Respuesta :
Answer:
Explanation:
1. Find spring constant k. From a free body diagram, you will get the forces on the 10 kg mass, with a displacement d. It will be gravity pulling the mass down and the spring force pulling the mass up. The 10 kg mass is in equilibrium. The resulting equation will be:
[tex]F = m_1a = kd - m_1g = 0 => m_1g = kd => k = \frac{m_1g}{d}[/tex]
2. Use the result from 1. to find the equations of motion. In general they are given by:
[tex]x(t) = Acos(\omega t + \phi), v(t) = -\omega Asin(\omega t + \phi)[/tex], where ω is:[tex]\omega = \sqrt{\frac{k}{m_2}}=\sqrt{\frac{m_1g}{m_2d}}[/tex]
To find the amplitude A and the phase angle Ф, use the given initial conditions:
m₂ = 2 kg, x(0) = -0.25 m, v(0) = 2m/s
[tex]-0.25 = Acos(\phi)\\2 = -\omega Asin(\phi)= -\sqrt{\frac{m_1g}{m_2d}}Asin(\phi)\\-0.24 = Asin(\phi)[/tex]
Solving for Ф:
[tex]\frac{0.24}{0.25}=tan(\phi) => \phi = 0.76[/tex]
Solving for A:
[tex]-0.25 = Acos(\phi) => A = -\frac{0.25}{cos(\phi)}=-0.35[/tex]
The equation for x(t) is now:
[tex]x(t) = -0.35 cos(\sqrt{\frac{m_1g}{m_2d}t} +0.76)[/tex]
The frequency f is given by:[tex]f = \frac{\omega}{2\pi}[/tex]
The period T is given by:[tex]T = \frac{2\pi}{\omega}[/tex]
For the stretched spring, frequency is √70 rad/sec, period is (2π)/√70, amplitude is 0.3458, and phase angle of the motion is 0.7628 rad.
What is equations of motion for a spring?
The equation of wave for the spring motion can be given as,
[tex]x(t)=A\cos(\omega t+\phi)[/tex]
Here, (A) is the amplitude.
Suppose a 2 kg mass is attached to the spring, initially displaced 25 cm below equilibrium, and given an upward velocity of 2 m/s.
The weight of the body is equal to the product of displacement of spring and spring constant. Thus,
[tex]W=kx\\mg=kx[/tex]
Here, (k) is the spring constant. As the mass of the spring is 10 kg and the displacement of the spring is 70 cm.
[tex]10\times9.81=0.7\timesk\\k=140[/tex]
Now by the spring mass system, the equation can be given as,
[tex]m\dfrac{d^2x}{dt^2}+kx=0[/tex]
Put the values of mass and the spring constant in the above equation as,
[tex]2\dfrac{d^2x}{dt^2}+(140)x=0\\\dfrac{d^2x}{dt^2}+70x=0[/tex]
General solution of the above differential equation is,
[tex]x(t)=c_1\cos\sqrt{70}t+c_2\sin\sqrt{70}t\\[/tex]
Put t equal to zero,
[tex]x(0)=-0.25, c_1=\dfrac{-1}{4}\\x'(0)=2, c_2=\dfrac{2}{\sqrt{70}}\\[/tex]
Thus, the above equation, can be given as,
[tex]x(t)=\dfrac{-1}{4}\cos\sqrt{70}t+\dfrac{2}{\sqrt{70}}\sin\sqrt{70}t\\\\x(t)=-(0.25\cos \sqrt{70}t-0.239\sin \sqrt{70}t)\\x(t)=-\sqrt{0.25^2+0.239^2}\cos ( \sqrt{70}t+\tan^{-1}\dfrac{0.239}{0.25})\\x(t)=-0.3458\cos ( \sqrt{70}t+0.762)[/tex]
Comparing it with the standard equation, we get,
Thus, for the stretched spring, frequency is √70/2π rad/sec, period is (2π)/√70, amplitude is 0.3458, and phase angle of the motion is 0.7628 rad.
Learn more about equations of motion for a spring here;
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