A stationary wave is given by `y = 5 sin (pi x)/(3) cos 40 pi t ` where x and y are in cm and t is in seconds
What is the velocity of a particle of the string at the position x = 1.5 cm when `t = 9/8 s `

Options

A. 0
B. `1ms^(-1)`
C. `2ms^(-1)`
D. `3ms^(-1)`

Answer

To find the velocity of a particle of the string at the position \( x = 1.5 \, \text{cm} \) when \( t = \frac{9}{8} \, \text{s} \), we will follow these steps:

### Step 1: Write down the equation of the stationary wave
The equation of the stationary wave is given as:
\[ y = 5 \sin\left(\frac{\pi x}{3}\right) \cos(40 \pi t) \]

### Step 2: Find the velocity of the particle
The velocity \( v \) of a particle in a wave is given by the time derivative of the displacement \( y \):
\[ v = \frac{dy}{dt} \]

### Step 3: Differentiate the equation with respect to time \( t \)
Using the product rule, we differentiate \( y \):
\[ v = 5 \sin\left(\frac{\pi x}{3}\right) \frac{d}{dt}[\cos(40 \pi t)] \]
The derivative of \( \cos(40 \pi t) \) is:
\[ \frac{d}{dt}[\cos(40 \pi t)] = -40 \pi \sin(40 \pi t) \]
Thus, substituting this back, we have:
\[ v = 5 \sin\left(\frac{\pi x}{3}\right)(-40 \pi \sin(40 \pi t)) \]
This simplifies to:
\[ v = -200 \pi \sin\left(\frac{\pi x}{3}\right) \sin(40 \pi t) \]

### Step 4: Substitute \( x = 1.5 \, \text{cm} \) and \( t = \frac{9}{8} \, \text{s} \)
Now, we substitute \( x = 1.5 \) and \( t = \frac{9}{8} \):
\[ v = -200 \pi \sin\left(\frac{\pi \cdot 1.5}{3}\right) \sin\left(40 \pi \cdot \frac{9}{8}\right) \]

### Step 5: Calculate \( \sin\left(\frac{\pi \cdot 1.5}{3}\right) \)
Calculating:
\[ \frac{\pi \cdot 1.5}{3} = \frac{1.5\pi}{3} = 0.5\pi \]
Thus,
\[ \sin(0.5\pi) = 1 \]

### Step 6: Calculate \( \sin\left(40 \pi \cdot \frac{9}{8}\right) \)
Calculating:
\[ 40 \pi \cdot \frac{9}{8} = 45 \pi \]
Since \( \sin(45 \pi) = 0 \) (as sine is zero at integer multiples of \( \pi \)), we have:
\[ \sin(45 \pi) = 0 \]

### Step 7: Substitute back to find \( v \)
Now substituting back into the velocity equation:
\[ v = -200 \pi \cdot 1 \cdot 0 = 0 \]

### Final Answer
The velocity of the particle at the position \( x = 1.5 \, \text{cm} \) when \( t = \frac{9}{8} \, \text{s} \) is:
\[ \boxed{0 \, \text{cm/s}} \]