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Nostrana [21]
3 years ago
15

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) di

splacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?
Physics
2 answers:
nlexa [21]3 years ago
8 0

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)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)

Also, you may know these fundamental formulas:

f=\frac{\omega}{2 \pi} \\\\T=\frac{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))+\frac{\pi}{5} ) \approx -0.70534m

(b)

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

(c)

a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\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 = \frac{\pi}{5}

(e)

f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz

(f)

T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s

riadik2000 [5.3K]3 years ago
7 0

Answer:

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

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:

<em>x = 1.2*cos((3πrad/s)*t+π/5)</em>

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

<u>Frequency (f):</u>

f = w/(2*π)

Then reeplacing values:

f = 3*π/2*π = 1,5 Hz

<u>Period:</u>

T = 1/f = 0,666 seconds

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<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>

In this case, a frictional force F_{r} is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma    -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F -  F_{r}

m = 50kg

a = 0   [At constant velocity, acceleration is zero]

<em>Substitute these values into equation (i) as follows;</em>

F -  F_{r} = m x a

F -  F_{r} = 50 x 0

F -  F_{r} = 0

F =  F_{r}            -------------------(ii)

<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>

In this case, the same frictional force F_{r} is opposing the movement of the box.

∑F = 1.5F -  F_{r}

m = 50kg

a =  0.1m/s²

<em>Substitute these values into equation (i) as follows;</em>

1.5F -  F_{r} = m x a

1.5F -  F_{r} = 50 x 0.1

1.5F -  F_{r} = 5            ---------------------(iii)

<em>Substitute </em>F_{r}<em> = F from equation (ii) into equation (iii) as follows;</em>

1.5F - F = 5            

0.5F = 5            

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

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