When two mutually perpendicular simple harmonic motions of same frequency, amplitude and phase are superimposed.

Options

A. A. the resulting motion is uniform circular motion.
B. B. the resulting motion is a linear simple harmonic motions along a straight ine inclined equally to the straight lines of motion of component ones.
C. C. the resulting motion is an elliptical motion, symmetrical about the lines of motion of the compounents.
D. D. the two `S.H.M.` will cancel each other.

Explanation

To solve the problem of superimposing two mutually perpendicular simple harmonic motions (SHMs) of the same frequency, amplitude, and phase, we can follow these steps:

### Step-by-step Solution:

1. **Understanding the SHMs**:
We have two SHMs, one along the x-axis and the other along the y-axis. Let's denote the amplitude of both SHMs as \( A \) and the angular frequency as \( \omega \).

2. **Equations of Motion**:
The equations for the two SHMs can be expressed as:
- For the x-direction: \( x(t) = A \sin(\omega t) \)
- For the y-direction: \( y(t) = A \sin(\omega t) \)

3. **Phase Difference**:
Since the problem states that the phase difference is zero (both SHMs are in phase), we can directly use the above equations.

4. **Resultant Motion**:
To find the resultant motion, we can express the relationship between \( x \) and \( y \). Since both equations are identical, we can set them equal to each other:
\[ x = y \]

5. **Geometric Interpretation**:
The equation \( x = y \) represents a straight line that passes through the origin with a slope of 1. This indicates that the resultant motion is along the line \( y = x \).

6. **Conclusion**:
Therefore, the resultant motion is a linear simple harmonic motion along a straight line inclined equally to the axes of the component motions. This corresponds to option B in the question.

### Final Answer:
The correct option is **B**: The resultant motion is linear simple harmonic motion along a straight line inclined equally to the axes of the component motions.

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