Question

And object is undergoing simple harmonic motion along the x-axis. Its position is described as a function of time by x(t) = 1.4 cos(3.5t – 1.4), where x is in meters, the time, t, is in seconds, and the argument of the cosine is in radians.

(a) What is the object’s velocity, in meters per second, at time t = 0?
(b) Calculate the object’s acceleration, in meters per second squared, at time t = 0.

Answer 1

(a) 4.8285m/s

(b) -2.9155m/s²

Step-by-step explanation:

The equation of the object's position is given as;

x(t) = 1.4 cos(3.5t - 1.4) -------------------------(i)

(a) Calculate the object's velocity;

Since velocity is the time rate of change of position, differentiating the above equation with respect to t will give the equation of the object's velocity, v(t). i.e

v(t) = Δx / Δt ≅ dx(t) / dt = 1.4 x 3.5(-sin(3.5t - 1.4))

v(t) = - 4.9 sin (3.5t - 1.4) -----------------(ii)

Now, at time t = 0;

v(t) = v(0) = -4.9 sin(3.5(0) - 1.4)

v(0) = - 4.9 sin(-1.4)

v(0) = - 4.9 x -0.9854

v(0) = 4.8285

At time t = 0, the velocity of the object is 4.8285m/s

(b) Calculate the object's acceleration

Since acceleration is the time rate of change in velocity, differentiating the equation of velocity, equation (ii), will give the equation of the object's acceleration, a(t). i.e

a(t) = dv(t) / dt = - 4.9 x 3.5 cos(3.5t - 1.4)

a(t) = - 17.15 cos (3.5t - 1.4)

Now, at t = 0, the object's acceleration is given as

a(t) = a(0) = -17.15 cos (3.5(0) - 1.4)

a(0) = -17.15 cos(0 - 1.4)

a(0) = -17.15 cos(-1.4)

a(0) = -17.15 x 0.170

a(0) = -2.9155

Therefore, the acceleration of the object at t = 0 is -2.9155m/s²

Note: The angles are in radians and not degrees. Therefore, be sure your calculations are in radians.

Answer 2

The velocity at t=0 is 0.12m/s

The acceleration at t=0 is -17.145m/s²

Step-by-step explanation:

The velocity is obtained from the first derivative of the position with respect to time i.e dx/dt

Since x=1.4cos(3.5t-1.4)

Let u=3.5t-14

Then, x= 1.4cos u

dx/dt=dx/du ×du/dt

dx/du= -1.4sin u

du/dt= 3.5

dx/dt= -1.4×3.5sin(3.5t-1.4)

Putting t=0

dx/dt= -1.4×3.5sin(3.5×0 - 1.4)

dx/dt= -4.9× sin(-1.4)

dx/dt=0.12m/s

Like wise for the acceleration, a=d(v)/dt or d/dx×dx/dt

This is the second derivative of the position with respect to time. We basically need to differentiate the velocity to obtain acceleration

Let p=v=-4.9sin(3.5t-1.4) and q= 3.5t-1.4

dp/dt=dp/dq×dq/dt

To obtain -4.9×3.5cos(3.5t-1.4)

a=-17.15cos(3.5t-1.4)

a=-17.15×0.9997

a-=17.145m/s² (negative because the direction of force is opposite displacement)