The (x, y) coordinates of the corners of a square plate are (0,0), (L,0), (L,L) and (0,L). The edges

Q: Why can't we have cosine terms in the displacement equation for a clamped plate?

A clamped edge is a fixed boundary, which physically acts as a node where displacement must be zero. Since $\cos(0) = 1$, any term containing $\cos(kx)$ or $\cos(ky)$ will not vanish at $x=0$ or $y=0$, violating the boundary conditions.

Q: What does 'clamped' mean physically compared to 'free'?

A clamped boundary is rigidly fixed, meaning the displacement $u$ is zero at all times (Dirichlet boundary condition). A free boundary allows the edge to move freely, meaning the slope or derivative of displacement is zero (Neumann boundary condition).

Q: Can we have fractional coefficients of $\pi$ in the sine arguments, like $\sin(1.5 \pi x / L)$?

No, because at the boundary $x=L$, the term would become $\sin(1.5\pi) = -1 \neq 0$. To satisfy the boundary condition at both ends, the argument must be an integer multiple of $\pi$ when $x=L$, i.e., $\sin(n\pi x / L)$ where $n$ is an integer.

Q: Is this problem related to 1D standing waves on a string?

Yes! This is a direct 2D generalization of a string fixed at both ends. For a 1D string, the displacement is $y(x) = A \sin(n\pi x / L)$. For a 2D plate, the displacement is a product of two such independent 1D modes: $u(x,y) = A \sin(n_x \pi x / L) \sin(n_y \pi y / L)$.

Q: Why is the amplitude $a$ a positive constant?

The constant $a$ represents the maximum amplitude of the standing wave at its antinodes. It must be positive because amplitude is defined as a magnitude, and any phase shifts are handled by the sine functions themselves.

Question

The $(x, y)$ coordinates of the corners of a square plate are $(0,0), (L,0), (L,L)$ and $(0,L)$. The edges of the plate are clamped and transverse standing waves are set-up in it. If $u(x,y)$ denotes the displacement of the plate at the point $(x,y)$ at some instant of time, the possible expression(s) for $u$ is (are) ($a = $ positive constant)

Options:

(A) a \cos \left(\frac{\pi x}{2L}\right) \cos \left(\frac{\pi y}{2L}\right)

(B) a \sin \left(\frac{\pi x}{L}\right) \sin \left(\frac{\pi y}{L}\right)

(C) a \sin \left(\frac{\pi x}{L}\right) \sin \left(\frac{2\pi y}{L}\right)

(D) a \cos \left(\frac{2\pi x}{L}\right) \sin \left(\frac{\pi y}{L}\right)

Sigma Percentile · Accepted Answer

### Introduction to 2D Standing Waves

Imagine a tightly stretched drumhead or a metal plate clamped firmly at its edges.

When you strike it, waves travel outward, reflect off the boundaries, and interfere with one another.

Under specific conditions, this interference produces **standing waves**, where certain points remain completely stationary while others vibrate with maximum amplitude.

This problem asks us to find the mathematically valid expressions for the transverse displacement $u(x,y)$ of a square plate clamped at all its four edges.

Let's dive into the physics of boundaries and discover how physical constraints dictate mathematical equations.

---

### The Power of Boundary Conditions

In physics, the behavior of a system at its limits is governed by **boundary conditions**.

For a square plate clamped at its edges, the physical constraint is simple: the edges are held completely rigid and cannot move.

Mathematically, this means the displacement $u(x,y)$ must be exactly zero at all points along the boundaries.

Let's define the boundaries of our square plate of side length $L$:

1. The bottom edge $OA$ lies along the line $y = 0$ for $0 \le x \le L$.

2. The right edge $AB$ lies along the line $x = L$ for $0 \le y \le L$.

3. The top edge $BC$ lies along the line $y = L$ for $0 \le x \le L$.

4. The left edge $OC$ lies along the line $x = 0$ for $0 \le y \le L$.

Therefore, any valid standing wave expression $u(x,y)$ must satisfy the following four conditions simultaneously:

$$u(0, y) = 0 \quad 	ext{and} \quad u(L, y) = 0 \quad 	ext{for all } y \in [0, L]$$

$$u(x, 0) = 0 \quad 	ext{and} \quad u(x, L) = 0 \quad 	ext{for all } x \in [0, L]$$

These boundary conditions act as a filter, immediately eliminating any mathematical functions that do not vanish at these boundaries.

