Question
A transverse harmonic wave on a string is described by
$\text{y}(\text{x, t})=3.0\sin\big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\big)$
where x and y are in cm and t in s. The positive direction of x is from left to right.
What is the least distance between two successive crests in the wave?

Answer

3.49m
Explanation:
Given,
$\text{y}(\text{x, t})=3\sin\big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\big)$
the smallest distance between two adjacent crests in the wave, $\lambda=\frac{2\pi}{\text{k}}=\frac{2\pi}{0.018}=349\text{cm}$

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