Question

The position of an OAS (Simple Harmonic Oscillator) as a function of time is given by x = 3.8m cos (1.25t + 0.52) where t is in seconds and x is in meters
Find
a) Period (s)
b) Acceleration (m / s^2) at t = 2.0s

Answer 1

1. The above problem is related to simple harmonic motion.
2. All the steps are clearly mentioned and easy to understand.
3. Hope the answer will help you. Thank you for your support.
4. 
   Solution ! Given the position of an OAS (Simple Harmonic Oscillator) as a function of time is given by X = 3.8 m cos (4.25+ 0.52) Chere t is in seconde and in meters X is (a) Period (s) (2 we get Comparing Ed Co with general S.H.M. X= A.cos (cut + w = 1.25 rad /sec 25 Therefore Period = 5.02 1.25 T=5.02 seconds 24 (6) Acceleration (m/32) at t = 2:0 S t get - sa de dt Now differentiating & writ. velocity ie= 3.8 x 1.25 sim (1.25 +0.52) velocity Again differentiating i wirit 't' , we get acceleration daxi 1.25€ + 0,52) du at² Q = -3.8x 1.250 1.25 cos
   
   Now at time t=2.0 sec the acceleration a = - 3.8% 1.25 x 1,25 cons 1.25x 2.0 +0.52 a =- 5.9375 Cos(3.02) Q = 5.99 m /