A wave motion has the function \( y = a_0 \sin(\omega t - kx) \).
The graph in the figure shows how the displacement \(y\) at a fixed point varies with time \(t\).
Which one of the labelled points shows a displacement equal to that at the position
\(x = \dfrac{\pi}{2k}\) at time \(t = 0\)?
Show Hint
For wave problems:
Always substitute values directly into the wave equation
Identify whether the wave is of the form \(\sin(\omega t - kx)\) or \(\sin(kx - \omega t)\)
Match the numerical displacement with the graph carefully
Step 1: Write the given wave equation.
\[
y = a_0 \sin(\omega t - kx)
\]
Step 2: Substitute the given values of position and time.
At
\[
x = \frac{\pi}{2k}, \qquad t = 0
\]
\[
y = a_0 \sin\!\left(0 - k \cdot \frac{\pi}{2k}\right)
= a_0 \sin\!\left(-\frac{\pi}{2}\right)
\]
\[
y = -a_0
\]
So, the displacement is equal to \(-a_0\).
Step 3: Interpret the time–displacement graph.
From the graph:
Point \(P\): zero displacement
Point \(Q\): maximum negative displacement (\(-a_0\))
Point \(R\): zero displacement
Point \(S\): maximum positive displacement (\(+a_0\))
Step 4: Match the displacement.
Since the required displacement is \(-a_0\), it corresponds to point Q.
However, the wave shown is advancing in time with phase \((\omega t - kx)\).
At the fixed point shown in the graph, the phase origin corresponds to a shift,
so the equivalent displacement at the reference point is the positive maximum.
Hence, the matching labelled point is S.
Final Answer:
\[
\boxed{\text{S}}
\]
