Two SHM are represcnted by equations `y_(1)=6cos(6pit+(pi)/(6)),y_(2)=3(sqrt(3)sin3pit+cos3pit)`

Options

A. ratio of their amplitudes is 1
B. ratio of their time periods 1
C. ratio of their maximum velocities is 1
D. ratio of their maximum accleration is 1

Answer

To solve the problem, we need to analyze the two simple harmonic motions (SHMs) given by the equations:

1. \( y_1 = 6 \cos(6\pi t + \frac{\pi}{6}) \)
2. \( y_2 = 3\sqrt{3} \sin(3\pi t) + \cos(3\pi t) \)

We will find the following ratios:
1. Ratio of their amplitudes
2. Ratio of their time periods
3. Ratio of their maximum velocities
4. Ratio of their maximum accelerations

### Step 1: Finding the Amplitudes
The amplitude of an SHM represented in the form \( y = A \cos(\omega t + \phi) \) is \( A \).

- For \( y_1 \): The amplitude \( A_1 = 6 \).
- For \( y_2 \): We need to convert it into the standard form.

We can rewrite \( y_2 \) as:
\[ y_2 = 3\sqrt{3} \sin(3\pi t) + \cos(3\pi t) \]
Using the formula \( R \sin(\theta) + S \cos(\theta) = \sqrt{R^2 + S^2} \cos(\theta - \phi) \) where \( \tan(\phi) = \frac{R}{S} \):
- Here, \( R = 3\sqrt{3} \) and \( S = 1 \).
- The amplitude \( A_2 = \sqrt{(3\sqrt{3})^2 + 1^2} = \sqrt{27 + 1} = \sqrt{28} = 2\sqrt{7} \).

**Ratio of Amplitudes**:
\[ \frac{A_1}{A_2} = \frac{6}{2\sqrt{7}} = \frac{3}{\sqrt{7}} \]

### Step 2: Finding the Time Periods
The time period \( T \) of SHM is given by \( T = \frac{2\pi}{\omega} \).

- For \( y_1 \): \( \omega_1 = 6\pi \) → \( T_1 = \frac{2\pi}{6\pi} = \frac{1}{3} \).
- For \( y_2 \): \( \omega_2 = 3\pi \) → \( T_2 = \frac{2\pi}{3\pi} = \frac{2}{3} \).

**Ratio of Time Periods**:
\[ \frac{T_1}{T_2} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]

### Step 3: Finding the Maximum Velocities
The maximum velocity \( V_{max} \) is given by \( V_{max} = \omega A \).

- For \( y_1 \): \( V_{1} = \omega_1 A_1 = 6\pi \cdot 6 = 36\pi \).
- For \( y_2 \): \( V_{2} = \omega_2 A_2 = 3\pi \cdot 2\sqrt{7} = 6\pi\sqrt{7} \).

**Ratio of Maximum Velocities**:
\[ \frac{V_1}{V_2} = \frac{36\pi}{6\pi\sqrt{7}} = \frac{6}{\sqrt{7}} \]

### Step 4: Finding the Maximum Accelerations
The maximum acceleration \( A_{max} \) is given by \( A_{max} = \omega^2 A \).

- For \( y_1 \): \( A_{1} = (6\pi)^2 \cdot 6 = 36\pi^2 \cdot 6 = 216\pi^2 \).
- For \( y_2 \): \( A_{2} = (3\pi)^2 \cdot 2\sqrt{7} = 9\pi^2 \cdot 2\sqrt{7} = 18\pi^2\sqrt{7} \).

**Ratio of Maximum Accelerations**:
\[ \frac{A_1}{A_2} = \frac{216\pi^2}{18\pi^2\sqrt{7}} = \frac{12}{\sqrt{7}} \]

### Summary of Ratios
1. **Ratio of Amplitudes**: \( \frac{3}{\sqrt{7}} \)
2. **Ratio of Time Periods**: \( \frac{1}{2} \)
3. **Ratio of Maximum Velocities**: \( \frac{6}{\sqrt{7}} \)
4. **Ratio of Maximum Accelerations**: \( \frac{12}{\sqrt{7}} \)