A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cm/s.

Q: Why is the particle velocity vertical if the wave is moving horizontally?

In a transverse wave, the particles of the medium oscillate perpendicular to the direction of wave propagation. Since the wave is traveling horizontally along the $x$-axis, the particles must oscillate vertically along the $y$-axis.

Q: How do we know if the slope at point P is positive or negative?

Look at the wave curve around point $P$. As you move from left to right (increasing $x$), the curve goes downwards. This means the tangent line at $P$ tilts downwards, indicating a negative slope ($\frac{\partial y}{\partial x} < 0$).

Q: Where does the formula $v_p = -v \left(\frac{\partial y}{\partial x}\right)$ come from?

For a wave described by $y = f(x - vt)$, taking the partial derivative with respect to time gives $v_p = \frac{\partial y}{\partial t} = -v f'(x - vt)$. Taking the partial derivative with respect to position gives the slope $\frac{\partial y}{\partial x} = f'(x - vt)$. Substituting the second into the first yields $v_p = -v \left(\frac{\partial y}{\partial x}\right)$.

Q: What would change if the wave was moving in the negative x-direction?

If the wave was moving in the negative $x$-direction, the wave speed $v$ in the formula $v_p = -v \left(\frac{\partial y}{\partial x}\right)$ would be negative. This would flip the sign of the particle velocity, making it point downwards ($-\hat{j}$).

Q: Can we use the standard SHM equation $v = \omega \sqrt{A^2 - x^2}$ here?

Yes, but be careful with the variables! In wave motion, the displacement from equilibrium is represented by $y$, and the amplitude is $A$. So the formula becomes $|v_p| = \omega \sqrt{A^2 - y^2}$, where $y$ is the vertical displacement of the particle.

Question

A transverse sinusoidal wave moves along a string in the positive $x$-direction at a speed of $10	ext{ cm/s}$. The wavelength of the wave is $0.5	ext{ m}$ and its amplitude is $10	ext{ cm}$. At a particular time $t$, the snap-shot of the wave is shown in figure. The velocity of point $P$ when its displacement is $5	ext{ cm}$ is

Options:

(A) $\frac{\sqrt{3}\pi}{50} \hat{j}	ext{ m/s}$

(B) $-\frac{\sqrt{3}\pi}{50} \hat{j}	ext{ m/s}$

(C) $\frac{\sqrt{3}\pi}{50} \hat{i}	ext{ m/s}$

(D) $-\frac{\sqrt{3}\pi}{50} \hat{i}	ext{ m/s}$

Sigma Percentile · Accepted Answer

### Introduction to Wave Kinematics

When we look at a wave propagating along a string, it is easy to get confused between the motion of the wave itself and the motion of the individual particles of the string.

The wave travels horizontally, carrying energy and momentum across space.

However, the particles of the string do not travel with the wave; they simply oscillate up and down in **Simple Harmonic Motion (SHM)** about their equilibrium positions.

This problem beautifully tests our understanding of both these motions and how they are mathematically linked.

---

### Analyzing the Setup

Let's list down the given parameters from the problem:

*   Wave speed, $v = 10	ext{ cm/s} = 0.1	ext{ m/s}$ (moving in the $+x$ direction)
*   Wavelength, $\lambda = 0.5	ext{ m}$
*   Amplitude, $A = 10	ext{ cm} = 0.1	ext{ m}$
*   Displacement of point $P$, $y = 5	ext{ cm} = 0.05	ext{ m}$

We need to find the velocity vector of point $P$ at this instant.

Since the wave is transverse and propagates along the $x$-axis, the particles of the string oscillate along the $y$-axis.

Therefore, the velocity of point $P$ must be purely vertical, pointing either in the $+\hat{j}$ or $-\hat{j}$ direction.

This immediately eliminates options (c) and (d), which suggest horizontal motion.

---

### Finding the Magnitude of Particle Velocity

Since the particles execute SHM, we can use the standard relation between velocity and displacement in SHM:

$$|v_p| = \omega \sqrt{A^2 - y^2}$$

Here, $\omega$ is the angular frequency of the wave.

We can find $\omega$ using the relation between wave speed, wavelength, and frequency:

$$v = f \lambda \implies f = \frac{v}{\lambda}$$

Since $\omega = 2\pi f$, we have:

$$\omega = \frac{2\pi v}{\lambda}$$

Substituting the given values:

$$\omega = \frac{2\pi 	imes 0.1}{0.5} = 0.4\pi	ext{ rad/s}$$

Now, let's substitute $\omega$, $A$, and $y$ into our velocity magnitude formula:

$$|v_p| = 0.4\pi \sqrt{(0.1)^2 - (0.05)^2}$$

$$|v_p| = 0.4\pi \sqrt{0.01 - 0.0025} = 0.4\pi \sqrt{0.0075}$$

We can write $0.0075$ as a fraction:

$$0.0075 = \frac{75}{10000} = \frac{3}{400}$$

Taking the square root:

$$\sqrt{\frac{3}{400}} = \frac{\sqrt{3}}{20}$$

Now, substitute this back into the magnitude expression:

$$|v_p| = 0.4\pi 	imes \frac{\sqrt{3}}{20} = \frac{4\pi}{10} 	imes \frac{\sqrt{3}}{20} = \frac{\sqrt{3}\pi}{50}	ext{ m/s}$$

---

### Determining the Direction of Particle Velocity

To find the direction, we use the fundamental relationship between particle velocity ($v_p$) and wave velocity ($v$):

$$v_p = -v \left(\frac{\partial y}{\partial x}ight)$$

Here, $\frac{\partial y}{\partial x}$ represents the slope of the wave profile at point $P$.

Let's analyze the signs of the terms on the right-hand side:

1.  The wave is moving in the positive $x$-direction, so $v$ is positive ($v > 0$).
2.  Looking at the snapshot, point $P$ lies on the downward slope of the wave (as $x$ increases, $y$ decreases). Therefore, the slope at $P$ is negative ($\frac{\partial y}{\partial x} < 0$).

Substituting these signs into our relation:

$$v_p = -(	ext{positive}) 	imes (	ext{negative}) = 	ext{positive}$$

Since $v_p$ is positive, the particle at $P$ is moving upwards, in the direction of $+\hat{j}$.

---

### Conclusion

Combining the magnitude and direction, we get the final velocity vector of point $P$:

$$\vec{v}_p = \frac{\sqrt{3}\pi}{50} \hat{j}	ext{ m/s}$$

This matches **Option (a)** perfectly.

Select Answer:

Visualized Solution

Visualizing the Wave Snapshot

The Particle Velocity Formula

Finding Angular Frequency $ω$

Calculating $ω$

Setting up the Velocity Magnitude

Computing the Magnitude

Determining the Direction of Velocity

Final Vector Velocity

The Way Forward

The Sigma Insight: Wave Equation and Wave Speed

Introduction to Wave Kinematics

Analyzing the Setup

Finding the Magnitude of Particle Velocity

Determining the Direction of Particle Velocity

Conclusion

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