A transverse wave along a string is given by
`y = 2 sin (2pi (3t - x) + (pi)/(4))`
where x and y are in cm and t in second. Find acceleration of a particle located at x = 4 cm at t = 1s.

Options

A. `36 sqrt(2)pi^(2) cm//s^(2)`
B. `36 pi^(2) cm//s^(2)`
C. `-36 sqrt(2)pi^(2)cm//s^(2)`
D. `-36 pi^(2) cm//s^(2)`

Answer

To find the acceleration of a particle located at \( x = 4 \, \text{cm} \) at \( t = 1 \, \text{s} \) for the given wave equation \( y = 2 \sin(2\pi(3t - x) + \frac{\pi}{4}) \), we will follow these steps:

### Step 1: Identify the wave equation
The wave equation is given as:
\[ y = 2 \sin(2\pi(3t - x) + \frac{\pi}{4}) \]

### Step 2: Differentiate the wave equation to find velocity
The velocity \( v \) of a particle in the wave can be found by taking the partial derivative of \( y \) with respect to time \( t \):
\[ v = \frac{\partial y}{\partial t} = 2 \cdot \frac{\partial}{\partial t} \left( \sin(2\pi(3t - x) + \frac{\pi}{4})\right) \]
Using the chain rule, we get:
\[ v = 2 \cdot \cos(2\pi(3t - x) + \frac{\pi}{4}) \cdot \frac{\partial}{\partial t}(2\pi(3t - x)) \]
\[ = 2 \cdot \cos(2\pi(3t - x) + \frac{\pi}{4}) \cdot (6\pi) \]
Thus, the velocity equation becomes:
\[ v = 12\pi \cos(2\pi(3t - x) + \frac{\pi}{4}) \]

### Step 3: Differentiate the velocity to find acceleration
Now, we differentiate the velocity \( v \) with respect to time \( t \) to find the acceleration \( a \):
\[ a = \frac{\partial v}{\partial t} = 12\pi \cdot \frac{\partial}{\partial t} \left( \cos(2\pi(3t - x) + \frac{\pi}{4})\right) \]
Using the chain rule again:
\[ a = 12\pi \cdot (-\sin(2\pi(3t - x) + \frac{\pi}{4})) \cdot \frac{\partial}{\partial t}(2\pi(3t - x)) \]
\[ = 12\pi \cdot (-\sin(2\pi(3t - x) + \frac{\pi}{4})) \cdot (6\pi) \]
Thus, the acceleration equation becomes:
\[ a = -72\pi^2 \sin(2\pi(3t - x) + \frac{\pi}{4}) \]

### Step 4: Substitute \( x = 4 \, \text{cm} \) and \( t = 1 \, \text{s} \)
Now we substitute \( x = 4 \) and \( t = 1 \) into the acceleration equation:
\[ a = -72\pi^2 \sin(2\pi(3(1) - 4) + \frac{\pi}{4}) \]
Calculating the argument of the sine function:
\[ = -72\pi^2 \sin(2\pi(3 - 4) + \frac{\pi}{4}) = -72\pi^2 \sin(2\pi(-1) + \frac{\pi}{4}) \]
\[ = -72\pi^2 \sin(-2\pi + \frac{\pi}{4}) = -72\pi^2 \sin(\frac{\pi}{4}) \]
Since \( \sin(-\theta) = -\sin(\theta) \):
\[ = -72\pi^2 \cdot \frac{1}{\sqrt{2}} = -72\pi^2 \cdot \frac{1}{\sqrt{2}} = -36\sqrt{2}\pi^2 \]

### Step 5: Final result
Thus, the acceleration of the particle located at \( x = 4 \, \text{cm} \) at \( t = 1 \, \text{s} \) is:
\[ a = -36\sqrt{2}\pi^2 \, \text{cm/s}^2 \]