The vibrations of a string of length 60 cm fixed at both ends are represented by the equation y = 4

Q: Why is the maximum displacement of a particle equal to its amplitude?

The displacement equation is $y(x,t) = A(x) \cos(\omega t)$. Since the maximum value of $\cos(\omega t)$ is $1$, the maximum displacement that a particle at position $x$ can ever achieve is $A(x)$.

Q: How do we know the boundary conditions of the string?

The problem states that the string is fixed at both ends and has a length of $60\text{ cm}$. This means the ends at $x = 0$ and $x = 60\text{ cm}$ must be nodes, which is confirmed by our node calculation where $x=0$ and $x=60$ are indeed solutions.

Q: Why is the particle velocity zero at $t = 0.25\text{ s}$ even though the point $x = 7.5\text{ cm}$ is an antinode?

At $t = 0.25\text{ s}$, the temporal term $\cos(96\pi t)$ becomes $\cos(24\pi) = 1$, meaning the string is at its maximum displacement. At the extreme position of any oscillation, the velocity of the particles momentarily becomes zero before they reverse direction.

Q: What is the difference between wave velocity and particle velocity?

Wave velocity is the speed with which the wave profile (or energy) propagates through the medium, given by $v = \frac{\omega}{k}$. Particle velocity is the physical speed of an individual medium element oscillating about its equilibrium position, given by $\frac{\partial y}{\partial t}$.

Q: How do we determine the direction of travel of the component progressive waves?

In a progressive wave equation of the form $y = A \sin(kx \pm \omega t)$, if the signs of the $kx$ and $\omega t$ terms are opposite (e.g., $kx - \omega t$), the wave travels in the positive $x$-direction. If they are the same (e.g., $kx + \omega t$), the wave travels in the negative $x$-direction.

Question

The vibrations of a string of length $60	ext{ cm}$ fixed at both ends are represented by the equation
$$y = 4 \sin\left(\frac{\pi x}{15}ight) \cos(96 \pi t)$$
where, $x$ and $y$ are in cm and $t$ is in second.
(a) What is the maximum displacement of a point at $x = 5	ext{ cm}$?
(b) Where are the nodes located along the string?
(c) What is the velocity of the particle at $x = 7.5	ext{ cm}$ at $t = 0.25	ext{ s}$?
(d) Write down the equations of the component waves whose superposition gives the above wave.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a guitar string stretched tight and clamped firmly at both ends. When plucked, waves travel back and forth, reflecting off the boundaries and interfering with one another. Under the right conditions, this interference creates a **standing wave**—a wave that appears to vibrate in place without propagating.

In this problem, we are given the mathematical description of such a standing wave on a string of length $L = 60	ext{ cm}$:

$$y(x, t) = 4 \sin\left(\frac{\pi x}{15}ight) \cos(96\pi t)$$

Here, $x$ and $y$ are measured in centimeters, and $t$ is in seconds. Our goal is to dissect this equation to understand the physical state of the string at different points and times.

---

### Part (a): Maximum Displacement at $x = 5	ext{ cm}$

In any standing wave, the displacement of a particle at position $x$ and time $t$ is the product of a spatial amplitude function $A(x)$ and a temporal oscillation function:

$$y(x, t) = A(x) \cos(\omega t)$$

Comparing this with our given equation, the amplitude of vibration at any position $x$ is:

$$A(x) = 4 \sin\left(\frac{\pi x}{15}ight)$$

The **maximum displacement** of any particle is simply the peak value of its oscillation, which is equal to the amplitude $A(x)$ at that point (since the maximum value of the cosine term is $1$).

To find this for a particle at $x = 5	ext{ cm}$, we substitute $x = 5$ into our amplitude function:

$$A(5) = 4 \sin\left(\frac{\pi \cdot 5}{15}ight) = 4 \sin\left(\frac{\pi}{3}ight)$$

Since $\sin\left(\frac{\pi}{3}ight) = \frac{\sqrt{3}}{2}$, we get:

$$A(5) = 4 	imes \frac{\sqrt{3}}{2} = 2\sqrt{3}	ext{ cm}$$

Thus, the particle at $x = 5	ext{ cm}$ oscillates back and forth with a maximum displacement of **$2\sqrt{3}	ext{ cm}$** from its equilibrium position.

