Question #160920
A particle executes simple harmonic motion with amplitude of 3.00cm. At what position does its speed equal half of its maximum speed?
1
Expert's answer
2021-02-11T17:11:15-0500

x=Acosωt,x=Acos\omega t,

v=Aωsinωt,v=-A\omega sin\omega t,

vmax=Aω,v_{max}=A\omega,

v=vmaxsinωt,v=-v_{max}sin \omega t,

vvmax=sinωt=12,\frac{v}{v_{max}}=-sin \omega t=\frac 12,

sinωt=12,  sin\omega t=-\frac 12,~~   \Rightarrow~~ cosωt=32,cos\omega t=\frac{\sqrt 3}{2},

x1=A322.60 cm.x_1=\frac{A\sqrt3}{2}\approx2.60~\text{cm}.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS