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.

mm

(b) Find the maximum blade speed.

m/s

(c) Find the magnitude of the maximum blade acceleration.

m/s2

- 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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