A progressive wave is represented by y = 5 `sin(100pit - 2pix)` where x and y are in m and t is in s. The maximum particle velocity is

Options

A. `100pims^(-1)`
B. `200pims^(-1)`
C. `300pims^(-1)`
D. `500pims^(-1)`

Explanation

To find the maximum particle velocity of the given progressive wave represented by the equation:

\[ y = 5 \sin(100\pi t - 2\pi x) \]

we will follow these steps:

### Step 1: Identify the wave function
The wave function is given as:

\[ y = 5 \sin(100\pi t - 2\pi x) \]

Here, the amplitude \( A = 5 \) m, the angular frequency \( \omega = 100\pi \) rad/s, and the wave number \( k = 2\pi \) rad/m.

### Step 2: Determine the particle velocity
The particle velocity \( v_p \) is given by the time derivative of the displacement \( y \):

\[ v_p = \frac{dy}{dt} \]

### Step 3: Differentiate the wave function
We differentiate \( y \) with respect to \( t \):

\[ v_p = \frac{d}{dt} [5 \sin(100\pi t - 2\pi x)] \]

Using the chain rule, we get:

\[ v_p = 5 \cdot \cos(100\pi t - 2\pi x) \cdot \frac{d}{dt}(100\pi t - 2\pi x) \]

Since \( x \) is constant with respect to \( t \), we have:

\[ \frac{d}{dt}(100\pi t - 2\pi x) = 100\pi \]

Thus, the expression for particle velocity becomes:

\[ v_p = 5 \cdot 100\pi \cdot \cos(100\pi t - 2\pi x) \]

### Step 4: Find the maximum particle velocity
The maximum value of \( \cos \) function is 1. Therefore, the maximum particle velocity \( v_{p_{max}} \) is:

\[ v_{p_{max}} = 5 \cdot 100\pi \cdot 1 = 500\pi \text{ m/s} \]

### Step 5: Calculate the numerical value
Now, we can calculate the numerical value of \( 500\pi \):

\[ v_{p_{max}} \approx 500 \cdot 3.14 \approx 1570 \text{ m/s} \]

### Final Answer
The maximum particle velocity is:

\[ \boxed{500\pi \text{ m/s}} \quad \text{or approximately } \boxed{1570 \text{ m/s}} \]

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