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.
- StephenLv 41 year agoFavorite 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
- billrussell42Lv 71 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