Two oscillations `x_(1)=Asinwt` and `x_(2)=Acoswt` superimpose at right angles in `x` and `y` axis respectively. What will be the resultant wave form ?

Options

A. ellipse
B. straight line
C. circle
D. parabola

Answer

To find the resultant waveform when two oscillations \( x_1 = A \sin(\omega t) \) and \( x_2 = A \cos(\omega t) \) superimpose at right angles, we can follow these steps:

### Step 1: Understand the Oscillations
The two oscillations are given as:
- \( x_1 = A \sin(\omega t) \) (along the x-axis)
- \( x_2 = A \cos(\omega t) \) (along the y-axis)

These oscillations are perpendicular to each other.

### Step 2: Relate the Two Oscillations
We can express \( x_2 \) in terms of \( x_1 \). Since \( \sin(\omega t) \) and \( \cos(\omega t) \) are related through the Pythagorean identity, we can write:
\[ x_2 = A \cos(\omega t) = A \sqrt{1 - \sin^2(\omega t)} = A \sqrt{1 - \left(\frac{x_1}{A}\right)^2} \]

### Step 3: Square the Equations
To eliminate \( t \), we square both equations:
\[ x_1^2 = A^2 \sin^2(\omega t) \]
\[ x_2^2 = A^2 \cos^2(\omega t) \]

### Step 4: Add the Squared Equations
Now, we add these two equations:
\[ x_1^2 + x_2^2 = A^2 \sin^2(\omega t) + A^2 \cos^2(\omega t) \]
Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ x_1^2 + x_2^2 = A^2 (\sin^2(\omega t) + \cos^2(\omega t)) = A^2 \]

### Step 5: Resultant Waveform
The equation \( x_1^2 + x_2^2 = A^2 \) represents a circle in the x-y plane with a radius \( A \). Thus, the resultant waveform is circular.

### Final Result
The resultant waveform formed by the superposition of the two oscillations is a circular motion in the x-y plane.

---