The length of a sonometer wire is `90cm` and the stationary wave setup in the wire is represented by an equation `y=6sin((pix)/(30))cos(250t)` where `x` , `y` are in cm and `t` is in second. The distances of successive antinodes from one end of the wire are

Options

A. `22.5cm`, `67.5cm`
B. `15cm`, `30cm`, `60cm`
C. `15cm`, `45cm`, `75cm`
D. `30cm`, `45cm`, `60cm`

Explanation

To solve the problem, we need to find the distances of successive antinodes from one end of the sonometer wire. The given wave equation is:

\[ y = 6 \sin\left(\frac{\pi}{30} x\right) \cos(250t) \]

### Step 1: Identify the Wavelength
From the wave equation, we can identify the wave number \( k \) which is given by:

\[ k = \frac{\pi}{30} \]

The relationship between the wave number \( k \) and the wavelength \( \lambda \) is:

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

By rearranging this, we can find \( \lambda \):

\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{\frac{\pi}{30}} = 60 \text{ cm} \]

### Step 2: Determine the Distance of Antinodes
In a stationary wave setup, the distance between successive antinodes is given by:

\[ \text{Distance between successive antinodes} = \frac{\lambda}{2} \]

Since we have the wavelength \( \lambda = 60 \) cm, the distance between successive antinodes is:

\[ \frac{\lambda}{2} = \frac{60}{2} = 30 \text{ cm} \]

### Step 3: Calculate the Positions of Antinodes
The positions of the antinodes can be calculated from the starting point (one end of the wire). The positions of the antinodes can be expressed as:

1. First Antinode: \( x_1 = \frac{\lambda}{4} = \frac{60}{4} = 15 \text{ cm} \)
2. Second Antinode: \( x_2 = \frac{3\lambda}{4} = \frac{3 \times 60}{4} = 45 \text{ cm} \)
3. Third Antinode: \( x_3 = \frac{5\lambda}{4} = \frac{5 \times 60}{4} = 75 \text{ cm} \)
4. Fourth Antinode: \( x_4 = \frac{7\lambda}{4} = \frac{7 \times 60}{4} = 105 \text{ cm} \) (This exceeds the length of the wire, which is 90 cm)

### Step 4: List the Valid Antinode Positions
The valid positions of the antinodes that fall within the length of the wire (90 cm) are:

- First Antinode: 15 cm
- Second Antinode: 45 cm
- Third Antinode: 75 cm

### Final Answer
The distances of successive antinodes from one end of the wire are:

- 15 cm
- 45 cm
- 75 cm