Answer:
Shortest distance = 0.1914 m
Step-by-step explanation:
Given that,
Amplitude of the wave is 0.0957 m
Frequency of the wave is 3.75 Hz
Wavelength of the wave is 1.97 m
We need to find the shortest transverse distance between a maximum and a minimum of the wave.
The distance between maximum point in positive axis and the baseline is equal to amplitude.
Shortest distance = 2 A
D = 2 × 0.0957
D = 0.1914 m
So, the shortest transverse distance between a maximum and a minimum of the wave is 0.1914 m.