---

### Testing the Candidates

Let's systematically test each option against our boundary conditions.

#### Testing Option (a)

The first candidate is:

$$u(x,y) = a \cos \left(\frac{\pi x}{2L}ight) \cos \left(\frac{\pi y}{2L}ight)$$

Let's check the boundary $x = 0$:

$$u(0, y) = a \cos(0) \cos \left(\frac{\pi y}{2L}ight) = a \cos \left(\frac{\pi y}{2L}ight)$$

Since $\cos \left(\frac{\pi y}{2L}ight)$ is not zero for all $y \in [0, L]$, this option violates the boundary condition $u(0,y) = 0$.

Thus, **option (a) is incorrect**.

---

#### Testing Option (b)

The second candidate is:

$$u(x,y) = a \sin \left(\frac{\pi x}{L}ight) \sin \left(\frac{\pi y}{L}ight)$$

Let's check the $x$-boundaries:

- At $x = 0$: $u(0, y) = a \sin(0) \sin \left(\frac{\pi y}{L}ight) = 0$ (Satisfied)

- At $x = L$: $u(L, y) = a \sin(\pi) \sin \left(\frac{\pi y}{L}ight) = 0$ (Satisfied)

Now, let's check the $y$-boundaries:

- At $y = 0$: $u(x, 0) = a \sin \left(\frac{\pi x}{L}ight) \sin(0) = 0$ (Satisfied)

- At $y = L$: $u(x, L) = a \sin \left(\frac{\pi x}{L}ight) \sin(\pi) = 0$ (Satisfied)

Since all four boundary conditions are perfectly satisfied, **option (b) is a possible expression**.

---

#### Testing Option (c)

The third candidate is:

$$u(x,y) = a \sin \left(\frac{\pi x}{L}ight) \sin \left(\frac{2\pi y}{L}ight)$$

Let's check the $x$-boundaries:

- At $x = 0$: $u(0, y) = a \sin(0) \sin \left(\frac{2\pi y}{L}ight) = 0$ (Satisfied)

- At $x = L$: $u(L, y) = a \sin(\pi) \sin \left(\frac{2\pi y}{L}ight) = 0$ (Satisfied)

Now, let's check the $y$-boundaries:

- At $y = 0$: $u(x, 0) = a \sin \left(\frac{\pi x}{L}ight) \sin(0) = 0$ (Satisfied)

- At $y = L$: $u(x, L) = a \sin \left(\frac{\pi x}{L}ight) \sin(2\pi) = 0$ (Satisfied)

Since all boundary conditions are satisfied, **option (c) is also a possible expression**.

---

#### Testing Option (d)

The final candidate is:

$$u(x,y) = a \cos \left(\frac{2\pi x}{L}ight) \sin \left(\frac{\pi y}{L}ight)$$

Let's check the boundary $x = 0$:

$$u(0, y) = a \cos(0) \sin \left(\frac{\pi y}{L}ight) = a \sin \left(\frac{\pi y}{L}ight)$$

This is not zero for all $y \in [0, L]$, violating the boundary condition $u(0,y) = 0$.

Thus, **option (d) is incorrect**.

---

### Conclusion

Only options **(b)** and **(c)** satisfy all the boundary conditions of the clamped square plate.

This elegant result shows how physical constraints act as mathematical filters, allowing only specific harmonic modes to exist on the plate.

Select Answer:

Visualized Solution

Visualizing the Clamped Square Plate

Formulating the Boundary Conditions

Testing Option (a)

Testing Option (b) - $x$ -boundaries

Testing Option (b) - $y$ -boundaries

Testing Option (c) - $x$ -boundaries

Testing Option (c) - $y$ -boundaries

Testing Option (d)

Identifying the Correct Options

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction to 2D Standing Waves

The Power of Boundary Conditions

Testing the Candidates

# Testing Option (a)

# Testing Option (b)

# Testing Option (c)

# Testing Option (d)

Conclusion

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