A wave disturbance in a medium is described by y(x, t) = 0.02 cos(50πt + π/2) cos(10πx), where x and

Q: Why is this a standing wave and not a travelling wave?

In a travelling wave, the position $x$ and time $t$ are combined inside a single trigonometric function, like $\sin(kx - \omega t)$. In a standing wave, the spatial part $\cos(kx)$ and temporal part $\cos(\omega t)$ are multiplied as separate functions, meaning the wave profile oscillates in place without propagating.

Q: Can a node or antinode position be negative?

Yes, mathematically nodes and antinodes exist on both sides of the origin. Since the cosine function is symmetric, nodes also occur at negative values like $x = -0.05\text{ m}, -0.15\text{ m}$, etc.

Q: What is the physical significance of the phase constant $\pi/2$ in the time term?

The phase constant $\frac{\pi}{2}$ simply shifts the starting time of the oscillation. Since $\cos(\theta + \pi/2) = -\sin(\theta)$, it means that at $t=0$, the particles at the antinodes are passing through their mean positions with maximum velocity.

Q: How do we know if the wave is clamped at the ends?

At $x=0$, the spatial term is $\cos(0) = 1$, which is an antinode. This indicates that the boundary at $x=0$ is free to vibrate (like an open end of a pipe) rather than being clamped (which would require a node at $x=0$).

Q: Is the speed of a standing wave really $5\text{ m/s}$ if it doesn't travel?

A standing wave itself does not propagate energy, so its net wave velocity is zero. However, the term 'speed of wave' in this context refers to the speed of the individual component progressive waves ($v = \omega/k$) that interfere to create the standing wave pattern.

Question

A wave disturbance in a medium is described by $y(x, t) = 0.02 \cos \left( 50\pi t + \frac{\pi}{2} ight) \cos (10\pi x)$, where $x$ and $y$ are in metre and $t$ is in second.

Options:

(A) A node occurs at $x = 0.15	ext{ m}$

(B) An antinode occurs at $x = 0.3	ext{ m}$

(C) The speed of wave is $5	ext{ ms}^{-1}$

(D) The wavelength of wave is $0.2	ext{ m}$

Sigma Percentile · Accepted Answer

### Introduction to Standing Waves

Imagine a guitar string plucked at both ends, or air vibrating inside a flute.

What you are witnessing is the beautiful physics of **standing waves** (or stationary waves).

Unlike progressive waves that travel through space carrying energy from one point to another, standing waves are formed by the superposition of two identical waves travelling in opposite directions.

They trap energy in localized pockets, creating points of absolute stillness called **nodes** and points of maximum thrashing called **antinodes**.

Let's dive deep into a classic JEE problem that tests our fundamental understanding of these wave structures.

---

### Deconstructing the Wave Equation

We are given the wave equation:

$$y(x, t) = 0.02 \cos \left( 50\pi t + \frac{\pi}{2} ight) \cos (10\pi x)$$

Notice how the spatial variable $x$ and the temporal variable $t$ are completely separated into their own cosine functions.

This is the mathematical hallmark of a standing wave.

To extract the physical properties of this wave, we compare it with the standard standing wave equation:

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

By direct comparison, we can immediately read off two crucial parameters:
- **Wave number ($k$):** $k = 10\pi 	ext{ rad/m}$
- **Angular frequency ($\omega$):** $\omega = 50\pi 	ext{ rad/s}$

Let's use these parameters to verify each option step-by-step.

---

### Finding the Wavelength (Option d)

The wave number $k$ is intimately related to the wavelength $\lambda$ by the formula:

$$k = \frac{2\pi}{\lambda}$$

Rearranging this to solve for $\lambda$:

$$\lambda = \frac{2\pi}{k}$$

Substituting our identified value of $k = 10\pi$:

$$\lambda = \frac{2\pi}{10\pi} = 0.2 	ext{ m}$$

The wavelength of the component waves is exactly $0.2 	ext{ m}$.

This confirms that **Option (d)** is mathematically correct!

---

### Determining the Wave Speed (Option c)

The speed $v$ of the component progressive waves is given by the ratio of angular frequency to wave number:

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

Substituting our values:

$$v = \frac{50\pi}{10\pi} = 5 	ext{ m/s}$$

The speed of the wave is indeed $5 	ext{ m/s}$.

