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.
mm
(b) Find the maximum blade speed.
m/s
(c) Find the magnitude of the maximum blade acceleration.
m/s2
5 answers
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
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
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
then a= w^2*xm = 549.697
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