Question

The x component of the velocity of an object vibrating along the x-axis obeys the equation

vx(t) = (0.445 m/s) sin[(25.4 rad/s)t + 0.223]

What is the object’s acceleration when its velocity has a maximum positive value? What is the object’s position x when it has a velocity of -0.200 m/s and a positive acceleration value?

Answer 1

The velocity of particle in SHM is V = (0.445 m/s) sin[(25.4 rad/s)t + 0.223]

a) If the velocity has maximum positive value then the acceleration of the particle is ZERO.

The acceleration of the particle is a = dV/dt = (0.445 x 25.4 m/s2) cos[(25.4 rad/s)t + 0.223]

If V is maximum then [(25.4 rad/s)t + 0.223] = 90o then cos 90o = 0

b) Given velocity is V = - 0.2 m/s and acceleration is positive.

since V = (0.445 m/s) sin[(25.4 rad/s)t + 0.223]

- 0.2 m/s = (0.445 m/s) sin[(25.4 rad/s)t + 0.223]

- 0.45 = sin[(25.4 rad/s)t + 0.223]

- 0.467 rad = [(25.4 rad/s)t + 0.223]

The acceleration and displacement are always opposite. Hence the position of particle is negative.

  

Then X = - (0.445/25.4 m) cos[(25.4 rad/s)t + 0.223]

X = - 0.01752 cos[(25.4 rad/s)t + 0.223]

X = - 0.01752 cos [ - 0.467 rad]

X = - 0.01752 x 0.893 = - 0.0156 m

The position of the object when V = - 0.2 m/s and acceleration is positive is X = - 1.56 cm