The (x, y) coordinates of the corners of a sq | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Since the edges of the square plate are clamped,the displacement $u(x, y)$ must be zero at all boundaries:$1$. At $y = 0$ (edge $OA$),$u(x, 0) = 0

Vedclass · Accepted Answer

Both $(b)$ and $(c)$

Vedclass · Answer

(D) Since the edges of the square plate are clamped,the displacement $u(x, y)$ must be zero at all boundaries:$1$. At $y = 0$ (edge $OA$),$u(x, 0) = 0$ for $0 \le x \le L$.$2$. At $x = L$ (edge $AB$),$u(L, y) = 0$ for $0 \le y \le L$.$3$. At $y = L$ (edge $BC$),$u(x, L) = 0$ for $0 \le x \le L$.$4$. At $x = 0$ (edge $OC$),$u(0, y) = 0$ for $0 \le y \le L$.Evaluating the options:- For option $(a)$: $u(0, y) = a \cos(0) \cos(\frac{\pi y}{2L}) = a \cos(\frac{\pi y}{2L}) 
eq 0$ for all $y$.- For option $(b)$: $u(x, y) = a \sin(\frac{\pi x}{L}) \sin(\frac{\pi y}{L})$. Here,$u = 0$ at $x=0, x=L, y=0, y=L$. This satisfies the boundary conditions.- For option $(c)$: $u(x, y) = a \sin(\frac{\pi x}{L}) \sin(\frac{2\pi y}{L})$. Here,$u = 0$ at $x=0, x=L, y=0, y=L/2, y=L$. This also satisfies the boundary conditions.Thus,both $(b)$ and $(c)$ are valid expressions.

Similar Questions

$A$ wave represented by the equation $y_1 = a \cos(kx - \omega t)$ is superimposed with another wave to form a stationary wave such that the point $x = 0$ is a node. The equation for the other wave is

Which property distinguishes between progressive and stationary waves?

$A$ progressive wave travelling in the positive $x$-direction given by $y=a \cos (k x-\omega t)$ meets a denser surface at $x=0, t=0$. The reflected wave is then given by

The stationary wave produced on a string is represented by the equation $y = 5 \cos (\pi x / 3) \sin (40 \pi t)$. Where $x$ and $y$ are in $cm$ and $t$ is in seconds. The distance between consecutive nodes is .... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module