Answer
(b) there will be no standing waves at the lower frequencies.
Work Step by Step
The antinode spacing $d$ is $λ/2$.
Using $v = fλ$, we arrive at $d = λ/2 = v/2f$.
Now plugging in the data, we have $d = \frac{1540}{2(1)} = 0.77 m = 77 cm.$
The cranium is much smaller than 77 cm. As a result, there will be no standing waves within it at 1.0 kHz. The correct answer, therefore, is choice (b).
