Question

A sinusoidal wave travelling in the positive direction on stretched string has amplitude 20 cm, wavelength 1 m and wave velocity 5 m/s. At \(x = 0\) and \(t = 0\), it is given that \(y = 0\) and \(\dfrac{dy}{dt}<0\). Find the wave function \(y(x,t)\).

• A. \(y(x,t) = (0.02\text{ m})\sin\!\left[(2\pi\text{ m}^{-1})x + (10\pi\text{ s}^{-1})t\right]\)
• B. \(y(x,t) = (0.02\text{ m})\cos\!\left[(10\pi\text{ s}^{-1})t + (2\pi\text{ m}^{-1})x\right]\)
• C. \(y(x,t) = (0.02\text{ m})\sin\!\left[(2\pi\text{ m}^{-1})x - (10\pi\text{ s}^{-1})t\right]\)
• D. \(y(x,t) = (0.02\text{ m})\sin\!\left[(\pi\text{ m}^{-1})x + (5\pi\text{ s}^{-1})t\right]\)

Hint:

Use the initial conditions \(y=0\) and sign of \(dy/dt\) at \((0,0)\) to uniquely determine the phase constant \(\phi\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A wave travelling in the +x direction: \(y = A\sin(kx - \omega t + \phi)\).
Step 2: Detailed Explanation:
Given: \(A = 0.02 \text{ m}\), \(\lambda = 1 \text{ m}\). Wave number: \(k = \frac{2\pi}{\lambda} = 2\pi \text{ m}^{-1}\). Angular frequency: \(\omega = vk = 5 \times 2\pi = 10\pi \text{ rad/s}\). At \(x=0, t=0\): \(y = 0 \Rightarrow \sin\phi = 0 \Rightarrow \phi = n\pi\). Also: \(\frac{dy}{dt} = -A\omega\cos\phi<0 \Rightarrow \cos\phi>0 \Rightarrow \phi = 0\).
Step 3: Final Answer:
\[ \boxed{y(x,t) = (0.02\text{ m})\sin\!\left[(2\pi\text{ m}^{-1})x - (10\pi\text{ s}^{-1})t\right]} \]