A particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motion is a. `0^@, b. 60^@, c. 90^@`

Explanation

To solve the problem of finding the resultant amplitude when a particle is subjected to two simple harmonic motions (SHM) with the same time period and in the same direction, we will use the formula for the resultant amplitude based on the phase difference between the two motions.

Given:
- Amplitude of the first motion, \( r_1 = 3.0 \, \text{cm} \)
- Amplitude of the second motion, \( r_2 = 4.0 \, \text{cm} \)
- Phase difference, \( \phi \)

The formula for the resultant amplitude \( R \) when two SHMs are combined is:

\[ R = \sqrt{r_1^2 + r_2^2 + 2 r_1 r_2 \cos(\phi)} \]

Now, we will calculate \( R \) for three different phase differences: \( 0^\circ \), \( 60^\circ \), and \( 90^\circ \).

### Step 1: Calculate for \( \phi = 0^\circ \)

1. Substitute \( r_1 = 3 \, \text{cm} \), \( r_2 = 4 \, \text{cm} \), and \( \cos(0^\circ) = 1 \) into the formula:
\[ R = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot 1} \]
2. Calculate each term:
- \( 3^2 = 9 \)
- \( 4^2 = 16 \)
- \( 2 \cdot 3 \cdot 4 \cdot 1 = 24 \)
3. Combine the results:
\[ R = \sqrt{9 + 16 + 24} = \sqrt{49} = 7 \, \text{cm} \]

### Step 2: Calculate for \( \phi = 60^\circ \)

1. Substitute \( r_1 = 3 \, \text{cm} \), \( r_2 = 4 \, \text{cm} \), and \( \cos(60^\circ) = \frac{1}{2} \):
\[ R = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot \frac{1}{2}} \]
2. Calculate each term:
- \( 3^2 = 9 \)
- \( 4^2 = 16 \)
- \( 2 \cdot 3 \cdot 4 \cdot \frac{1}{2} = 12 \)
3. Combine the results:
\[ R = \sqrt{9 + 16 + 12} = \sqrt{37} \approx 6.1 \, \text{cm} \]

### Step 3: Calculate for \( \phi = 90^\circ \)

1. Substitute \( r_1 = 3 \, \text{cm} \), \( r_2 = 4 \, \text{cm} \), and \( \cos(90^\circ) = 0 \):
\[ R = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot 0} \]
2. Calculate each term:
- \( 3^2 = 9 \)
- \( 4^2 = 16 \)
- \( 2 \cdot 3 \cdot 4 \cdot 0 = 0 \)
3. Combine the results:
\[ R = \sqrt{9 + 16 + 0} = \sqrt{25} = 5 \, \text{cm} \]

### Final Result

- For \( \phi = 0^\circ \), \( R = 7 \, \text{cm} \)
- For \( \phi = 60^\circ \), \( R \approx 6.1 \, \text{cm} \)
- For \( \phi = 90^\circ \), \( R = 5 \, \text{cm} \)