An object oscillates with simple harmonic motion along the x-axis. Its displacem

Question

An object oscillates with simple harmonic motion along the x-axis. Its displacement from the origin varies with time according the equation: x = (4.00m) cos (xt + pi/4) Where t is in seconds and the angles in the parentheses are in radians. Determine the amplitude. frequency and period of the motion. Calculate the velocity and acceleration of the object at any time t. Using the results of part (b), determine the position, velocity and acceleration of the object at t = 1.00s. Determine the maximum speed and maximum acceleration of the object. Find the displacement of the object between t= 0 and t = 1.00s. Calculate the kinetic energy of the object at any time t. Calculate the potential energy of the object at any time t. Deduce the total energy remains constant.

Explanation / Answer

The object oscillates about the x-axis with the condition x = 4 Cos (pi.t + pi/5) like x = A cos (omega x t + phi)

(a) Hence amplitude A = 4 m,

We see here Angular velocity omega =pi, rad/s.

We know omega = 2 x pi x f . So frequency, f = omega/(2 pi) = pi/(2 pi) = 1/2

and time period T = 1/f = 2 second.

(b) Velocity at time t = first derivative of position with respect to time = -A omega x sin (omega x t + phi)

= -4 x pi x sin (pi x t +pi/5)

Acceleration at time t = second derivative of position wrt time = -omega 2 x A cos (omega x t + phi)

= -(pi)2 x 4 x cos (pi x t + pi/5)

(c) Position at t = 1 is x = 4 cos (pi x 1 + pi/5) = 4 x 0.2 = 0.8

Velocity at t = 1 is v = -4 x omega x sin (pi + pi/5) = -4 x pi x (sin 6 pi/5) = - 4 x 3.14 x 0.696 = -8.74 m/s

Acceleration at t = 1 is -4 x pi2 x cos (6 x pi/5) = -4 x 3.142 x 0.7179 = 28.312 m/sec2

(d)

  

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