The diagram above illustrates a wave setup between two fixed ends 4.0m apart. If the speed of the wave is 1.0\(ms^{-1}\), its wavelength and frequency respectively are
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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