A 10-kg mass is attached to a spring, stretching it 0.7 m from its natural length. The mass is started in motion from the equilibrium position with an initial velocity of 1 m/sec in the upward direction. Find the subsequent motion, if the force due to air resistance is -90x N.

Question

A 10-kg mass is attached to a spring, stretching it 0.7 m from its natural length. The mass is started in motion from the equilibrium position with an initial velocity of 1 m/sec in the upward direction. Find the subsequent motion, if the force due to air resistance is -90[latex]\dot{x} N[/latex].

Accepted Answer

Taking [latex]g=9.8 m / sec ^{2}[/latex], we have [latex]w=m g=98 N ext { and } k=w / l=140 N / m[/latex]. Furthermore, [latex]a=90 ext { and } F(t) \equiv 0[/latex] (there is no external force). Equation (14.1) becomes

[latex]\ddot{x}+9 \dot{x}+14 x=0[/latex] (1)

The roots of the associated characteristic equation are [latex]\lambda_{1}=-2 ext { and } \lambda_{2}=-7[/latex], which are real and distinct; hence this problem is an example of overdamped motion. The solution of (1) is

[latex]x=c_{1} e^{-2 t}+c_{2} e^{-7 t}[/latex]

The initial conditions are [latex]x(0)=0[/latex] (the mass starts at the equilibrium position) and [latex]\dot{x}(0)=-1[/latex] (the initial velocity is in the negative direction). Applying these conditions, we find that [latex]c_{1}=-c_{2}=-\frac{1}{5}[/latex], so that [latex]x=\frac{1}{5}\left(e^{-7 t}-e^{-2 t} ight)[/latex]. Note that [latex]x ightarrow 0 ext { as } t ightarrow \infty[/latex]; thus, the motion is transient

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Eq. (14.1) : [latex]\ddot{x}+\frac{a}{m} \dot{x}+\frac{k}{m} x=\frac{F(t)}{m}[/latex]

Question 14.5: A 10-kg mass is attached to a spring, stretching it 0.7 m fr...

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