Extension of Spring from Mass of 50N

[Dustinsfl]

A weight of \(50\) N is suspended from a spring of stiffness \(4000\) N/m and is subjected to a harmonic force of amplitude \(60\) N and frequency \(6\) Hz.

Since \(W = mg = 50\), we have that the mass, \(m = 5.10204\), and we know that \(f = \frac{\omega}{2\pi} = 6\) so \(\omega = 12\pi\). The harmonic forcing term is then
\[
F(t) = 60\cos(12\pi t)
\]
and our equation of motion is
\[
\ddot{x} + \frac{4000}{5.10204}x = \frac{60}{5.10204}\cos(12\pi t).
\]
Solving the transient and steady solution, we obtain
\[
x(t) = A\cos(28t) + B\sin(28t) - 0.0184551\cos(12\pi t)
\]
How do I determine the extension of spring from the suspended mass? This value would then be \(x(0) = x_0\). Additionally, I will assume any motion starts from rest so \(\dot{x}(0) = 0\) which leads to \(B = 0\) and \(A\) can be defined as \(x_0 - \frac{F_0}{k - m\omega^2}\) where \(\omega = 12\pi\)
\[
x(t) = (x_0 + 0.0184551)\cos(28t) - 0.0184551\cos(12\pi t)
\]
Would the extension of the spring simply be, \(F = kx\) where \(F = 50\) so
\[
x = \frac{F}{k} = \frac{1}{80}\mbox{?}
\]

 

[topsquark]

Typically \(\displaystyle x_0\) is defined to be the origin of the motion so \(\displaystyle x_0 = 0\). If this is not the case then, as you correctly stated, the value of A will depend on \(\displaystyle x_0\). In this situation you have to define where the origin is and usually that would involve knowing the length of the spring before the mass is applied if you want to set the origin at the top or bottom of the spring. Otherwise you have to have some other related point on the spring to measure from.

-Dan

 

[Dustinsfl]

Typically \(\displaystyle x_0\) is defined to be the origin of the motion so \(\displaystyle x_0 = 0\). If this is not the case then, as you correctly stated, the value of A will depend on \(\displaystyle x_0\). In this situation you have to define where the origin is and usually that would involve knowing the length of the spring before the mass is applied if you want to set the origin at the top or bottom of the spring. Otherwise you have to have some other related point on the spring to measure from.

-Dan

I have already solved the problem. I should have marked it solved sooner.

 

[Jameson]

How about copying your solution to this thread so others can use it for future reference? :)

 

[CellCoree]

Yes, that is correct. The extension of the spring, or displacement of the mass from its equilibrium position, can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement. In this case, the force exerted by the spring is equal to the weight of the mass, \(50\) N, and the stiffness of the spring is \(4000\) N/m. Therefore, the displacement can be calculated as \(x = \frac{F}{k} = \frac{50}{4000} = \frac{1}{80}\) m.