Question

A tuning fork oscillates at a frequency of 441 Hz and the tip of each prong moves 1.5 mm to either side of center. Calculate (a) the maximum speed and (b) the maximum acceleration of the tip of a prong.

Answer 1

To find the maximum speed and acceleration of a prong of the tuning fork, the formulas for maximum speed and acceleration in harmonic motion were used, with amplitude converted to meters. The maximum speed was found to be 4.146 m/s, and the maximum acceleration 3867.691 m/s^2.

Step-by-step explanation:

The question involves a tuning fork oscillating at a frequency of 441 Hz, where each prong moves 1.5 mm to either side of the center.

To calculate the maximum speed (vmax), we use the formula vmax = 2π•f•A, where f is the frequency and A is the amplitude (maximum displacement from the center). Given that A is 1.5 mm (or 0.0015 meters since we need to work in SI units), and f is 441 Hz, we can plug in these values to get the maximum speed.

vmax = 2π•(441 Hz)•(0.0015 m) = 4.146 m/s.

To calculate the maximum acceleration (amax), we use the formula amax = (2π•f)2•A. Plugging in the given values:

amax = (2π•441 Hz)2•(0.0015 m) = 3867.691 m/s2.

The units of amplitude were converted from mm to meters, as the SI unit for displacement is meters (m), and the SI units for speed and acceleration are m/s and m/s2, respectively.