Posted by johnny on Thursday, December 11, 2008 at 8:08pm.
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.
- physics - bobpursley, Thursday, December 11, 2008 at 8:12pm
I will be happy to critique your thinking.
- physics - Christiaan, Thursday, December 11, 2008 at 8:42pm
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
- physics - Christiaan, Thursday, December 11, 2008 at 8:48pm
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
- physics - johnny, Thursday, December 11, 2008 at 9:07pm
nope, we need to find w then find v= w*xm = 741.416*0.001= .741416
then a= w^2*xm = 549.697
- cxor pizoc - cxor pizoc, Saturday, January 17, 2009 at 3:06am
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