A particle performing simple harmonic motion having time period 3 s is in phase with another particle which also undergoes simple harmonic motion at `t=0`. The time period of second particle is T (less that 3s). If they are again in the same phase for the third time after 45 s, then the value of T will be

Options

A. 2.8 s
B. 2.7 s
C. 2.5 s
D. none of these

Explanation

To solve the problem, we need to find the time period \( T \) of the second particle, given that it is less than 3 seconds and that both particles are in the same phase for the third time after 45 seconds.

### Step-by-Step Solution:

1. **Identify the Time Periods**:
- Let the time period of the first particle be \( T_1 = 3 \) seconds.
- Let the time period of the second particle be \( T_2 = T \) seconds (where \( T < 3 \)).

2. **Understanding Phase Condition**:
- Both particles start in phase at \( t = 0 \).
- The condition for the two particles to be in phase again is given by:
\[ \phi_2 - \phi_1 = n \cdot 2\pi \]
where \( n \) is an integer.

3. **Expressing Phase in Terms of Angular Frequency**:
- The angular frequency \( \omega \) is related to the time period by:
\[ \omega = \frac{2\pi}{T} \]
- For the first particle:
\[ \phi_1 = \omega_1 t = \frac{2\pi}{T_1} t = \frac{2\pi}{3} t \]
- For the second particle:
\[ \phi_2 = \omega_2 t = \frac{2\pi}{T_2} t = \frac{2\pi}{T} t \]

4. **Setting Up the Equation**:
- At the third occurrence of being in phase, we have \( n = 3 \):
\[ \phi_2 - \phi_1 = 3 \cdot 2\pi \]
- Substituting the expressions for \( \phi_1 \) and \( \phi_2 \):
\[ \frac{2\pi}{T} t - \frac{2\pi}{3} t = 6\pi \]

5. **Simplifying the Equation**:
- Factor out \( 2\pi t \):
\[ 2\pi t \left(\frac{1}{T} - \frac{1}{3}\right) = 6\pi \]
- Dividing both sides by \( 2\pi \):
\[ t \left(\frac{1}{T} - \frac{1}{3}\right) = 3 \]

6. **Substituting \( t = 45 \) seconds**:
- Substitute \( t = 45 \):
\[ 45 \left(\frac{1}{T} - \frac{1}{3}\right) = 3 \]
- Rearranging gives:
\[ \frac{1}{T} - \frac{1}{3} = \frac{3}{45} = \frac{1}{15} \]

7. **Solving for \( T \)**:
- Rearranging the equation:
\[ \frac{1}{T} = \frac{1}{15} + \frac{1}{3} \]
- Finding a common denominator (15):
\[ \frac{1}{T} = \frac{1}{15} + \frac{5}{15} = \frac{6}{15} = \frac{2}{5} \]
- Taking the reciprocal:
\[ T = \frac{5}{2} = 2.5 \text{ seconds} \]

### Final Answer:
The time period \( T \) of the second particle is \( 2.5 \) seconds.

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