A travelling wave on a string is given by `y = A sin [alphax + betat + (pi)/(6)]`. If `alpha = 0.56 //cm, beta = 12//sec, A = 7.5 cm`, then find the position and velocity of particle at `x = 1 cm` and `t = 1s` is

Options

A. `4.6 cm, 46.5 cm s^(-1)`
B. `3.75 cm, 77.94 cm s^(-1)`
C. `1.76 cm, 7.5 cm s^(-1)`
D. `7.5 cm, 75 cm s^(-1)`

Explanation

To solve the problem, we need to find the position and velocity of a particle on a string described by the wave equation:

\[ y = A \sin(\alpha x + \beta t + \frac{\pi}{6}) \]

where:
- \( \alpha = 0.56 \, \text{cm}^{-1} \)
- \( \beta = 12 \, \text{s}^{-1} \)
- \( A = 7.5 \, \text{cm} \)

We will evaluate this at \( x = 1 \, \text{cm} \) and \( t = 1 \, \text{s} \).

### Step 1: Calculate the position \( y \)

1. Substitute the values into the wave equation:
\[ y = 7.5 \sin(0.56 \cdot 1 + 12 \cdot 1 + \frac{\pi}{6}) \]

2. Calculate the argument of the sine function:
\[ 0.56 \cdot 1 + 12 \cdot 1 + \frac{\pi}{6} = 0.56 + 12 + \frac{\pi}{6} \]
- Convert \( \frac{\pi}{6} \) to a decimal (approximately \( 0.5236 \)):
\[ 0.56 + 12 + 0.5236 \approx 12.56 \]

3. Now, calculate \( y \):
\[ y = 7.5 \sin(12.56) \]

4. Recognize that \( 12.56 \) can be simplified:
\[ 12.56 = 4\pi + \frac{\pi}{6} \]
- Using the periodic property of sine:
\[ \sin(4\pi + \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2} \]

5. Substitute back to find \( y \):
\[ y = 7.5 \cdot \frac{1}{2} = 3.75 \, \text{cm} \]

### Step 2: Calculate the velocity \( v \)

1. The velocity of the particle is given by the derivative of the displacement with respect to time:
\[ v = \frac{dy}{dt} = A \frac{d}{dt} \left( \sin(\alpha x + \beta t + \frac{\pi}{6}) \right) \]

2. Using the chain rule:
\[ v = A \cos(\alpha x + \beta t + \frac{\pi}{6}) \cdot \beta \]

3. Substitute the values:
\[ v = 7.5 \cos(0.56 \cdot 1 + 12 \cdot 1 + \frac{\pi}{6}) \cdot 12 \]

4. We already calculated the argument:
\[ 0.56 + 12 + \frac{\pi}{6} \approx 12.56 \]

5. Calculate \( v \):
\[ v = 7.5 \cos(12.56) \cdot 12 \]

6. Recognize that:
\[ \cos(12.56) = \cos(4\pi + \frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \]

7. Substitute back to find \( v \):
\[ v = 7.5 \cdot \frac{\sqrt{3}}{2} \cdot 12 \]
\[ v = 7.5 \cdot 6 \cdot \sqrt{3} = 45 \sqrt{3} \]
- Using \( \sqrt{3} \approx 1.732 \):
\[ v \approx 45 \cdot 1.732 \approx 77.86 \, \text{cm/s} \]

### Final Results

- The position of the particle at \( x = 1 \, \text{cm} \) and \( t = 1 \, \text{s} \) is \( y = 3.75 \, \text{cm} \).
- The velocity of the particle at the same point is \( v \approx 77.86 \, \text{cm/s} \).