Question

a particle executes S.H.M of period 12s and of amplitude 8cm.what time will it take to travel 4cm from the extreme point?​

Answer 1

Answer:

X=Acosωt

X=Acosωt5=8cosωt

X=Acosωt5=8cosωtωt=cos

X=Acosωt5=8cosωtωt=cos −1

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t=

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 =

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9 =

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9 = 2π

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9 = 2π9T

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9 = 2π9T

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9 = 2π9T =

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9 = 2π9T = 2π

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9 = 2π9T = 2π0.9×1.2

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9 = 2π9T = 2π0.9×1.2

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9 = 2π9T = 2π0.9×1.2

X=Acosωt5=8cosωtωt=cos −1 (5/8)=0.9∴ t= ω0.9 = (2π/T)0.9 = 2π9T = 2π0.9×1.2 =0.17 s