Waves and its Characteristics: Practice Problem & Solution
For a wave $y = y_0 \sin(\omega t - kx)$, for what value of $\lambda$ is the maximum particle velocity equal to two times the wave velocity: (1998)
Solution Explained:
To solve this problem, we apply the core principles of Waves and its Characteristics. Understanding the underlying formula is key to arriving at the correct answer below:
Maximum particle velocity $v_{max} = A\omega = y_0\omega$. Wave velocity $v = \frac{\omega}{k} = \frac{\omega\lambda}{2\pi}$. Given $y_0\omega = 2(\frac{\omega\lambda}{2\pi})$, resolving gives $\lambda = \pi y_0$.
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