An SHM is give by `y=5["sin"(3pit)+sqrt(3)"cos"(3pit)]`. What is the amplitude of the motion of y in metre ?

Options

A. 10
B. 20
C. 1
D. 5

Explanation

To find the amplitude of the simple harmonic motion (SHM) given by the equation \( y = 5 \sin(3\pi t) + \sqrt{3} \cos(3\pi t) \), we can follow these steps:

### Step 1: Identify the Amplitudes
The equation consists of two components:
- The first component is \( 5 \sin(3\pi t) \), where the amplitude \( A_1 = 5 \).
- The second component is \( \sqrt{3} \cos(3\pi t) \), where the amplitude \( A_2 = \sqrt{3} \).

### Step 2: Calculate the Resultant Amplitude
When two SHM motions are superimposed, the resultant amplitude \( A_{\text{net}} \) can be calculated using the formula:
\[ A_{\text{net}} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta \phi)} \]
where \( \Delta \phi \) is the phase difference between the sine and cosine components.

### Step 3: Determine the Phase Difference
The phase difference \( \Delta \phi \) between \( \sin \) and \( \cos \) functions is \( \frac{\pi}{2} \) (90 degrees).

### Step 4: Substitute Values into the Formula
Now, substituting the values into the formula:
- \( A_1 = 5 \)
- \( A_2 = \sqrt{3} \)
- \( \Delta \phi = \frac{\pi}{2} \)

We calculate:
\[ A_{\text{net}} = \sqrt{5^2 + (\sqrt{3})^2 + 2 \cdot 5 \cdot \sqrt{3} \cdot \cos\left(\frac{\pi}{2}\right)} \]

### Step 5: Simplify the Expression
Calculating each term:
- \( A_1^2 = 25 \)
- \( A_2^2 = 3 \)
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)

Thus, the expression simplifies to:
\[ A_{\text{net}} = \sqrt{25 + 3 + 0} = \sqrt{28} \]

### Step 6: Final Calculation
Calculating \( \sqrt{28} \):
\[ A_{\text{net}} = \sqrt{4 \cdot 7} = 2\sqrt{7} \]

### Step 7: Approximate the Result
To find the approximate value of \( 2\sqrt{7} \):
\[ \sqrt{7} \approx 2.64575 \implies 2\sqrt{7} \approx 5.2915 \]

### Final Answer
The amplitude of the motion \( y \) is approximately \( 5.2915 \) meters, which can be rounded to \( 5 \) meters.

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