A mechanical wave propagating along a string is defined by the displacement equation y = 0… — JAMB UTME

A mechanical wave propagating along a string is defined by the displacement equation $y = 0.05 \sin\left(160\pi t - 8\pi x + \frac{\pi}{3}\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. The wave transitions into a different section of string where its propagation speed drops by $25\%$ . Calculate the minimum distance (in meters) between two points in this second section that have a phase difference of $\frac{2\pi}{3}\text{ rad}$ .

Answer: 0.0625 m

Answer

The minimum distance between the two points in the second section is

0.0625\text{ m}

The correct calculation gives

0.0625\text{ m}

. Comparing

y = 0.05 \sin(160\pi t - 8\pi x + \pi/3)

with the general form

y = A \sin(\omega t - k x + \phi_0)

identifies

\omega = 160\pi\text{ rad/s}

and

k_1 = 8\pi\text{ rad/m}

, yielding an initial speed of

v_1 = \omega / k_1 = 20\text{ m/s}

. Upon transitioning into the second string section, the speed drops by

25\%

v_2 = 15\text{ m/s}

. Since the angular frequency

\omega = 160\pi\text{ rad/s}

remains invariant during refraction, the wave number in the second section is

k_2 = \omega / v_2 = 160\pi / 15 = 32\pi / 3\text{ rad/m}

. Substituting

k_2

and the given phase difference

\Delta \phi = 2\pi / 3\text{ rad}

into

\Delta \phi = k_2 \Delta x

yields

\Delta x = (2\pi / 3) / (32\pi / 3) = 2/32 = 0.0625\text{ m}

Step-by-Step Solution

Extract angular frequency and wave number from the displacement equation

\omega = 160\pi\text{ rad/s}

and

k_1 = 8\pi\text{ rad/m}

The standard progressive wave equation is formatted as

y = A \sin(\omega t - k x + \phi_0)

Calculate the initial wave propagation speed

v_1

v_1 = \frac{\omega}{k_1} = \frac{160\pi}{8\pi} = 20\text{ m/s}

Wave speed is equal to the ratio of angular frequency to wave number.

Determine the wave speed

v_2

in the second section

v_2 = 20 \times (1 - 0.25) = 15\text{ m/s}

The wave speed decreases by

25\%

, making

v_2 = 0.75 v_1

Find the new wave number

k_2

in the second section

k_2 = \frac{\omega}{v_2} = \frac{160\pi}{15} = \frac{32\pi}{3}\text{ rad/m}

Frequency and angular frequency remain invariant when a wave passes from one medium to another.

Calculate the spatial separation

\Delta x

for the specified phase difference

\Delta x = \frac{\Delta \phi}{k_2} = \frac{2\pi / 3}{32\pi / 3} = \frac{2}{32} = 0.0625\text{ m}

Phase difference relates to spatial distance via

\Delta \phi = k \Delta x

Key Concept

Wave Equation Parameter Extraction and Invariance of Frequency in Refraction

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