question_answer 12)

Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial \[(t=0)\] position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise m every case; (x is in cm and t is in s). (a) \[x=-2\sin \] \[+\pi /2\] (b) \[x=\cos \,(\pi /6-t)\] (c) \[x=3\sin \,(2\pi +\pi /4)\] (d) \[x=2\cos \pi t\]

Answer:

If we express each function in the form \[x=a\cos \pi t\] where \[\phi =+\pi .\] is the initial phase, i.e., \[\text{x=aco}s\left( \frac{2\pi t}{T}+\phi \right)\text{=2cos}\left( \frac{\text{2 }\!\!\pi\!\!\text{ t}}{\text{4}}\text{+ }\!\!\pi\!\!\text{ } \right)\text{=-2cos}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{t} \right)\] represents the angle which the initial radius vector of the particle makes with the positive direction of x-axis. (a)\[\left( \text{3t+ }\!\!\pi\!\!\text{ /3} \right)\]\[\left( \text{ }\!\!\pi\!\!\text{ /6-t} \right)\] or \[\left( \text{2 }\!\!\pi\!\!\text{ t+ }\!\!\pi\!\!\text{ /4} \right)\] (i) Comparing it with equation (0, we note that \[a=2,\,\,\omega =3\] and = 5 71/6 Hence, the reference circle will be as shown in Fig. 10(NCT).5(a). (b)\[\pi t\] Comparing it with equation (i), we note that \[\phi \] The reference circle will be as shown in Fig. 10(NCT).5(b). (c) x = 3 \[\phi \] Comparing it with equation (i), we note that \[a=3,\] \[\text{x=-2sin}\left( \text{3t+ }\!\!\pi\!\!\text{ /3} \right)\text{=2cos}\] The reference circle will be as shown in Fig. 10 (NCT). 5(d).