A particle executes SHMof period 22 second and amplitude 10 cm. Find the distance it travels in 3.5 second starting from the point of zero displacement.

A particle executes SHMof period 22 second and amplitude 10 cm. Find t

1.	A particle executes SHMof period 22 second and amplitude 10 cm. Find the distance it travels in 3.5 second starting from the point of zero displacement.
Answer» Solution :Amplitude of the particle = 10 cm=0.01 m, Time period `t=22s` and `phi_(0)=0` The angular velocity `OMEGA=(2pi)/T=(2xx22)/(7xx22)=2/7` rad/s The equation for displacement of the particle is `x=Asin (omega t+phi)=0.10sin(2/7t)` when `t=3.5s=7/2`s the distance it travels BECOMES equal to the displacement because the GIVES time is less than `T//4`. `:. x=0.10sin(2/7xx7/2)=0.10sin(1)` `=0.10sin (57^(@)17.) ( :. 1"radian"=57^(@)17.)` `=0.10xx0.8414=0.8414m`