One end of a taut string of length 3 m along the X-axis is fixed at x = 0. The speed of the waves in

Q: Why is the vibrating end at $x = 3\text{ m}$ considered an antinode instead of a node?

The problem states that the end at $x = 3\text{ m}$ is vibrating to set up stationary waves. For a driver to efficiently transfer energy and maintain a standing wave, the driven end must be a point of maximum displacement (an antinode) or very close to it. In idealized textbook problems, we treat the driven end as a displacement antinode.

Q: How do we find the wave speed from the wave equation?

For any wave equation of the form $y = A \sin(kx) \cos(\omega t)$, the wave speed $v$ is given by the ratio of the angular frequency (coefficient of $t$, which is $\omega$) to the wave number (coefficient of $x$, which is $k$), i.e., $v = \omega/k$.

Q: What if $\sin(3k) = -1$? Is that still an antinode?

Yes! An antinode is a point of maximum displacement amplitude, which means the magnitude of displacement is maximum. Since $|\sin(3k)| = |-1| = 1$, it is indeed an antinode.

Q: Why does option (b) fail?

Option (b) has $k = \pi/3$. At $x = 3\text{ m}$, the spatial term becomes $\sin(\pi) = 0$. This represents a node (zero displacement), which contradicts the fact that this end is actively vibrating.

Q: Can a string have multiple different standing wave frequencies at the same time?

Yes, through superposition, a string can vibrate in multiple modes simultaneously. However, each option represents a single pure harmonic mode of standing wave.

Question

One end of a taut string of length $3	ext{ m}$ along the X-axis is fixed at $x = 0$. The speed of the waves in the string is $100	ext{ ms}^{-1}$. The other end of the string is vibrating in the y-direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary wave is (are)

Options:

(A) y(t) = A \sin \frac{\pi x}{6} \cos \frac{50\pi t}{3}

(B) y(t) = A \sin \frac{\pi x}{3} \cos \frac{100\pi t}{3}

(C) y(t) = A \sin \frac{5\pi x}{6} \cos \frac{250\pi t}{3}

(D) y(t) = A \sin \frac{5\pi x}{2} \cos 250\pi t

Sigma Percentile · Accepted Answer

### Introduction: The Magic of Standing Waves

Imagine a tightly stretched string, fixed firmly at one end and driven dynamically at the other.

When waves travel down the string and reflect off the boundaries, they interfere with the incoming waves.

Under specific conditions, this interference produces a beautiful, stationary pattern of loops where some points remain completely still, while others vibrate with maximum amplitude.

These are **standing waves**, and they form the physical basis of musical instruments, from violins to pipe organs.

In this problem, we are tasked with finding the mathematically valid waveforms for a string of length $L = 3	ext{ m}$ with a fixed end at $x = 0$ and a vibrating end at $x = 3	ext{ m}$.

---

### The Physics of Boundary Conditions

To solve this problem, we must first establish the physical constraints imposed by the boundaries of the string.

First, the string is fixed at $x = 0$.

Because it is clamped, the displacement at this point must always be zero.

This means we must have a **displacement node** at $x = 0$:

$$y(0, t) = 0$$

Looking at the general form of a standing wave, $y(x, t) = A \sin(kx) \cos(\omega t)$, we see that the spatial term $\sin(kx)$ naturally becomes zero at $x = 0$ for any value of $k$.

Thus, the node condition at $x = 0$ is satisfied by all given options.

Second, the other end of the string at $x = 3	ext{ m}$ is vibrating.

For a stable standing wave to be maintained by an external vibrator, the driven end must act as a **displacement antinode** (a point of maximum amplitude):

$$|\sin(3k)| = 1$$

This means that the argument of the sine function at $x = 3	ext{ m}$ must be an odd multiple of $\pi/2$:

$$3k = (2n+1)\frac{\pi}{2} \implies k = (2n+1)\frac{\pi}{6}$$

where $n$ is an integer.

---

### The Wave Speed Constraint: Our First Filter

We are given that the speed of the waves in the string is $v = 100	ext{ ms}^{-1}$.

For any wave equation, the wave speed is related to the angular frequency $\omega$ and the wave number $k$ by the fundamental relation:

$$v = \frac{\omega}{k}$$

Therefore, for any option to be correct, the ratio of the coefficient of $t$ (which is $\omega$) to the coefficient of $x$ (which is $k$) must be exactly $100$:

$$\frac{\omega}{k} = 100	ext{ m/s}$$

This is a powerful filter that allows us to quickly test each option.

---

### Systematically Testing the Waveforms

Let's apply our two filters—the wave speed constraint and the boundary condition at $x = 3	ext{ m}$—to each option.

