A transverse sinusoidal wave of amplitude a, wavelength λ and frequency f is travelling on a stretch

Q: Why is the wave velocity constant while the particle velocity varies?

Wave velocity depends only on the properties of the medium (tension and linear mass density of the string) and remains constant in a uniform medium. Particle velocity, however, represents the physical up-and-down motion of the string elements executing Simple Harmonic Motion, which naturally varies from zero at the crests/troughs to a maximum at the equilibrium position.

Q: What is the difference between angular frequency $\omega$ and frequency $f$?

Frequency $f$ is the number of complete oscillations per second (measured in Hertz), while angular frequency $\omega = 2\pi f$ represents the rate of change of phase angle (measured in radians per second).

Q: Why did we use partial differentiation $\frac{\partial y}{\partial t}$ instead of total differentiation $\frac{dy}{dt}$ to find particle velocity?

The displacement $y(x,t)$ depends on both position $x$ and time $t$. To find the velocity of a specific particle, we must look at a fixed position $x$ and see how its displacement changes over time. Keeping $x$ constant while differentiating with respect to $t$ is mathematically represented by a partial derivative.

Q: Does the maximum particle velocity depend on the wavelength?

Directly, no. The maximum particle velocity is $v_{p,\text{max}} = a \omega = 2\pi f a$, which depends only on amplitude and frequency. However, because frequency and wavelength are linked by wave speed ($f = v/\lambda$), changing the wavelength for a wave of constant speed will indirectly affect the frequency and thus the maximum particle velocity.

Q: What would happen to the maximum particle velocity if the tension in the string were increased?

Increasing the tension increases the wave speed ($v = \sqrt{T/\mu}$). If the frequency of the source remains constant, the maximum particle velocity ($v_{p,\text{max}} = 2\pi f a$) will remain unchanged because it depends only on the source frequency and amplitude, not on the medium's properties.

Question

A transverse sinusoidal wave of amplitude $a$, wavelength $\lambda$ and frequency $f$ is travelling on a stretched string. The maximum speed of any point on the string is $v/10$, where $v$ is the speed of propagation of the wave. If $a = 10^{-3}	ext{ m}$ and $v = 10	ext{ m/s}$, then $\lambda$ and $f$ are given by

Options:

(A) $\lambda = 2\pi 	imes 10^{-2}	ext{ m}$

(B) $\lambda = 10^{-3}	ext{ m}$

(C) $f = \frac{10^3}{2\pi}	ext{ Hz}$

(D) $f = 10^4	ext{ Hz}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When a transverse sinusoidal wave travels along a stretched string, we must carefully distinguish between two entirely different physical velocities:

1. **The Wave Velocity ($v$):** This is the speed at which the wave pattern (the phase) propagates along the string. It depends solely on the tension $T$ and the linear mass density $\mu$ of the string, expressed as $v = \sqrt{T/\mu}$.

2. **The Particle Velocity ($v_p$):** This is the velocity of individual string elements as they oscillate up and down in Simple Harmonic Motion (SHM) perpendicular to the direction of wave propagation.

Let the wave propagate in the positive $x$-direction. The displacement $y$ of any particle at position $x$ and time $t$ is given by the standard wave equation:

$$y(x, t) = a \sin(kx - \omega t)$$

where:
- $a$ is the amplitude of the wave,
- $k = \frac{2\pi}{\lambda}$ is the wave number,
- $\omega = 2\pi f$ is the angular frequency.

---

### Deriving Particle Velocity

To find the velocity of a particle at a specific position $x$, we differentiate the displacement equation with respect to time $t$, keeping $x$ constant:

$$v_p = \frac{\partial y}{\partial t} = \frac{\partial}{\partial t} [a \sin(kx - \omega t)]$$

Using the chain rule of differentiation:

$$v_p = -a \omega \cos(kx - \omega t)$$

The maximum speed of any particle on the string occurs when the cosine term reaches its maximum magnitude of $1$:

$$v_{p,	ext{max}} = a \omega$$

Substituting $\omega = 2\pi f$ into this expression gives:

$$v_{p,	ext{max}} = 2\pi f a$$

---

### Applying the Given Condition

The problem states that the maximum speed of any point on the string is one-tenth of the wave propagation velocity:

$$v_{p,	ext{max}} = \frac{v}{10}$$

Substituting our derived expression for $v_{p,	ext{max}}$:

$$2\pi f a = \frac{v}{10}$$

We are given the following numerical values:
- Amplitude, $a = 10^{-3}	ext{ m}$
- Wave velocity, $v = 10	ext{ m/s}$

Let's substitute these values into our equation:

$$2\pi f (10^{-3}) = \frac{10}{10}$$

$$2\pi f (10^{-3}) = 1$$

---

### Calculating Frequency and Wavelength

Now, we solve for the frequency $f$:

$$f = \frac{1}{2\pi 	imes 10^{-3}}$$

$$f = \frac{10^3}{2\pi}	ext{ Hz}$$

This matches **Option (c)**.

Next, we use the fundamental wave relation $v = f \lambda$ to find the wavelength $\lambda$:

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

Substitute the values of $v$ and $f$:

$$\lambda = \frac{10}{\frac{10^3}{2\pi}}$$

$$\lambda = \frac{10 	imes 2\pi}{10^3}$$

$$\lambda = 2\pi 	imes 10^{-2}	ext{ m}$$

This matches **Option (a)**.

Thus, the correct options are **(a)** and **(c)**.

Select Answer:

Visualized Solution

Visualizing the Wave Motion

Connecting Wave Velocity, Frequency, and Wavelength

Understanding Particle Velocity

Finding Maximum Particle Velocity

Setting Up the Given Condition

Substituting Known Values

Calculating the Frequency $f$

Setting Up the Wavelength Calculation

Calculating the Wavelength $λ$

Conclusion

The Sigma Insight: Wave Equation and Wave Speed

Analyzing the Setup

Deriving Particle Velocity

Applying the Given Condition

Calculating Frequency and Wavelength

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