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JAMB Physics 2024 Paper

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Question : 30

Total: 125

The distance between two successive crests of a water wave is 0.25m. If a particle on the surface of the water makes four complete vertical oscillations in one second. Calculate the speed of the wave.

4ms $^{-1}$
3ms $^{-1}$
2ms $^{-1}$
1ms $^{-1}$

Solution:

Concept:

The speed of a wave is the product of its frequency and wavelength.

Explanation:

Distance between two successive crests is the wavelength,

\lambda = 0.25 \, \text{m}

.

The particle makes four complete vertical oscillations in one second, so frequency

f = 4 \, \text{Hz}

.

Wave speed is given by

v = f \lambda

.

Substitute the values:

v = 4 \times 0.25 = 1 \, \text{m s}^{-1}

.

Answer:

The correct option is D.

1 \, \text{ms}^{-1}

.

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