Two mutually perpendicular simple harmonic vibrations have same amplitude, frequency and phase. When they superimpose, the resultant form of vibration will be

Options

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

Answer

To solve the problem of finding the resultant form of vibration when two mutually perpendicular simple harmonic motions (SHM) with the same amplitude, frequency, and phase superimpose, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the SHM equations**:
Let the first simple harmonic motion be represented as:
\[ Y_1 = A \sin(\omega t) \]
Since the second SHM is mutually perpendicular and has the same amplitude, frequency, and phase, we can represent it as:
\[ Y_2 = A \sin(\omega t + 90^\circ) \]

2. **Convert the second SHM**:
The sine function can be expressed in terms of cosine for a phase shift of 90 degrees:
\[ Y_2 = A \cos(\omega t) \]

3. **Superimpose the two SHMs**:
The resultant displacement \( Y \) when these two motions are superimposed can be expressed as:
\[ Y = Y_1 + Y_2 = A \sin(\omega t) + A \cos(\omega t) \]

4. **Factor out the amplitude**:
We can factor out \( A \):
\[ Y = A (\sin(\omega t) + \cos(\omega t)) \]

5. **Use the trigonometric identity**:
To combine the sine and cosine terms, we can use the identity:
\[ \sin(x) + \cos(x) = \sqrt{2} \sin\left(x + 45^\circ\right) \]
Therefore, we can rewrite \( Y \) as:
\[ Y = A \sqrt{2} \sin\left(\omega t + 45^\circ\right) \]

6. **Conclusion**:
The resultant form of the vibration when the two SHMs superimpose is:
\[ Y = A \sqrt{2} \sin\left(\omega t + 45^\circ\right) \]

### Final Result:
The resultant form of vibration will be a simple harmonic motion with an amplitude of \( A \sqrt{2} \) and a phase shift of \( 45^\circ \).