In an electric shaver, the blade moves back and forth over a distance of 2.00 mm

Question

In an electric shaver, the blade moves back and forth over a distance of 2.00 mm in simple harmonic motion, with a frequency of 120 Hz. The blade has a mass of 1.20 grams. Find the amplitude of the SHM the maximum blade speed the maximum blade acceleration the maximum kinetic energy the spring constant k. Wheels A and B shown below are connected by a belt that does not slip. The times the radius of A. The moment of inertia of B is 15 kg m^2; that of A is 2 kg m^2. of B is 10 radians per second. Find

Explanation / Answer

f=120Hz, m = 0.00120kg

a) Amplitude , A = crest to trough distance /2 = 2mm/2 = 1mm

b) Vmax = A*w = A*2f = 1mm*2*3.14*120 = 0.75m/s = 75cm/s

c) a(max) = Aw^2 = 1*10^-3*(2*3.14*120)^2 = 567.9 m/s^2

d) KEmax = ME(total) = ½*m*A^2 = ½*0.00120*(1.0*10^-3)^2 = 6.0*10^-10 J

e) w= sqrt(k/m)

2f = sqrt(k/m)

2*3.14*120 = sqrt(k/0.00120)    => k = 681.5 N/m

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