Two SHMs `s_(1) = a sin omega t` and `s_(2) = b sin omega t` are superimposed on a particle. The `s_(1)` and `s_(2)` are along the direction which makes `37^(@)` to each other

Options

A. the particle will perform SHM
B. the path of particle is straight line
C. Both (a) and (b) are correct
D. Both (a) and (b) are wrong

Explanation

To solve the problem of two superimposed simple harmonic motions (SHMs) given by \( s_1 = a \sin(\omega t) \) and \( s_2 = b \sin(\omega t) \) at an angle of \( 37^\circ \) to each other, we can follow these steps:

### Step 1: Understand the Superposition of SHMs
When two SHMs are superimposed, the resultant displacement can be expressed as a vector sum. The two displacements \( s_1 \) and \( s_2 \) can be represented as vectors in a two-dimensional plane.

### Step 2: Resolve \( s_2 \) into Components
Since \( s_2 \) makes an angle of \( 37^\circ \) with \( s_1 \), we can resolve \( s_2 \) into its components:
- The x-component of \( s_2 \) is \( s_2 \cos(37^\circ) \)
- The y-component of \( s_2 \) is \( s_2 \sin(37^\circ) \)

### Step 3: Write the Components
Using the given equations:
- \( s_1 = a \sin(\omega t) \)
- \( s_2 = b \sin(\omega t) \)

The components become:
- \( x = s_1 + s_2 \cos(37^\circ) = a \sin(\omega t) + b \cos(37^\circ) \sin(\omega t) \)
- \( y = s_2 \sin(37^\circ) = b \sin(37^\circ) \sin(\omega t) \)

### Step 4: Substitute the Values of \( \cos(37^\circ) \) and \( \sin(37^\circ) \)
Using the trigonometric values:
- \( \cos(37^\circ) \approx 0.8 \)
- \( \sin(37^\circ) \approx 0.6 \)

We can rewrite the components:
- \( x = a \sin(\omega t) + 0.8b \sin(\omega t) = (a + 0.8b) \sin(\omega t) \)
- \( y = 0.6b \sin(\omega t) \)

### Step 5: Express \( \sin(\omega t) \) in Terms of \( y \)
From the equation for \( y \):
\[ \sin(\omega t) = \frac{y}{0.6b} \]

### Step 6: Substitute \( \sin(\omega t) \) into the \( x \) Equation
Substituting into the \( x \) equation gives:
\[ x = (a + 0.8b) \left(\frac{y}{0.6b}\right) = \frac{(a + 0.8b)}{0.6b} y \]

### Step 7: Identify the Resultant Motion
This equation represents a straight line in the \( xy \)-plane, indicating that the particle moves in a straight line.

### Conclusion
Both statements in the question are correct:
1. The particle performs SHM.
2. The path of the particle is a straight line.