Question

The function
x = (8.5 m) cos[(5?rad/s)t + ?/3 rad]
gives the simple harmonic motion of a body. At t = 7.3 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?

Answer 1

The position is given by,

                  x = (8.5 m) cos[(5(pi)rad/s)t + (pi)/3 rad]

(a)

at t = 7.3 s, the displacement is,

                  x = (8.5 m) cos[(5(pi)rad/s)(7.3 s) + (pi)/3 rad]

                     = -7.36 m

(b)

The velocity of the oscillations is,

                  v = dx/dt

                     = d/dt [(8.5 m) cos[(5(pi)rad/s)t + (pi)/3 rad]]

                     = -(8.5 m)(5)(pi)sin[(5(pi)rad/s)t + (pi)/3 rad]

At t = 7.3 s, the velocity is,

                    v = -(8.5 m)(5)(pi)sin[(5(pi)rad/s)(7.3 s) + (pi)/3 rad]

                       = -66.75 m/s

(c)

The acceleration is,

                   a = dv/dt

= d/dt{-(8.5 m)(5)(pi)sin[(5(pi)rad/s)t + (pi)/3 rad]}

                      = -(8.5 m)[(5)(pi)]²cos[(5(pi)rad/s)t + (pi)/3 rad]

At t = 7.3 s, the acceleration is,

                    a = -(8.5 m)[(5)(pi)]²cos[(5(pi)rad/s)(7.3 s) + (pi)/3 rad]

                       = 1.816x10³ m/s²

(d)

The phase angle is,

                     angle = (pi)/3 rad

                              = 1.047 rad

(e)

The frequency is,

                     f = w/2(pi)

   = [(5)(pi) rad/s]/2(pi)

   = 2.5 Hz

(f)

the period is,

             T = 1/f

                = 1 / 2.5 Hz

                = 0.4 s