Examkin

Question

Difficulty: MediumSimple Harmonic Motion

An object of mass 0.20 kg0.20\text{ kg} executes simple harmonic motion with an amplitude of 0.05 m0.05\text{ m} and a maximum acceleration of 20 m/s220\text{ m/s}^2. What is the speed of the object when its displacement from the equilibrium position is 0.03 m0.03\text{ m}?

  1. A
    0.16 m/s0.16\text{ m/s}
  2. B
    0.60 m/s0.60\text{ m/s}
  3. 0.80 m/s0.80\text{ m/s}Answer
  4. D
    1.00 m/s1.00\text{ m/s}

Answer

The speed of the object at a displacement of 0.03 m0.03\text{ m} is 0.80 m/s0.80\text{ m/s}.
The maximum acceleration in simple harmonic motion is given by amax=ω2Aa_{\text{max}} = \omega^2 A. Substituting amax=20 m/s2a_{\text{max}} = 20\text{ m/s}^2 and A=0.05 mA = 0.05\text{ m} gives ω2=400 rad2/s2\omega^2 = 400\text{ rad}^2/\text{s}^2, so ω=20 rad/s\omega = 20\text{ rad/s}. The speed at any displacement xx is given by v=ωA2−x2v = \omega \sqrt{A^2 - x^2}. For x=0.03 mx = 0.03\text{ m}, v=200.052−0.032=20×0.04=0.80 m/sv = 20 \sqrt{0.05^2 - 0.03^2} = 20 \times 0.04 = 0.80\text{ m/s}.

Step-by-Step Solution

1
Determine the angular frequency (ω\omega) of the simple harmonic motion from the maximum acceleration formula.
ω=20 rad/s\omega = 20\text{ rad/s}
Maximum acceleration is given by amax=ω2Aa_{\text{max}} = \omega^2 A. Rearranging gives ω2=amaxA=200.05=400 rad2/s2\omega^2 = \frac{a_{\text{max}}}{A} = \frac{20}{0.05} = 400\text{ rad}^2/\text{s}^2, so ω=20 rad/s\omega = 20\text{ rad/s}.
2
Calculate the speed (vv) at the given displacement (x=0.03 mx = 0.03\text{ m}) using the SHM velocity formula.
v=0.80 m/sv = 0.80\text{ m/s}
The speed at displacement xx is v=ωA2−x2=20×0.052−0.032=20×0.0016=20×0.04=0.80 m/sv = \omega \sqrt{A^2 - x^2} = 20 \times \sqrt{0.05^2 - 0.03^2} = 20 \times \sqrt{0.0016} = 20 \times 0.04 = 0.80\text{ m/s}.

Key Concept

Simple Harmonic Motion Velocity and Acceleration Relationships
Estimated Time:1m 30s
Rate this question

Topics

Simple Harmonic Motion

More questions from this topic