Question

A spring with a spring constant k of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. It is then stretched an additional 1 inch and released. Find the equation of motion, the amplitude, and the period. Neglect friction.
y(t)= ?


where t is time and y(t) is displacement in time.

Amplitude: ? inch(es)

Period: ? second(s).

Answer 1

Answer:

Equation of movement y(t)

$y(t)=sin(10t-\pi/2)$

Amplitude: 1 inch

Period: 0.628 seconds

Step-by-step explanation:

If there is no friction, the amplitude will be the length it was streched from the equilibrium. In this case, this value is 1 inch

$A=1\,inch$

The period depends on the mass and the spring constant.

The formula for the period is:

$T=\frac{1}{f} =\frac{2/pi}{\omega} =2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{1}{100}}=2\pi*0.1=0.628\, s$

The model y(t) for the movement of the mass-spring system is

$y(t)=A\cdot sin(\omega t +\phi)\\\\A=1\\\\\omega=\sqrt{k/m} =\sqrt{100/1}=10\\\\y(0)=Asin(\phi)=-A\\\\ \phi=arcsin(-1)=-1.571=-\pi/2\\\\\\y(t)=sin(10t-\pi/2)$