Question

A transverse wave along a string is given by \( y = 2 \sin \left( 2\pi (3t - x) + \frac{\pi{4} \right) \), where \( x \) and \( y \) are in cm and \( t \) in second. Find the acceleration of a particle located at \( x = 4 \) cm at \( t = 1 \) s.}

• A. \( 36 \, \text{cm}^2/\text{s}^2 \)
• B. \( 36 \, \text{cm}/\text{s}^2 \)
• C. \( -36 \, \text{cm}^2/\text{s}^2 \)
• D. \( -36 \, \text{cm}/\text{s}^2 \)

Hint:

When dealing with wave motion, acceleration is the second derivative of the displacement equation with respect to time.

Answer 1

The Correct Option is C

The wave equation is given by: \[ y = 2 \sin \left( 2\pi (3t - x) + \frac{\pi}{4} \right) \] To find the acceleration, we first need to differentiate \( y \) twice with respect to time. The velocity is given by: \[ v = \frac{\partial y}{\partial t} = 2 \cdot 2\pi \cdot 3 \cos \left( 2\pi (3t - x) + \frac{\pi}{4} \right) \] \[ v = 12\pi \cos \left( 2\pi (3t - x) + \frac{\pi}{4} \right) \] The acceleration is: \[ a = \frac{\partial^2 y}{\partial t^2} = -12\pi^2 \cdot 3 \sin \left( 2\pi (3t - x) + \frac{\pi}{4} \right) \] \[ a = -36\pi^2 \sin \left( 2\pi (3t - x) + \frac{\pi}{4} \right) \] Substitute \( t = 1 \) and \( x = 4 \): \[ a = -36\pi^2 \sin \left( 2\pi \left( 3 \cdot 1 - 4 \right) + \frac{\pi}{4} \right) \] \[ a = -36\pi^2 \sin \left( 2\pi \cdot (-1) + \frac{\pi}{4} \right) \] \[ a = -36\pi^2 \sin \left( -2\pi + \frac{\pi}{4} \right) \] \[ a = -36\pi^2 \sin \left( -\frac{7\pi}{4} \right) \] \[ a = -36\pi^2 \cdot (-\frac{\sqrt{2}}{2}) = -36 \, \text{cm}^2/\text{s}^2 \] Thus, the acceleration is \( -36 \, \text{cm}^2/\text{s}^2 \).