Question

In the given wave equation $y=0.05 \sin \frac{2\pi}{\lambda}(x-200t)\text{ m}$, the velocity of the wave (in $\text{ms}^{-1}$) is

• A. $2\sqrt{200}$
• B. 400
• C. $200\sqrt{2}$
• D. $2\sqrt{300}$
• E. 200

Hint:

In any wave equation of the form $y = f(ax \pm bt)$, the wave speed is always the coefficient of time divided by the coefficient of position: $v = \frac{b}{a}$. Here, $v = \frac{(2\pi/\lambda) \times 200}{2\pi/\lambda} = 200$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We are given a 1-dimensional traveling wave equation. We can extract the wave speed by comparing it to the standard mathematical form of a traveling wave.
Step 2: Key Formula or Approach:
The standard traveling wave equation is \(y = A \sin(kx - \omega t)\), where velocity \(v = \frac{\omega}{k}\).
Another common standard form is \(y = A \sin\left(\frac{2\pi}{\lambda}(x - vt)\right)\).
By directly comparing the given equation to this second form, we can identify \(v\).
Step 3: Detailed Explanation:
The given equation is:
\[ y = 0.05 \sin \left[ \frac{2\pi}{\lambda} (x - 200t) \right] \] Comparing this to the standard format:
\[ y = A \sin \left[ \frac{2\pi}{\lambda} (x - vt) \right] \] We can clearly see the mapping:
\(A = 0.05\)
\(v = 200\)
Therefore, the wave velocity is 200 m/s.
Step 4: Final Answer:
The velocity of the wave is 200.