A particle is subjected to two SHMs simultaneously
`X_(1) = a_(1) sinomegat` and `X_(2) = a_(2)sin(omegat + phi)`
Where `a_(1) = 3.0 cm, a_(2) = 4.0 cm`
Find resultant amplitude if the phase difference `phi` has value `30^(0)`

Explanation

To find the resultant amplitude of a particle subjected to two simultaneous simple harmonic motions (SHMs), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- For the first SHM:
\[ X_1 = a_1 \sin(\omega t) \]
where \( a_1 = 3.0 \, \text{cm} \).
- For the second SHM:
\[ X_2 = a_2 \sin(\omega t + \phi) \]
where \( a_2 = 4.0 \, \text{cm} \) and \( \phi = 30^\circ \).

2. **Determine the Phase Difference:**
- The phase of the first SHM is \( \phi_1 = \omega t \).
- The phase of the second SHM is \( \phi_2 = \omega t + 30^\circ \).
- The phase difference \( \Delta \phi \) is calculated as:
\[ \Delta \phi = \phi_2 - \phi_1 = (\omega t + 30^\circ) - (\omega t) = 30^\circ. \]

3. **Use the Formula for Resultant Amplitude:**
- The resultant amplitude \( A \) when two SHMs are combined is given by:
\[ A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos(\Delta \phi)}. \]

4. **Substitute the Values:**
- Substitute \( a_1 = 3 \, \text{cm} \), \( a_2 = 4 \, \text{cm} \), and \( \Delta \phi = 30^\circ \):
\[ A = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot \cos(30^\circ)}. \]

5. **Calculate Each Term:**
- Calculate \( 3^2 = 9 \).
- Calculate \( 4^2 = 16 \).
- Calculate \( 2 \cdot 3 \cdot 4 = 24 \).
- Calculate \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \).
- Thus, \( 24 \cdot \cos(30^\circ) = 24 \cdot \frac{\sqrt{3}}{2} = 12\sqrt{3} \).

6. **Combine the Results:**
- Now, substitute back into the amplitude formula:
\[ A = \sqrt{9 + 16 + 12\sqrt{3}} = \sqrt{25 + 12\sqrt{3}}. \]

7. **Final Result:**
- The resultant amplitude is:
\[ A = \sqrt{25 + 12\sqrt{3}} \, \text{cm}. \]