Question

A body oscillates with simple harmonic motion according to the equation x= (6.12 m) cos [(8.38 rad/s)t +1.92 rad] Find (a) the displacement, (b) he velocity, and (c) the accel- eration at the time t =1.90 s. Find also (d) the frequency and (e) the period of the motion.

Answer 1

Answer:

(a) 3.267 m

(b) 43.36 m/s

(c) - 229.47 m/s^2

Explanation:

x = 6.12 Cos[8.38t + 1.92]

Compare with the standard equation

x = A Cos (ωt + Ф)

ω = 8.38 rad/s

Ф = 1.92 rad

(A) at t = 1.90 s

x = 6.12 Cos (8.38 x 1.90 + 1.92)

x = 6.12 Cos (17.842)

x = 6.12 (0.5339)

x = 3.267 m

Thus, the displacement at t = 1.9 s is 3.267 m.

(B) Velocity is v.

$v = \frac{dx}{dt}$

So, differentiate the displacement function with respect to time

v = - 6.12 (8.38) Sin (8.38 t + 1.92) = - 51.286 Sin (8.38 t + 1.92)

Now substitute t = 1.9 s

v = - 51.286 Sin (17.842)

v = - 51.286 (- 0.8455)

v = 43.36 m/s

Thus, the velocity of the body at t = 1.9 s is 43.36 m/s.

(c) Acceleration is a.

$a = \frac{dv}{dt}$

So, differentiate the velocity function with respect to time

a = - 51.286 (8.38) Cos (8.38 t+ 1.92)

Now substitute t = 1.9 s

a = - 429.78 Cos (17.842)

a = - 229.47 m/s^2

Thus, the acceleration of the body at t = 1.9 s is - 229.47 m/s^2.