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. `36sqrt(2)pi^2 cm s^(-2)`
B. `36 pi^2cm s^(-2)`
C. `-36 sqrt(2) pi^2 cm s^(-2)`
D. `-36 pi^2cms ^(-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: Differentiate the wave equation with respect to time \( t \)

The acceleration of a particle is given by the second derivative of \( y \) with respect to \( t \):

\[ a = \frac{d^2y}{dt^2} \]

First, we differentiate \( y \) with respect to \( t \):

\[ y = 2 \sin(2\pi(3t - x) + \frac{\pi}{4}) \]

Using the chain rule, we differentiate:

\[ \frac{dy}{dt} = 2 \cdot \cos(2\pi(3t - x) + \frac{\pi}{4}) \cdot \frac{d}{dt}(2\pi(3t - x)) \]

Here, \( x \) is constant while differentiating with respect to \( t \):

\[ \frac{d}{dt}(2\pi(3t - x)) = 2\pi \cdot 3 = 6\pi \]

Thus, we have:

\[ \frac{dy}{dt} = 2 \cdot 6\pi \cdot \cos(2\pi(3t - x) + \frac{\pi}{4}) = 12\pi \cos(2\pi(3t - x) + \frac{\pi}{4}) \]

### Step 2: Differentiate again to find acceleration

Now, we differentiate \( \frac{dy}{dt} \) with respect to \( t \):

\[ \frac{d^2y}{dt^2} = \frac{d}{dt}(12\pi \cos(2\pi(3t - x) + \frac{\pi}{4})) \]

Using the chain rule again:

\[ \frac{d^2y}{dt^2} = -12\pi \cdot \sin(2\pi(3t - x) + \frac{\pi}{4}) \cdot \frac{d}{dt}(2\pi(3t - x)) \]

Substituting \( \frac{d}{dt}(2\pi(3t - x)) = 6\pi \):

\[ \frac{d^2y}{dt^2} = -12\pi \cdot \sin(2\pi(3t - x) + \frac{\pi}{4}) \cdot 6\pi = -72\pi^2 \sin(2\pi(3t - x) + \frac{\pi}{4}) \]

### Step 3: Substitute \( x = 4 \, \text{cm} \) and \( t = 1 \, \text{s} \)

Now we need to evaluate \( \frac{d^2y}{dt^2} \) at \( x = 4 \, \text{cm} \) and \( t = 1 \, \text{s} \):

\[ 2\pi(3t - x) = 2\pi(3 \cdot 1 - 4) = 2\pi(3 - 4) = 2\pi(-1) = -2\pi \]

Thus,

\[ \frac{d^2y}{dt^2} = -72\pi^2 \sin(-2\pi + \frac{\pi}{4}) = -72\pi^2 \sin(\frac{\pi}{4}) = -72\pi^2 \cdot \frac{1}{\sqrt{2}} = -\frac{72\pi^2}{\sqrt{2}} \]

### Final Result

The acceleration of the particle located at \( x = 4 \, \text{cm} \) at \( t = 1 \, \text{s} \) is:

\[ a = -36\sqrt{2}\pi^2 \, \text{cm/s}^2 \]