if a 170 g mass on a 2.0-m-long string is pulled 9.0 ∘ to one side and released.then How long does it take for the pendulum to reach 3.0 ∘ on the opposite side?
since at t=0 ,the position x=A (where A is the amplitude)
A=l (length of pendulam,since it is pulled 900)
so we can consider the equation of SHM as x=Acos(ωt) -because at t=0,x=A
now time period of motion is given by T=2π (l/g)1/2 where l-length of pendulam, g- gravitational constant
T=2π * (2/10)1/2 keep in mind that time period is irrespective of mass bob
now time to reach 300 in opposite side=time to cover (900+300)=T/4+t
t is the time travelling for 300
from the fig its clear that x corresponds to 30 is
sin α=x/l
x=sin 30*l=0.5*2=1
sub in equation x=a cos(ωt)
1=2 cos(ωt)
ωt=cos-1(1/2)=π /3 where ω=2π/T
t=T/6
so total time is T/4+T/6=5T/12
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