Question

In an electric shaver, the blade moves back and forth over a distance of 2.1 mm in simple harmonic motion, with frequency 116 Hz. Find (a) the amplitude, (b) the maximum blade speed, and (c) the magnitude of the maximum blade acceleration.

Answer 1

Answer:

amplitude =  1.05×$10^{-3}$ m

maximum speed = 76.49 cm/s

acceleration = 557.22 m/s²

Explanation:

given data

distance = 2.1 mm

frequency = 116 Hz

to find out

amplitude , maximum speed and acceleration

solution

we  will apply here amplitude formula that is

amplitude = wavelength distance / 2

so

amplitude = 2.1 /2

amplitude = 1.05 mm =  1.05×$10^{-1}$ cm

amplitude =  1.05×$10^{-3}$ m

and

maximum speed = amplitude × angular frequency

so angular frequency = 2πf

angular frequency = 2π(116)

angular frequency = 728.48 rad/s

so

maximum speed = amplitude × angular frequency

maximum speed = 1.05×$10^{-1}$ × 728.48

maximum speed = 76.49 cm/s

and

acceleration = amplitude × (angular frequency)²

acceleration = 1.05×$10^{-3}$ × (728.48)²

acceleration = 557.22 m/s²