The displacement `xi` in centimetres of a particle is `xi = 3 sin 314 t + 4 cos 314 t`. Amplitude and initial phase are

Options

A. ` 5 cm , tan^(-1) (4/3)`
B. ` 3 cm , tan^(-1) (3)/(4)`
C. ` 4 cm , tan ^(-1) (4)/(9)`
D. ` 4 cm , 0`

Explanation

To find the amplitude and initial phase of the given displacement \( x_i = 3 \sin(314t) + 4 \cos(314t) \), we can follow these steps:

### Step 1: Identify the components of the displacement
The displacement can be expressed as two components:
- \( x_1 = 3 \sin(314t) \)
- \( x_2 = 4 \cos(314t) \)

### Step 2: Convert cosine to sine
To combine these two components, we can express the cosine term in terms of sine:
\[ x_2 = 4 \cos(314t) = 4 \sin\left(314t + 90^\circ\right) \]
This allows us to rewrite the displacement as:
\[ x_i = 3 \sin(314t) + 4 \sin\left(314t + 90^\circ\right) \]

### Step 3: Use vector addition
We can represent these components as vectors in a phasor diagram:
- The vector for \( x_1 \) has a magnitude of 3 and is along the x-axis (0 degrees).
- The vector for \( x_2 \) has a magnitude of 4 and is along the y-axis (90 degrees).

### Step 4: Calculate the resultant amplitude
Using the Pythagorean theorem, the resultant amplitude \( A_R \) can be calculated as:
\[ A_R = \sqrt{(3^2) + (4^2)} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm} \]

### Step 5: Calculate the initial phase
To find the initial phase \( \theta \), we use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{3} \]
Thus, we can find \( \theta \) as:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]
Calculating this gives:
\[ \theta \approx 53.13^\circ \]

### Final Result
The amplitude \( A_R \) is 5 cm and the initial phase \( \theta \) is approximately \( 53^\circ \).

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