A particle of mass `50 g` participates in two simple harmonic oscillations simultaneously as given by `x_(1) = 10 (cm) cos [80 pi(s^(-1))t]` and `x_(2) = 5(cm) sin [(80 pi(s^(-1))t + pi//6]`. The amplitude of particle's oscillations is given by 'A'. Find the value of `A^(2) (in cm^(2))`.

Options

A. 175
B. 165
C. 275
D. 375

Explanation

To find the amplitude \( A \) of a particle participating in two simple harmonic oscillations, we can use the principle of superposition. The given equations for the oscillations are:

\[ x_1 = 10 \, \text{cm} \cos(80 \pi t) \]
\[ x_2 = 5 \, \text{cm} \sin(80 \pi t + \frac{\pi}{6}) \]

### Step 1: Identify the Amplitudes
From the equations, we identify the amplitudes:
- Amplitude of \( x_1 \), \( A_1 = 10 \, \text{cm} \)
- Amplitude of \( x_2 \), \( A_2 = 5 \, \text{cm} \)

### Step 2: Determine the Phase Difference
The second oscillation can be rewritten in terms of cosine using the sine addition formula:
\[ x_2 = 5 \, \text{cm} \left( \sin(80 \pi t) \cos\left(\frac{\pi}{6}\right) + \cos(80 \pi t) \sin\left(\frac{\pi}{6}\right) \right) \]
Using \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \) and \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \):
\[ x_2 = 5 \, \text{cm} \left( \sin(80 \pi t) \cdot \frac{\sqrt{3}}{2} + \cos(80 \pi t) \cdot \frac{1}{2} \right) \]

### Step 3: Combine the Two Oscillations
We can express the total displacement \( x \) as:
\[ x = x_1 + x_2 \]
Using the cosine form for both terms:
\[ x = 10 \, \text{cm} \cos(80 \pi t) + 5 \, \text{cm} \left( \frac{\sqrt{3}}{2} \sin(80 \pi t) + \frac{1}{2} \cos(80 \pi t) \right) \]
This can be rearranged to:
\[ x = \left(10 + \frac{5}{2}\right) \cos(80 \pi t) + \frac{5\sqrt{3}}{2} \sin(80 \pi t) \]

### Step 4: Calculate the Resultant Amplitude
The resultant amplitude \( A \) can be calculated using the formula:
\[ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]
Where \( \phi \) is the phase difference. Since \( x_1 \) is a cosine function and \( x_2 \) has a phase shift of \( \frac{\pi}{6} \), we can find \( \phi \) as:
\[ \phi = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \]
Thus, \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).

### Step 5: Substitute Values into the Amplitude Formula
Now substituting the values:
\[ A = \sqrt{10^2 + 5^2 + 2 \cdot 10 \cdot 5 \cdot \frac{1}{2}} \]
Calculating each term:
- \( 10^2 = 100 \)
- \( 5^2 = 25 \)
- \( 2 \cdot 10 \cdot 5 \cdot \frac{1}{2} = 50 \)

So,
\[ A = \sqrt{100 + 25 + 50} = \sqrt{175} \]

### Step 6: Find the Value of \( A^2 \)
Finally, we need \( A^2 \):
\[ A^2 = 175 \, \text{cm}^2 \]

### Final Answer
The value of \( A^2 \) is \( 175 \, \text{cm}^2 \).

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