Physics help?


A particle executes simple harmonic motion

with an amplitude of 1.4 cm.

At what positive displacement from the

midpoint of its motion does its speed equal

one half of its maximum speed?

Answer in units of cm.

2 Answers

  • 1 year ago
    Favorite Answer

    There are various ways to do this. Here's one. Velocity as a function of position in SHM is: v = ±ω√(A² - x²)

    Maximum value of speed occurs when ω√(A² - x²) is maximum (max value = ωA when x=0). ω is a constant so for half max. speed we need

    √(A² - x²) = A/2

    A² - x² = A²/4

    x² = (3/4)A²

    x = A√3/2 = 1.4√3/2 = 1.2cm

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  • 1 year ago

    assuming 1.4 cm is peak amplitude

    If you put the position as

    s = 1.4 sin ωt

    then speed = ds/dt = 1.4 ω cos ωt

    and max speed is when cos ωt = 1

    and half that is when cos ωt = 0.5

    so ωt = arccos 0.5 = 60º

    now you want the displacement when ωt = 60º

    s = 1.4 sin ωt = 1.4 sin 60 = 1.212 cm

    if 1.4 cm is peak-peak amplitude, answer would be 0.606 cm

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