A particle executes SHM of 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 SHM of period 22 second and amplitude 10 cm. Find

1.	A particle executes SHM of 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 = 10cm =0.01 m, Time period t= 22s and `phi= 0` The angular velocity , `omega= (2pi)/(T)= (2xx 22)/(7xx 22)= (2)/(7)rad//s` The equation for displacement of the particle is `x= ASIN(omegat+phi)= 0.10sin((2)/(7))t` When `t= 3.5 s= (7)/(2)s` the distance it TRAVEL becomes EQUAL to the displacement because the given time is less than `T//4` `:. x= 0.10sin((2)/(7) xx (7)/(2))= 0.10 sin(1)` `= 0.10 sin(57^(@)17.)( :. 1 "radian"= 57^(@)17.)= 0.10 xx 0.8414= 0.08414m`