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

Answer:

(A)  0.001 m  

(B) 0.75 m/s  

(C) $5.7 x 10^{2} m/s^{2}$      

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 = $(2 x 3.14 x 120)^{2}x0.001$ = $5.7 x 10^{2} m/s^{2}$