school
Acemy

Representing and Analyzing SHM

AP Physics 1· difficulty 3/5

For an oscillator with ω=4 rad/s\omega = 4~\text{rad/s} and amplitude A=0.10 mA = 0.10~\text{m}, the speed at x=0.06 mx = 0.06~\text{m} is

  • A

    0.24 m/s0.24~\text{m/s}

  • B

    0.32 m/s0.32~\text{m/s}

    check_circle
  • C

    0.16 m/s0.16~\text{m/s}

  • D

    0.40 m/s0.40~\text{m/s}

Explanation

v=ωA2−x2=40.01−0.0036=40.0064=4(0.08)=0.32 m/sv = \omega\sqrt{A^2 - x^2} = 4\sqrt{0.01 - 0.0036} = 4\sqrt{0.0064} = 4(0.08) = 0.32~\text{m/s}.

Want 10 more like this — adaptive to your weak spots?

Practice 10 like thisarrow_forwardPredict my AP score

Related questions

auto_awesomeAll Toolsquestion_answerHomework Q&Amenu_bookAP & SAT PreptimelineLearn Mode