Question

A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was released from rest at t = 0 from the position x = 0.0360 m. (a) Determine the location of the mass at t = 5.47 s? (b) Determine if the mass is moving in the positive or negative x direction at t = 5.47 s? positive x direction negative x direction

Answer 1

Answer:

$a)x=-0.2025$

b) The mass is moving in the positive x direction at t=5.47s

Explanation:

The position of an object that describes a simple harmonic movement as a function of time is given by the equation:

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

Where:

$x=Elongation\hspace{3}or\hspace{3}displacement\hspace{3}with\hspace{3}respect\hspace{3} to\hspace{3}the\hspace{3}point\hspace{3}of\hspace{3}equilibrium.$

$A= Amplitude$

$\omega=Angular\hspace{3}frequency$

$t=Time$

$\phi=Initial\hspace{3}phase\hspace{3}and\hspace{3}indicates\hspace{3}the\hspace{3}state\hspace{3}of\hspace{3}oscillation\hspace{3}or\hspace{3}vibration\hspace{3}(phase)\hspace{3}at\hspace{3}time\hspace{3}t = 0\hspace{3}of\hspace{3}the \hspace{3}oscillating\hspace{3}particle.$

Let's asume A=1. Now, we need to find ω. So, let's use the next equation:

$\omega=2\pi f$

Where:

$f=\frac{1}{T}$

So:

$\omega=\frac{2\pi}{T} =\frac{2\pi}{2.25} =\frac{8}{9}\pi$

We need to calculate Φ as well, so let's use the information provide by the problem:

$x(0)=0.0360\\x(0)=cos(\omega (0)+\phi)=0.0360\\x(0)=cos(\phi)=0.0360$

Arccosine to both sides:

$arccosine(\phi)=arccosine(0.0360)\\\phi \approx97.71\°$

We are ready to evaluate the function x(t) at t=5.47:

$x(5.47)=cos(\frac{8}{9}\pi*(5.47) +97.71 )\\x(5.47)=cos(112.9851216)\approx -0.2025584365m$

The minus sign in the answer tell us that the mass is moving  in the negative x direction.

Answer 2

(a) The location of the mass at t = 5.47 s is -0.032 m.

(b) The mass is moving in the negative x-direction.

The wave equation of the given spring is calculated as follows;

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

where;

• A is the amplitude = 0.036 m
• ω is the angular frequency

The angular frequency is calculated as follows;

$\omega = \frac{2\pi}{T} \\\\T = \frac{2\pi}{\omega} \\\\ \omega = \frac{2\pi}{T} \\\\ \omega = \frac{2\pi}{2.25} \\\\\omega = 2.79 \ rad/s$

The location of the mass at t = 5.47 s is determined from the wave equation as follows;

$x(t) = Acos(\omega t + 0)\\\\x(t) = 0.036cos (2.79 t) \\\\x(5.47) = 0.036\times cos(2.79 \times 5.47)\\\\x(5.47) = 0.036 \times cos(15.26)\\\\x(5.47) = 0.036 \times -0.901\\\\x(5.47) = - 0.032 \ m$

Thus, the mass is moving in the negative x-direction.

Learn more here:https://brainly.com/question/14182140