Question

In an electric shaver, the blade moves back and forth over a distance of 2.00 mm. The motion is simple harmonic, with frequency 120 Hz. Find (a) the amplitude, (b) the maximum blade speed, and (c) the maximum blade acceleration.

Answer 1

(A) 0.001 m

(B) 0.75 m/s

(C)
`5.7 x 10^(2) m/s^(2)`

Step-by-step explanation:

distance (d) = 2 mm

frequency (f) = 120 Hz

(A) amplitude (A) = half the distance moved by the blade

amplitude = 2 / 2 = 1 mm = 0.001 m

(B) maximum blade speed (v) = ωA = 2πfA

v = 2 x 3.14 x 120 x 0.001 = 0.75 m/s

(c) maximum blade acceleration (a) =
`w^(2)A`

a =
`(2 x 3.14 x 120)^(2)x0.001` =
`5.7 x 10^(2) m/s^(2)`