The equation of a travelling wave is,
`Y = A sin 2pi (pt-x//5)`
Then the ratio of maximum particle velocity to wave velocity is,

Options

A. `(pi a )/(5) `
B. `2sqrt(5pia ) `
C. `(2pi a )/(5) `
D. `(2pi a )/(sqrt(5))`

Answer

To solve the problem, we need to find the ratio of maximum particle velocity to wave velocity for the given wave equation \( Y = A \sin(2\pi(pt - \frac{x}{5})) \).

### Step 1: Identify the parameters from the wave equation
The given wave equation is:
\[ Y = A \sin(2\pi(pt - \frac{x}{5})) \]
From this equation, we can identify:
- The amplitude \( A \).
- The angular frequency \( \omega \) and the wave number \( k \).

### Step 2: Rewrite the wave equation in standard form
We can rewrite the argument of the sine function:
\[ Y = A \sin(2\pi pt - 2\pi \frac{x}{5}) \]
This can be compared with the standard wave equation:
\[ Y = A \sin(\omega t - kx) \]
From this comparison, we can identify:
- \( \omega = 2\pi p \)
- \( k = \frac{2\pi}{5} \)

### Step 3: Calculate maximum particle velocity
The maximum particle velocity \( v_{max} \) is given by the formula:
\[ v_{max} = A \omega \]
Substituting the value of \( \omega \):
\[ v_{max} = A (2\pi p) \]

### Step 4: Calculate wave velocity
The wave velocity \( v \) is given by:
\[ v = \frac{\omega}{k} \]
Substituting the values of \( \omega \) and \( k \):
\[ v = \frac{2\pi p}{\frac{2\pi}{5}} = 5p \]

### Step 5: Calculate the ratio of maximum particle velocity to wave velocity
Now, we find the ratio \( \frac{v_{max}}{v} \):
\[ \frac{v_{max}}{v} = \frac{A(2\pi p)}{5p} \]
The \( p \) terms cancel out:
\[ \frac{v_{max}}{v} = \frac{2\pi A}{5} \]

### Final Result
Thus, the ratio of maximum particle velocity to wave velocity is:
\[ \frac{v_{max}}{v} = \frac{2\pi A}{5} \]