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» <html><body><p></p>Solution :Amplitude of the particle = <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> cm=0.01 m, Time <br/> period `t=22s` and `phi_(0)=0` <br/> The angular velocity `<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>=(2pi)/T=(2xx22)/(7xx22)=2/7` rad/s <br/> The equation for displacement of the particle is <br/> `x=Asin (omega t+phi)=0.10sin(2/7t)` <br/> when `t=3.5s=7/2`s the distance it travels <a href="https://interviewquestions.tuteehub.com/tag/becomes-1994370" style="font-weight:bold;" target="_blank" title="Click to know more about BECOMES">BECOMES</a> equal to the displacement because the <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> time is less than `T//4`. <br/> `:. x=0.10sin(2/7xx7/2)=0.10sin(1)` <br/> `=0.10sin (57^(@)17.) ( :. 1"radian"=57^(@)17.)` <br/> `=0.10xx0.8414=0.8414m`</body></html>