A simple harmonic motion is represented by : `y = 5(sin 3pi t + sqrt(3) cos 3pi t)cm` The amplitude and time period of the motion are :

Options

A. `10 cm, (3)/(2)s`
B. `5 cm, (3)/(2)s`
C. `10 cm, (2)/(3)s`
D. `5cm, (2)/(3)s`

Explanation

To solve the problem, we need to find the amplitude and time period of the simple harmonic motion represented by the equation:

\[ y = 5(\sin(3\pi t) + \sqrt{3} \cos(3\pi t)) \, \text{cm} \]

### Step 1: Rewrite the equation in standard form
We can express the given equation in the form of \( y = A \sin(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

### Step 2: Identify coefficients
From the equation:
- Coefficient of \( \sin(3\pi t) \) is \( 5 \)
- Coefficient of \( \cos(3\pi t) \) is \( \sqrt{3} \)

### Step 3: Calculate the amplitude
The amplitude \( A \) can be calculated using the formula:
\[ A = \sqrt{(5)^2 + (\sqrt{3})^2} \]

Calculating this gives:
\[ A = \sqrt{25 + 3} = \sqrt{28} = 2\sqrt{7} \, \text{cm} \]

### Step 4: Determine the angular frequency
From the equation, we see that:
\[ \omega = 3\pi \]

### Step 5: Calculate the time period
The time period \( T \) is given by the formula:
\[ T = \frac{2\pi}{\omega} \]

Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{3\pi} = \frac{2}{3} \, \text{seconds} \]

### Final Results
- Amplitude \( A = 2\sqrt{7} \, \text{cm} \)
- Time period \( T = \frac{2}{3} \, \text{seconds} \)

### Summary
The amplitude of the motion is \( 2\sqrt{7} \, \text{cm} \) and the time period is \( \frac{2}{3} \, \text{seconds} \).

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