---

### Part (b): Locating the Nodes

**Nodes** are points along the medium that remain completely stationary at all times. Physically, this means the amplitude of vibration at these points is zero:

$$A(x) = 0 \implies 4 \sin\left(\frac{\pi x}{15}ight) = 0$$

For the sine function to be zero, its argument must be an integer multiple of $\pi$:

$$\frac{\pi x}{15} = n\pi \implies x = 15n	ext{ cm}$$

where $n$ is an integer ($n = 0, 1, 2, 3, \dots$). Since the string is fixed at both ends and has a total length of $60	ext{ cm}$, the physical domain of $x$ is restricted to $0 \le x \le 60	ext{ cm}$.

Substituting possible values of $n$:
- For $n = 0$: $x = 0	ext{ cm}$
- For $n = 1$: $x = 15	ext{ cm}$
- For $n = 2$: $x = 30	ext{ cm}$
- For $n = 3$: $x = 45	ext{ cm}$
- For $n = 4$: $x = 60	ext{ cm}$

Thus, the nodes are located at **$x = 0, 15, 30, 45, 	ext{ and } 60	ext{ cm}$**.

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### Part (c): Particle Velocity at $x = 7.5	ext{ cm}$ and $t = 0.25	ext{ s}$

The velocity of any particle on the string is the rate of change of its displacement with respect to time, which is given by the partial derivative $\frac{\partial y}{\partial t}$:

$$v_p(x, t) = \frac{\partial y}{\partial t} = \frac{\partial}{\partial t} \left[ 4 \sin\left(\frac{\pi x}{15}ight) \cos(96\pi t) ight]$$

Performing the differentiation:

$$v_p(x, t) = -384\pi \sin\left(\frac{\pi x}{15}ight) \sin(96\pi t)$$

Now, we substitute the given values $x = 7.5	ext{ cm}$ and $t = 0.25	ext{ s}$:

$$v_p(7.5, 0.25) = -384\pi \sin\left(\frac{\pi \cdot 7.5}{15}ight) \sin(96\pi \cdot 0.25)$$

Let's simplify the trigonometric terms:
1. Spatial term: $\sin\left(\frac{7.5\pi}{15}ight) = \sin\left(\frac{\pi}{2}ight) = 1$ (This point is an antinode!)
2. Temporal term: $\sin(24\pi) = 0$

Since the temporal term is zero, the overall velocity is:

$$v_p = -384\pi 	imes 1 	imes 0 = 0	ext{ cm/s}$$

At this exact instant, the entire string momentarily comes to rest as it reaches its maximum displacement. Therefore, the particle velocity is **$0	ext{ cm/s}$**.

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### Part (d): Decomposing into Component Waves

A standing wave is mathematically and physically the superposition of two identical progressive waves traveling in opposite directions. To find these component waves, we can use the product-to-sum trigonometric identity:

$$2 \sin A \cos B = \sin(A + B) + \sin(A - B)$$

Let's rewrite our standing wave equation to match this form:

$$y(x, t) = 2 \left[ 2 \sin\left(\frac{\pi x}{15}ight) \cos(96\pi t) ight]$$

Applying the identity with $A = \frac{\pi x}{15}$ and $B = 96\pi t$:

$$y(x, t) = 2 \sin\left(\frac{\pi x}{15} + 96\pi tight) + 2 \sin\left(\frac{\pi x}{15} - 96\pi tight)$$

Thus, the two component progressive waves are:

$$y_1 = 2 \sin\left(\frac{\pi x}{15} - 96\pi tight)$$
$$y_2 = 2 \sin\left(\frac{\pi x}{15} + 96\pi tight)$$

Here, $y_1$ represents a wave traveling in the positive $x$-direction, and $y_2$ represents an identical wave traveling in the negative $x$-direction.

Visualized Solution

Visualizing the Standing Wave

Identifying the Position-Dependent Amplitude

Part (a): Finding Maximum Displacement at $x = 5 cm$

Part (a): Calculating the Amplitude

Part (b): Condition for Nodes

Part (b): Solving for Node Locations

Part (c): Finding Particle Velocity

Part (c): Substituting $x = 7.5 cm$ and $t = 0.25 s$

Part (c): Evaluating the Velocity

Part (d): Decomposing into Component Waves

Part (d): Writing the Equations of Component Waves

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

Part (a)

Part (b)

Part (c)

Part (d)

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