Question #125357
1. A mass of 250 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 20 cms, and if there is no damping, determine the position u of the mass at any time t.

Enclose arguments of functions in parentheses. For example, sin(2x).

a. Assume g = 9.8 m/s^2. Enter ans exact answer.
u(t) = ______ m

b. When does the mass first return to its equilibrium position?
Enter an exact answer.

t = ______s
1
Expert's answer
2020-07-06T20:21:24-0400

m=0.25m=0.25 kg

L=0.01568L=0.01568 m

u(0)=0, u(0)=0.2u(0)=0, \ u^{\prime}(0)=0.2 m/s, where u(t)u(t) is position of the mass at any time t.t.


k=mgL=0.25×9.80.01568=156.25k=\frac{mg}{L}=\frac{0.25\times 9.8}{0.01568}=156.25 N/m


Now we can write the equation of the system: mu+ku=0mu^{\prime \prime }+ku =0

0.25u+156.25u=00.25u^{\prime \prime }+156.25u =0

u+625u=0u^{\prime \prime }+625u =0

The characteristic equation: r2+625=0,  r=±25ir^2+625=0, \ \ r=\pm25i .

The general solution is  u(t)=C1sin(25t)+C2cos(25t)\ u(t)=C_1\sin(25t)+C_2\cos(25t)

u(0)=0=C1×0+C2×1=C2u(0)=0=C_1\times 0+C_2\times 1=C_2 , C2=0C_2=0

u(0)=0.2=25C1u^{\prime }(0)=0.2=25C_1 , C1=0.008C_1=0.008

So, u(t)=0.008sin(25t)u(t)=0.008\sin (25t) m.

Equilibrium position: u(t)=0,  0.008sin(25t)=0,  sin(25t)=0,  25t=πn, nZu(t)=0, \ \ 0.008\sin(25t)=0, \ \ \sin(25t)=0, \ \ 25t=\pi n, \ n\in \mathbb{Z} .

The mass first return to its equilibrium position when n=1n=1 or t=π/25.t=\pi/25.


Answer:

a. u(t)=0.008sin(25t)u(t)=0.008\sin (25t) m.

b. t=π/25t=\pi/25 s.


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