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Q13.12: NCERT Physics - Part 2 OSCILLATIONS 13.12 Solution: Plot the corresponding reference circle for each of the…

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 in every case: (x is in cm and t is in s). (a) x = �2 sin (3t + �/3) (b) x = cos (�/6 � t) (c) x = 3 sin (2�t + �/4) (d) x = 2 cos �t

Step-by-Step Verified Solution

**Answer:** Reference circles with radius, angular speed, and initial position. **Step-by-Step Explanation:** (a) x=−2sin⁡(3t+π/3)=2cos⁡(3t+π/3+π/2)=2cos⁡(3t+5π/6)x = -2\sin(3t + \pi/3) = 2\cos(3t + \pi/3 + \pi/2) = 2\cos(3t + 5\pi/6). Radius = 2 cm, ω=3\omega = 3 rad/s, initial phase = 5π/65\pi/6 [citation:9]. (b) x=cos⁡(π/6−t)=cos⁡(t−π/6)x = \cos(\pi/6 - t) = \cos(t - \pi/6). Radius = 1 cm, ω=1\omega = 1 rad/s, initial phase = −π/6-\pi/6 [citation:9]. (c) x=3sin⁡(2πt+π/4)=3cos⁡(2πt+π/4+π/2)=3cos⁡(2πt+3π/4)x = 3\sin(2\pi t + \pi/4) = 3\cos(2\pi t + \pi/4 + \pi/2) = 3\cos(2\pi t + 3\pi/4). Radius = 3 cm, ω=2π\omega = 2\pi rad/s, initial phase = 3π/43\pi/4 [citation:9]. (d) x=2cos⁡(πt)x = 2\cos(\pi t). Radius = 2 cm, ω=π\omega = \pi rad/s, initial phase = 0.

All Questions in this Chapter

Q13.1 Q13.2 Q13.3 Q13.4 Q13.5 Q13.6 Q13.7 Q13.8 Q13.9 Q13.10 Q13.11 Q13.12 Q13.13 Q13.14 Q13.15 Q13.16 Q13.17 Q13.18

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