P16_026.PDF

26. (a) The textbook notes (in the discussion immediately after Eq. 16-7) that the acceleration amplitude

is a m = ω 2 x m ,where ω is the angular frequency ( ω =2 πf since there are 2 π radians in one cycle).

Therefore, in this circumstance, we obtain

a m =(2 π (1000Hz)) 2 (0 . 00040m) =1 . 6

10 4 m / s 2 .

(b) Similarly, in the discussion after Eq. 16-6, we ﬁnd v m = ωx m so that

v m =(2 π (1000Hz))(0 . 00040m) =2 . 5m / s .

=(2 π (1000Hz)) 2 (0 . 00020m) =7 . 9

10 3 m / s 2 .

16-4, but here we will use trigonometric

relations along with Eq. 16-3 and Eq. 16-6. Thus, allowing for both roots stemming from the

square root,

sin( ωt + φ )=

−

cos 2 ( ωt + φ )

−

ωx m

−

x 2

x 2 m

Taking absolute values and simplifying, we obtain

=2 πf x 2 m −

x 2 =2 π (1000) 0 . 00040 2

−

0 . 00020 2 =2 . 2m / s .

(d) This can be approached with the energy methods of