This confirms that **Option (c)** is also correct!

---

### Mapping the Nodes (Option a)

**Nodes** are points in the medium where the displacement is always zero, regardless of the time $t$.

For $y(x, t) = 0$ at all times, the spatial term must vanish:

$$\cos(10\pi x) = 0$$

We know that the cosine function is zero at odd multiples of $\frac{\pi}{2}$:

$$10\pi x = (2n + 1)\frac{\pi}{2} \quad 	ext{for } n = 0, 1, 2, \dots$$

Solving for $x$:

$$x = \frac{2n + 1}{20} 	ext{ m}$$

Let's find the positions of the first few nodes by substituting integer values for $n$:
- For $n = 0$: $x = \frac{1}{20} = 0.05 	ext{ m}$
- For $n = 1$: $x = \frac{3}{20} = 0.15 	ext{ m}$
- For $n = 2$: $x = \frac{5}{20} = 0.25 	ext{ m}$

We see that a node indeed occurs at $x = 0.15 	ext{ m}$.

Thus, **Option (a)** is correct!

---

### Locating the Antinodes (Option b)

**Antinodes** are points where the medium undergoes maximum displacement.

This happens when the spatial term has its maximum magnitude of $1$:

$$|\cos(10\pi x)| = 1$$

The cosine function has a magnitude of $1$ at integer multiples of $\pi$:

$$10\pi x = m\pi \quad 	ext{for } m = 0, 1, 2, \dots$$

Solving for $x$:

$$x = \frac{m}{10} 	ext{ m}$$

Let's find the positions of the first few antinodes by substituting values for $m$:
- For $m = 0$: $x = 0 	ext{ m}$
- For $m = 1$: $x = 0.1 	ext{ m}$
- For $m = 2$: $x = 0.2 	ext{ m}$
- For $m = 3$: $x = 0.3 	ext{ m}$

An antinode indeed occurs at $x = 0.3 	ext{ m}$.

Thus, **Option (b)** is also correct!

---

### Conclusion

By systematically applying the principles of wave mechanics, we have verified that:
1. A node occurs at $x = 0.15 	ext{ m}$ (Option a)
2. An antinode occurs at $x = 0.3 	ext{ m}$ (Option b)
3. The wave speed is $5 	ext{ m/s}$ (Option c)
4. The wavelength is $0.2 	ext{ m}$ (Option d)

All four options are correct, making this a beautiful multi-correct question that perfectly tests the anatomy of standing waves!

Animated Solution for Physics - Waves: A wave disturbance in a medium is described by $y (x, t) = 0.02 cos (50 π t + \frac{π}{2}) cos (10 π x)$ , where $x$ and $y$ are in metre and $t$ is in second.

Select Answer:

Visualized Solution

Understanding the Standing Wave Equation

Identifying Wave Parameters

Calculating Wavelength $λ$

Calculating Wave Speed $v$

Condition for Nodes

Verifying Node at $x = 0.15 m$

Condition for Antinodes

Verifying Antinode at $x = 0.3 m$

All Options are Correct

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction to Standing Waves

Deconstructing the Wave Equation

Finding the Wavelength (Option d)

Determining the Wave Speed (Option c)

Mapping the Nodes (Option a)

Locating the Antinodes (Option b)

Conclusion

Similar Questions

A wave $y = a sin (ω t - k x)$ on a string meets with another wave producing a node at $x = 0$ . Then, the equation of the unknown wave is

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

A $20 cm$ long string, having a mass of $1.0 g$ , is fixed at both the ends. The tension in the string is $0.5 N$ . The string is set into vibration using an external vibrator of frequency $100 Hz$ . Find the separation (in cm) between the successive nodes on the string.

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, $y (x, t) = (0.01 m) [sin (62.8 m^{- 1}) x] cos [(628 s^{- 1}) t]$ . Assuming $π = 3.14$ , the correct statement(s) is (are)

A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is $Y = 0.3 sin (0.157 x) cos (200 π t)$ . The length of the string is (All quantities are in SI units)

Length of a string tied to two rigid supports is $40 cm$ . Maximum length (wavelength in cm) of a stationary wave produced on it, is

A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to

Animated Solution for Physics - Waves: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A wave disturbance in a medium is described by y(x,t)=0.02cos(50πt+2π​)cos(10πx), where x and y are in metre and t is in second.