P16_010.PDF

(64 KB) Pobierz
Chapter 16 - 16.10
10. (a) The problem gives the frequency f = 440 Hz, where the SI unit abbreviation Hz stands for Hertz,
which means a cycle-per-second. The angular frequency ω is similar to frequency except that ω
is in radians-per-second. Recalling that 2 π radians are equivalent to a cycle, we have ω =2 πf
2800 rad/s.
(b) In the discussion immediatelyafter Eq. 16-6, the book introduces the velocity amplitude v m = ωx m .
With x m =0 . 00075 m and the above value for ω , this expression yields v m =2 . 1m/s.
(c) In the discussion immediately after Eq. 16-7, the book introduces the acceleration amplitude a m =
ω 2 x m , which (if the more precise value ω = 2765 rad/s is used) yields a m =5 . 7km/s.
Zgłoś jeśli naruszono regulamin