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WAEC Physics 2017 Paper

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Question : 30

Total: 50

The diagram above illustrates a wave setup between two fixed ends 4.0m apart. If the speed of the wave is 1.0

\mathrm{m\,s^{-1}}

, its wavelength and frequency respectively are

1m; 0.25Hz
2m; 0.25Hz
2m; 0.50Hz
4m; 0.50Hz

Solution:

For the 4th harmonic, $L = 2\lambda$

$4m = 2\lambda \implies \lambda = \frac{4m}{2}$

= $2m$

Recall, $v = f\lambda$

$1 = f \times 2 \implies f = \frac{1}{2}$

= 0.50Hz

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