Final answer:
The maximum velocity of particles in the described transverse wave motion on a string is 62.83 cm/s, calculated by the derivative of the wave function with respect to time.
Step-by-step explanation:
The student has asked for the maximum velocity of particles in a wave motion described by the equation y=(10cm)sinπ(0.02x−2.00t), where x is in meters and t in seconds. The equation represents a transverse wave moving through a string. To find the maximum velocity (vmax) of the particles, we determine the derivative of the wave function with respect to time (t), since this will give us the velocity of the particles in the string. The maximum value for velocity occurs when the time derivative of the sine function is at its maximum, which is π for sin(πt). Therefore:
vmax = Aπf
Where A is the amplitude, and f is the frequency. Given A = 10 cm and f (frequency) is equal to the coefficient of the t term divided by 2π, which in this case is 2.00 s−1, we have:
vmax = 10 cm × π × 2.00 s−1
vmax = 62.83 cm/s
Therefore, the maximum velocity of the particles in the transverse wave motion is 62.83 cm/s.