When a wave travels in a medium, the particle displacement is given by the equation
`y=asin 2pi(bt-cx), ` where `a,b` and `c` are constants. The maximum particle velocity will be twice the wave velocity. If

Options

A. b=ac
B. `v=(1)/(ac)`
C. `c=pia`
D. `c=(1)/(pia)`

Explanation

To solve the problem, we need to analyze the given wave equation and derive the relationships between the constants involved. The wave equation is given as:

\[ y = a \sin(2\pi(bt - cx)) \]

### Step 1: Identify the parameters
From the wave equation, we can identify:
- \( a \) is the amplitude of the wave.
- \( b \) is related to the frequency of the wave.
- \( c \) is related to the wave number.

### Step 2: Relate wave parameters
The angular frequency \( \omega \) and wave number \( k \) can be defined as:
- \( \omega = 2\pi b \) (angular frequency)
- \( k = \frac{2\pi}{\lambda} = \frac{2\pi c}{v} \) (wave number)

### Step 3: Determine wave velocity
The wave velocity \( v \) can be expressed in terms of \( b \) and \( c \):
\[ v = \frac{\lambda}{T} = \frac{c}{b} \]

### Step 4: Calculate maximum particle velocity
The maximum particle velocity \( v_{max} \) is given by:
\[ v_{max} = \omega \cdot a = (2\pi b) \cdot a \]

### Step 5: Set up the relationship
According to the problem, the maximum particle velocity is twice the wave velocity:
\[ v_{max} = 2v \]

Substituting the expressions we derived:
\[ 2\pi b a = 2 \cdot \frac{c}{b} \]

### Step 6: Simplify the equation
We can simplify this equation:
\[ 2\pi b a = \frac{2c}{b} \]

Multiplying both sides by \( b \):
\[ 2\pi b^2 a = 2c \]

Dividing both sides by 2:
\[ \pi b^2 a = c \]

### Step 7: Solve for \( c \)
Thus, we find:
\[ c = \pi b^2 a \]

### Final Answer
The relationship derived is:
\[ c = \pi b^2 a \]