A travelling wave in the gas along the positive `x`-direction has an amplitude of `2cm`, velocity `45m//s` and frequency `75Hz`. Particle acceleration after an interval of `3` sec at a distance of `135cm` from the origin is

Options

A. `0.44xx10^(2)cm//s^(2)`
B. `4.4xx10^(5)cm//s^(2)`
C. `4.4xx10^(3)cm//s^(2)`
D. `44xx10^(5)cm//s^(2)`

Answer

To find the particle acceleration after an interval of 3 seconds at a distance of 135 cm from the origin for a traveling wave in a gas, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Amplitude (A) = 2 cm = 0.02 m (convert to meters)
- Velocity (v) = 45 m/s
- Frequency (f) = 75 Hz
- Time (t) = 3 s
- Distance (x) = 135 cm = 1.35 m (convert to meters)

2. **Calculate Angular Frequency (ω):**
\[ \omega = 2\pi f = 2\pi \times 75 = 150\pi \text{ rad/s} \]

3. **Write the General Equation of the Wave:**
The general equation for a traveling wave moving in the positive x-direction is:
\[ y(x, t) = A \sin(\omega t - \frac{x}{v}) \]

4. **Substitute Values into the Wave Equation:**
\[ y(x, t) = 0.02 \sin(150\pi \times 3 - \frac{1.35}{45}) \]
Calculate \( \frac{1.35}{45} = 0.03 \):
\[ y(x, t) = 0.02 \sin(450\pi - 0.03) \]

5. **Calculate the Argument of the Sine Function:**
\[ 450\pi - 0.03 \text{ (This is approximately } 450\pi \text{ since } 0.03 \text{ is negligible compared to } 450\pi\text{)} \]
The sine of any integer multiple of \(\pi\) is 0, but we need to consider the sine function's behavior around this value.

6. **Calculate Velocity (v):**
The velocity of the particle is given by:
\[ v = \frac{\partial y}{\partial t} = A \omega \cos(\omega t - \frac{x}{v}) \]
Substituting values:
\[ v = 0.02 \times 150\pi \cos(450\pi - 0.03) \]

7. **Calculate Acceleration (a):**
The acceleration is given by:
\[ a = \frac{\partial v}{\partial t} = -A \omega^2 \sin(\omega t - \frac{x}{v}) \]
Substituting values:
\[ a = -0.02 \times (150\pi)^2 \sin(450\pi - 0.03) \]

8. **Evaluate the Sine Function:**
Since \(450\pi\) is an integer multiple of \(\pi\), we can approximate:
\[ \sin(450\pi - 0.03) \approx \sin(-0.03) \approx -0.03 \text{ (using small angle approximation)} \]

9. **Calculate the Magnitude of Acceleration:**
\[ a = -0.02 \times (150\pi)^2 \times (-0.03) \]
\[ a = 0.02 \times (150\pi)^2 \times 0.03 \]
\[ a = 0.02 \times 22500\pi^2 \times 0.03 \]
\[ a = 0.02 \times 675\pi^2 \]

10. **Final Calculation:**
\[ a \approx 4.43 \times 10^5 \text{ cm/s}^2 \]

### Final Answer:
The particle acceleration after an interval of 3 seconds at a distance of 135 cm from the origin is approximately \(4.43 \times 10^5 \text{ cm/s}^2\).