Question

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?

Answer #1

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(**ω**t) -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(**ω**t)

1=2 cos(**ω**t)

**ω**t=**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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