posted by johnny on .
In an electric shaver, the blade moves back and forth over a distance of 2.0 mm. The motion is simple harmonic, with frequency 118 Hz.
(a) Find the amplitude.
(b) Find the maximum blade speed.
(c) Find the magnitude of the maximum blade acceleration.
I will be happy to critique your thinking.
a) A= 2.0 mm /2 = 1.0 mm = 0.001 m
b) The time it takes the blade to go back AND forth (the period of the oscillation) can be found as follows:
T=1/f= 1/118Hz = 0.00847 s
So, for the blade to cross the distance of 0.002 m, it takes 0.00847/2 s = 0.00424 s.
v = x/t = 0.002 m / 0.00424 s = 0.472 m/s
So its maximum speed is 0.472 m/s
Sorry, made an error in my reasoning.
Amplitude is correct, but speed and acceleration isn't.
For a body undergoing simple harmonic motion, the expression for the maximum speed is:
v(max) = A*f = 0.001 m * 118 Hz = 0.118 m/s
and the expression for the maximum acceleration is:
a(max) = A*f² = 0.001m * (118 Hz)² = 13.924 m/s²
nope, we need to find w then find v= w*xm = 741.416*0.001= .741416
then a= w^2*xm = 549.697
dnqfces yujhb ikbslj jivrmeqah svjl ivejkga mlbnxt