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
I will gladly check your answers after you have domn some thinking about the proplem. They practically tell you what the amplitude is.
Maximum speeds in simple harmonic motion are w*Amplitude, and Maximum acclerations are w^2*Amplitude.
w is the angular frequency in radians per second. You must compute that from the frequency in Hz. You should know how to do that.
0.5
To solve this problem, we can use the formulas for simple harmonic motion.
(a) The amplitude of a simple harmonic oscillator is defined as half the distance between the maximum and minimum displacements. In this case, the maximum displacement is given as 2.0 mm. Therefore, the amplitude is 2.0 mm / 2 = 1.0 mm.
(b) The maximum blade speed can be found using the formula v_max = ω * A, where ω is the angular frequency and A is the amplitude. The angular frequency can be calculated using the formula ω = 2πf, where f is the frequency. Hence, ω = 2 * π * 118 Hz = 742.44 rad/s. Plugging this into the formula gives us v_max = 742.44 rad/s * 1.0 mm = 742.44 mm/s. To convert this to m/s, we divide by 1000, so the maximum blade speed is 742.44 mm/s / 1000 = 0.742 m/s.
(c) The magnitude of the maximum blade acceleration can be found using the formula a_max = ω^2 * A, where ω is the angular frequency and A is the amplitude. Substituting the values, we have a_max = (742.44 rad/s)^2 * 1.0 mm. To convert mm to m, we divide by 1000, so a_max = (742.44 rad/s)^2 * 1.0 mm / 1000 = 551.28 m/s^2. Therefore, the magnitude of the maximum blade acceleration is 551.28 m/s^2.