A standing wave is created on a string of length `120 m` and it is vibrating in `6 th` harmonic. Maximum possible amplitude of any particle is `10 cm` and maximum possible velocity will be `10 cm//s`. Choose the correct statement.

Options

A. Angular wave number of two waves will be `(pi)/(20)`
B. Time period of any particle's `SHM` will be `4 pi sec`.
C. Any particle will have same kinetic energy as potential energy
D. Amplitude of interfering waves are `10 cm` each.

Explanation

To solve the problem step by step, we will analyze the standing wave created on a string of length 120 m vibrating in the 6th harmonic.

### Step 1: Understanding the Harmonic
In a standing wave on a string, the length of the string (L) is related to the wavelength (λ) and the harmonic number (n). For the nth harmonic, the relationship is given by:

\[ L = \frac{n \lambda}{2} \]

Here, n = 6 (since it is the 6th harmonic) and L = 120 m.

### Step 2: Finding the Wavelength
Substituting the values into the equation:

\[ 120 = \frac{6 \lambda}{2} \]

This simplifies to:

\[ 120 = 3 \lambda \]

Now, solving for λ:

\[ \lambda = \frac{120}{3} = 40 \text{ m} \]

### Step 3: Calculating the Angular Wave Number (k)
The angular wave number (k) is given by the formula:

\[ k = \frac{2\pi}{\lambda} \]

Substituting the wavelength we found:

\[ k = \frac{2\pi}{40} = \frac{\pi}{20} \text{ rad/m} \]

### Step 4: Finding the Maximum Velocity (v_max)
The maximum velocity of a particle in a standing wave is given by the product of the amplitude (A) and the angular frequency (ω):

\[ v_{max} = A \cdot \omega \]

We know the maximum possible amplitude A = 10 cm = 0.1 m and the maximum velocity v_max = 10 cm/s = 0.1 m/s.

### Step 5: Calculating Angular Frequency (ω)
Rearranging the equation for maximum velocity to find ω:

\[ \omega = \frac{v_{max}}{A} \]

Substituting the values:

\[ \omega = \frac{0.1}{0.1} = 1 \text{ rad/s} \]

### Step 6: Finding the Time Period (T)
The time period (T) is related to the angular frequency (ω) by the formula:

\[ T = \frac{2\pi}{\omega} \]

Substituting the value of ω:

\[ T = \frac{2\pi}{1} = 2\pi \text{ seconds} \]

### Step 7: Evaluating the Options
Now we can evaluate the options based on our calculations:
- Option A: Angular wave number is \( \frac{\pi}{20} \) - **Correct**
- Option B: Time period of the particle is \( 4\pi \) - **Incorrect** (it is \( 2\pi \))
- Option C: Particle will have the same kinetic and potential energy - **Incorrect**
- Option D: Amplitude of the interfering waves is 10 cm - **Incorrect**

### Conclusion
The correct statement is Option A.

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