A progressive wave traveling along a stretched string has a frequency of 250\text{ Hz} and… — JAMB UTME

A progressive wave traveling along a stretched string has a frequency of $250\text{ Hz}$ and a speed of $300\text{ m/s}$ . What is the minimum distance, in meters, between two points on the string that differ in phase by $\frac{\pi}{3}\text{ rad}$ ?

Answer: 0.2 m

Answer

The minimum distance between the two points is

0.2\text{ m}

The wavelength is found using

\lambda = \frac{v}{f} = \frac{300}{250} = 1.2\text{ m}

. Substituting

\lambda = 1.2\text{ m}

and

\Delta \phi = \frac{\pi}{3}\text{ rad}

into the phase difference formula

\Delta \phi = \frac{2\pi \Delta x}{\lambda}

gives

\Delta x = \frac{(\pi / 3) \times 1.2}{2\pi} = 0.2\text{ m}

Step-by-Step Solution

Calculate the wavelength ($\lambda$) from wave speed ($v$) and frequency ($f$)

$\lambda = \frac{300}{250} = 1.2\text{ m}$

The fundamental wave equation relates wave speed, frequency, and wavelength as $v = f\lambda$.

Apply the phase difference formula to solve for spatial separation ($\Delta x$)

$\Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} = \frac{(\pi / 3) \cdot 1.2}{2\pi} = 0.2\text{ m}$

A full cycle of $2\pi\text{ radians}$ corresponds to a spatial displacement of one wavelength ($\lambda$).

Key Concept

Relationship between phase difference and spatial displacement

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