A wave has SHM (simple harmonic motion) whose period is 4s while another periods 3 s. If both are combined, then the resultant wave will have the period equal to

Options

A. 4 s
B. 5 s
C. 12 s
D. 3 s

Answer

To find the period of the resultant wave when two waves with different periods are combined, we can follow these steps:

### Step 1: Identify the periods of the two waves
Let the period of the first wave, \( T_1 \), be 4 seconds and the period of the second wave, \( T_2 \), be 3 seconds.

### Step 2: Calculate the frequencies of the waves
The frequency \( \nu \) of a wave is given by the formula:
\[ \nu = \frac{1}{T} \]
So, we calculate the frequencies of both waves:
- For the first wave:
\[ \nu_1 = \frac{1}{T_1} = \frac{1}{4} \text{ Hz} \]
- For the second wave:
\[ \nu_2 = \frac{1}{T_2} = \frac{1}{3} \text{ Hz} \]

### Step 3: Calculate the difference in frequencies
The difference in frequencies \( \Delta \nu \) is given by:
\[ \Delta \nu = |\nu_1 - \nu_2| = \left|\frac{1}{4} - \frac{1}{3}\right| \]
To perform this calculation, we need a common denominator:
\[ \Delta \nu = \left|\frac{3}{12} - \frac{4}{12}\right| = \left|-\frac{1}{12}\right| = \frac{1}{12} \text{ Hz} \]

### Step 4: Calculate the period of the resultant wave
The period \( T \) of the resultant wave is the reciprocal of the difference in frequencies:
\[ T = \frac{1}{\Delta \nu} = \frac{1}{\frac{1}{12}} = 12 \text{ seconds} \]

### Conclusion
The period of the resultant wave when the two waves are combined is **12 seconds**.

---