#### Testing Option (a):
$$y(t) = A \sin \frac{\pi x}{6} \cos \frac{50\pi t}{3}$$

Here, $k = \frac{\pi}{6}	ext{ m}^{-1}$ and $\omega = \frac{50\pi}{3}	ext{ rad/s}$.

1. **Wave Speed Check:**
   $$v = \frac{\omega}{k} = \frac{50\pi / 3}{\pi / 6} = \frac{50\pi}{3} 	imes \frac{6}{\pi} = 100	ext{ m/s}$$
   This perfectly matches the given wave speed!

2. **Boundary Condition Check at $x = 3	ext{ m}$:**
   $$\sin(3k) = \sin\left(3 	imes \frac{\pi}{6}ight) = \sin\frac{\pi}{2} = 1$$
   Since the magnitude is $1$, this represents an antinode.

Thus, **Option (a) is a valid waveform**.

---

#### Testing Option (b):
$$y(t) = A \sin \frac{\pi x}{3} \cos \frac{100\pi t}{3}$$

Here, $k = \frac{\pi}{3}	ext{ m}^{-1}$ and $\omega = \frac{100\pi}{3}	ext{ rad/s}$.

1. **Wave Speed Check:**
   $$v = \frac{\omega}{k} = \frac{100\pi / 3}{\pi / 3} = 100	ext{ m/s}$$
   The speed is correct.

2. **Boundary Condition Check at $x = 3	ext{ m}$:**
   $$\sin(3k) = \sin\left(3 	imes \frac{\pi}{3}ight) = \sin\pi = 0$$
   This yields a displacement of zero, which means a node would form at the vibrating end.

Since the vibrating end cannot be a node, **Option (b) is incorrect**.

---

#### Testing Option (c):
$$y(t) = A \sin \frac{5\pi x}{6} \cos \frac{250\pi t}{3}$$

Here, $k = \frac{5\pi}{6}	ext{ m}^{-1}$ and $\omega = \frac{250\pi}{3}	ext{ rad/s}$.

1. **Wave Speed Check:**
   $$v = \frac{\omega}{k} = \frac{250\pi / 3}{5\pi / 6} = \frac{250\pi}{3} 	imes \frac{6}{5\pi} = 100	ext{ m/s}$$
   The speed is correct.

2. **Boundary Condition Check at $x = 3	ext{ m}$:**
   $$\sin(3k) = \sin\left(3 	imes \frac{5\pi}{6}ight) = \sin\frac{5\pi}{2} = 1$$
   Since $|\sin(5\pi/2)| = 1$, this represents an antinode.

Thus, **Option (c) is a valid waveform**.

---

#### Testing Option (d):
$$y(t) = A \sin \frac{5\pi x}{2} \cos 250\pi t$$

Here, $k = \frac{5\pi}{2}	ext{ m}^{-1}$ and $\omega = 250\pi	ext{ rad/s}$.

1. **Wave Speed Check:**
   $$v = \frac{\omega}{k} = \frac{250\pi}{5\pi / 2} = 250\pi 	imes \frac{2}{5\pi} = 100	ext{ m/s}$$
   The speed is correct.

2. **Boundary Condition Check at $x = 3	ext{ m}$:**
   $$\sin(3k) = \sin\left(3 	imes \frac{5\pi}{2}ight) = \sin\frac{15\pi}{2} = -1$$
   Since $|\sin(15\pi/2)| = 1$, this represents an antinode.

Thus, **Option (d) is a valid waveform**.

---

### Conclusion: The Harmony of Math and Physics

By systematically applying the physical constraints of boundary conditions and wave speed, we have determined that the possible waveforms are those given in options **(a)**, **(c)**, and **(d)**.

This problem beautifully illustrates how physical constraints translate directly into mathematical filters, allowing us to find the exact harmonic modes of a vibrating system.

Select Answer:

Visualized Solution

Understanding the Physical Setup

Mathematical Formulation of Boundary Conditions

The Wave Speed Constraint

Testing Option (a)

Testing Option (b)

Testing Option (c)

Testing Option (d)

Summary of Correct Waveforms

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction

The Physics of Boundary Conditions

The Wave Speed Constraint

Systematically Testing the Waveforms

# Testing Option (a)

# Testing Option (b)

# Testing Option (c)

# Testing Option (d)

Conclusion

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Select Answer:

Visualized Solution

Understanding the Physical Setup

Mathematical Formulation of Boundary Conditions

The Wave Speed Constraint

Testing Option (a)

Testing Option (b)

Testing Option (c)

Testing Option (d)

Summary of Correct Waveforms

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction

The Physics of Boundary Conditions

The Wave Speed Constraint

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