Two simple harmonic motions are given by `y _(1) = 5 sin ( omegat- pi //3). y_(2) = 5 ( sin omegat+ sqrt(3) cos omegat)`. Ratio of their amplitudes is

Options

A. `1:2`
B. `1:4`
C. `1:5`
D. `1:8`

Answer

To solve the problem of finding the ratio of the amplitudes of two simple harmonic motions given by \( y_1 = 5 \sin(\omega t - \frac{\pi}{3}) \) and \( y_2 = 5 \left( \sin(\omega t) + \sqrt{3} \cos(\omega t) \right) \), we will follow these steps:

### Step 1: Identify the amplitude of \( y_1 \)
The first equation is given as:
\[ y_1 = 5 \sin(\omega t - \frac{\pi}{3}) \]
In this equation, the amplitude \( A_1 \) is simply the coefficient of the sine function, which is:
\[ A_1 = 5 \]

### Step 2: Simplify the expression for \( y_2 \)
The second equation is given as:
\[ y_2 = 5 \left( \sin(\omega t) + \sqrt{3} \cos(\omega t) \right) \]
To find the amplitude, we can rewrite this expression in the form of \( R \sin(\omega t + \phi) \), where \( R \) is the resultant amplitude.

### Step 3: Calculate \( R \) using the formula
The resultant amplitude \( R \) can be calculated using the formula:
\[ R = \sqrt{A^2 + B^2} \]
where \( A \) is the coefficient of \( \sin(\omega t) \) and \( B \) is the coefficient of \( \cos(\omega t) \).

Here, \( A = 5 \) and \( B = 5\sqrt{3} \). Thus,
\[ R = \sqrt{(5)^2 + (5\sqrt{3})^2} = \sqrt{25 + 75} = \sqrt{100} = 10 \]

### Step 4: Find the ratio of the amplitudes
Now that we have both amplitudes:
- \( A_1 = 5 \)
- \( A_2 = 10 \)

The ratio of the amplitudes \( \frac{A_1}{A_2} \) is:
\[ \frac{A_1}{A_2} = \frac{5}{10} = \frac{1}{2} \]

### Final Answer
The ratio of the amplitudes \( A_1 : A_2 \) is \( 1 : 2 \).

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