and does not oscillate about zero. Therefore, **\( y_1 \) does not represent SHM**.

### Step 3: Check if the second equation represents SHM
For the second equation \( y_2 = A(\sin(\omega t) + \sqrt{3} \cos(\omega t)) \), we can express this as:
\[ y_2 = A \left( \sin(\omega t) + \sqrt{3} \cos(\omega t) \right) \]
This can be rewritten using the amplitude-phase form:
\[ y_2 = A \sqrt{1^2 + (\sqrt{3})^2} \sin\left(\omega t + \phi\right) \]
where \( R = \sqrt{1 + 3} = 2A \) and \( \tan(\phi) = \frac{\sqrt{3}}{1} \) which gives \( \phi = \frac{\pi}{3} \).

Thus, \( y_2 \) is in the form of SHM. Therefore, **\( y_2 \) represents SHM**.

### Step 4: Calculate maximum speeds
The maximum speed \( V \) in SHM is given by:
\[ V_{\text{max}} = A \omega \]

For \( y_1 \):
- The maximum speed can be derived from \( V_1 = \frac{dy_1}{dt} \):
\[ V_1 = \frac{d}{dt}(2A \cos^2(\omega t)) = 2A \cdot 2\cos(\omega t)(-\sin(\omega t)) \cdot \omega = -4A \omega \sin(\omega t) \cos(\omega t) \]
The maximum value of \( V_1 \) occurs when \( \sin(\omega t) \cos(\omega t) = \frac{1}{2} \), thus:
\[ V_{1,\text{max}} = 2A \omega \]

For \( y_2 \):
- The maximum speed can be derived from \( V_2 = \frac{dy_2}{dt} \):
\[ V_2 = A(\omega \cos(\omega t) - \sqrt{3}\omega \sin(\omega t)) \]
The maximum value occurs when \( \sqrt{1^2 + (\sqrt{3})^2} = 2A \), thus:
\[ V_{2,\text{max}} = 2A \omega \]

### Step 5: Calculate the ratio of maximum speeds
The ratio of maximum speeds is:
\[ \frac{V_{1,\text{max}}}{V_{2,\text{max}}} = \frac{2A \omega}{2A \omega} = 1:1 \]

### Step 6: Calculate maximum accelerations
The maximum acceleration \( A \) in SHM is given by:
\[ A_{\text{max}} = A \omega^2 \]

For \( y_1 \):
\[ A_1 = \frac{dV_1}{dt} = -4A \omega^2 \cos(2\omega t) \]
Thus, the maximum value is:
\[ A_{1,\text{max}} = 4A \omega^2 \]

For \( y_2 \):
\[ A_2 = \frac{dV_2}{dt} = -A \omega^2 \sin(\omega t + \frac{\pi}{3}) \]
Thus, the maximum value is:
\[ A_{2,\text{max}} = 2A \omega^2 \]

### Step 7: Calculate the ratio of maximum accelerations
The ratio of maximum accelerations is:
\[ \frac{A_{1,\text{max}}}{A_{2,\text{max}}} = \frac{4A \omega^2}{2A \omega^2} = 2:1 \]

### Summary of Results
- Only one of the equations represents SHM: \( y_2 \).
- The ratio of maximum speeds is \( 1:1 \).
- The ratio of maximum accelerations is \( 2:1 \).