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Luda [366]
3 years ago
7

In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 124 Hz

. Find (a) the amplitude, (b) the maximum blade speed, and (c) the magnitude of the maximum blade acceleration.
Physics
1 answer:
bagirrra123 [75]3 years ago
8 0

Answer:

(a) A = 1 mm

(b) V_{max}=0.77872 m/s

(c) a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A

V_{max}=778.72 \times 0.001

V_{max}=0.77872 m/s

(c) The formula for the maximum acceleration is given by

a_{max}=\omega ^{2}A

a_{max}=778.72 ^{2}\times 0.001

[tex]a_{max}=606.4 m/s^{2}/tex]

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