Final answer:
The maximum speed of the vibrating particles in this transverse wave is approximately 0.314 m/s or 31.4 cm/s. This is calculated by using the amplitude of 0.02 m and the frequency of 2.5 Hz with the formula for maximum speed in wave motion.
Step-by-step explanation:
To calculate the maximum speed of the vibrating particles in a transverse wave, we need to understand wave characteristics. The question states that the distance between a crest and the next trough is 4.0 cm, meaning the wave's amplitude is half of that, which is 2.0 cm or 0.02 m. Since a particle moves from the highest point (crest) to the lowest point (trough) in half a period, we will use this to calculate the maximum speed.
The time for a crest to be replaced by the next crest at the same place is given as 0.4 s, which is the wave's period (T). Therefore, the frequency (f) of the wave is its reciprocal, 1/T, which is 1/0.4 s or 2.5 Hz. The maximum speed (Vmax) of a vibrating particle is given by the formula Vmax = 2πfA. Inserting the values, Vmax = 2π(2.5)(0.02), which gives an approximate maximum speed of 0.314 m/s or 31.4 cm/s.