Math Problem Statement

  1. In an electric shaver, the blade moves back and forth over 8 mm in simple harmonic motion, with a frequency of 120 Hz. Find (i) Displacement at t = 0.5 s (ii) the maximum blade speed. (iii) the magnitude of the maximum acceleration of blade

Solution

We can solve this problem using the principles of simple harmonic motion (SHM). Here's the given data:

  • Amplitude AA = 8mm=0.008m8 \, \text{mm} = 0.008 \, \text{m}
  • Frequency ff = 120Hz120 \, \text{Hz}
  • Angular frequency ω=2πf=2π120rad/s=240πrad/s\omega = 2 \pi f = 2 \pi \cdot 120 \, \text{rad/s} = 240 \pi \, \text{rad/s}
  • Time tt = 0.5s0.5 \, \text{s}

The general equation for SHM is: x(t)=Acos(ωt+ϕ)x(t) = A \cos(\omega t + \phi) Assuming no phase shift (ϕ=0\phi = 0):


(i) Displacement at t=0.5st = 0.5 \, \text{s}

Substitute A=0.008mA = 0.008 \, \text{m}, ω=240π\omega = 240 \pi, and t=0.5st = 0.5 \, \text{s}: x(0.5)=0.008cos(240π0.5)x(0.5) = 0.008 \cos(240 \pi \cdot 0.5)

Simplify 240π0.5=120π240 \pi \cdot 0.5 = 120 \pi, which is an integer multiple of 2π2 \pi: cos(120π)=cos(602π)=cos(0)=1\cos(120 \pi) = \cos(60 \cdot 2\pi) = \cos(0) = 1 Thus: x(0.5)=0.008m=8mmx(0.5) = 0.008 \, \text{m} = 8 \, \text{mm}


(ii) Maximum blade speed

The maximum speed occurs when the blade passes through the equilibrium point, and is given by: vmax=ωAv_{\text{max}} = \omega A Substitute ω=240π\omega = 240 \pi and A=0.008A = 0.008: vmax=240π0.008=1.92πm/sv_{\text{max}} = 240 \pi \cdot 0.008 = 1.92 \pi \, \text{m/s} Numerically: vmax6.03m/sv_{\text{max}} \approx 6.03 \, \text{m/s}


(iii) Magnitude of the maximum acceleration of the blade

The maximum acceleration occurs at the extreme positions and is given by: amax=ω2Aa_{\text{max}} = \omega^2 A Substitute ω=240π\omega = 240 \pi and A=0.008A = 0.008: amax=(240π)20.008=57600π20.008a_{\text{max}} = (240 \pi)^2 \cdot 0.008 = 57600 \pi^2 \cdot 0.008 amax=4608π2m/s2a_{\text{max}} = 4608 \pi^2 \, \text{m/s}^2 Numerically: amax45453.2m/s2a_{\text{max}} \approx 45453.2 \, \text{m/s}^2


Final Results:

  1. Displacement at t=0.5st = 0.5 \, \text{s}: 8mm8 \, \text{mm}
  2. Maximum blade speed: 6.03m/s\approx 6.03 \, \text{m/s}
  3. Magnitude of maximum acceleration: 45453.2m/s2\approx 45453.2 \, \text{m/s}^2

Would you like detailed derivations or explanations of any specific steps?


Related Questions:

  1. How is the displacement in SHM derived?
  2. Why is the maximum speed proportional to ωA\omega A?
  3. How does the angular frequency affect the motion in SHM?
  4. Can the maximum acceleration be used to determine AA if unknown?
  5. What is the effect of doubling the frequency on vmaxv_{\text{max}} and amaxa_{\text{max}}?

Tip:

In SHM problems, always determine ω=2πf\omega = 2\pi f first, as it governs both speed and acceleration.

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Math Problem Analysis

Mathematical Concepts

Simple Harmonic Motion (SHM)
Trigonometric Functions
Frequency and Angular Frequency
Kinematics in SHM

Formulas

x(t) = A cos(ωt + φ)
v_max = ωA
a_max = ω²A
ω = 2πf

Theorems

Properties of Simple Harmonic Motion
Relationship between Angular Frequency and Linear Motion

Suitable Grade Level

Grades 11-12