Question

The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 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

a) -0.705 m ; b) -9,15 m/s ; c) a = -62,65 m/s² ; d) 92,05 rad ; e) 1,5 Hz ; f) 0,666 s

Step-by-step explanation:

By definition of sum of angles: sin(α-β) = sin(α)*cos(β) - sin(β)*cos(α)

In the particular case of: β = π/2, then: sin(α-π/2) =

In the expression:

x = 1.2*cos((3πrad/s)*t+π/5)

where: ∅ = (3πrad/s)*t+π/5 , es called PHASE.

By definition:

(velocity) v = dx/dt ; (aceleration) a = dv/dt = d²x/dt²

If, we express x like:

x = A*cos(w*t+π/5) (with A = 1.2 m , w = 3π rad/s)

Then:

v = dx/dt = -A*w*sin(w*t+π/5)

a = d²x/dt² = -A*w²*cos(w*t+π/5)

reemplacing in the expresions t = 9.7 seconds:

x = -0.705 m

v = -9,15 m/s

a = -62,65 m/s²

∅ = 92,05 rad

Frequency (f):

f = w/(2*π)

Then reeplacing values:

f = 3*π/2*π = 1,5 Hz

Period:

T = 1/f = 0,666 seconds

Answer 2

Answer and Explanation:

Let:

`x(t)=Acos(\omega t+ \phi)`

The equation representing a simple harmonic motion, where:

`x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time`

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

`v(t)=(dx(t))/(dt) =- \omega A sin(\omega t + \phi)\\\\a(t)=(dv(t))/(dt) =- \omega^2 A cos(\omega t + \phi)`

Also, you may know these fundamental formulas:

`f=(\omega)/(2 \pi) \\\\T=(2 \pi)/(\omega)`

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

`x(9.7)=1.2 cos((3 \pi *(9.7))+(\pi)/(5) ) \approx -0.70534m`

(b)

`v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+(\pi)/(5) ) \approx 9.1498 m/s`

(c)

`a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+(\pi)/(5) )\approx -62.653m/s^2`

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

`\phi = (\pi)/(5)`

(e)

`f=(\omega)/(2 \pi) =(3\pi)/(2 \pi) =(3)/(2) =1.5 Hz`

(f)

`T=(2 \pi)/(\omega) =(2 \pi)/(3 \pi) =(2)/(3) \approx 0.667s`