Question #45776

A spring of stiffness 2o KN/m supports a mass of 4 Kg. the mass is pulled down 8 mm and released to produce linear oscillations. calculate the frequency and periodic time. sketch the graphs of displacement, velocity and acceleration. calculate the displacement, velocity and acceleration 0.05 s after released.

note X= A sin(w t + phase angle)

X : displacement
A : amplitude
w : omega
t : periodic time
ceta : phase angle

Expert's answer

Answer on Question #45776, Physics, Mechanics | Kinematics | Dynamics

A spring of stiffness 20 KN/m supports a mass of 4 Kg. the mass is pulled down 8 mm and released to produce linear oscillations. calculate the frequency and periodic time. sketch the graphs of displacement, velocity and acceleration. calculate the displacement, velocity and acceleration 0.05 s after released.

note X = A sin(w t + phase angle)

X : displacement

A : amplitude

w : omega

t : periodic time

ceta : phase angle


ω=km=20103N/m4kg=1050s170.7s1\omega = \sqrt {\frac {k}{m}} = \sqrt {\frac {20 * 10^{3} N / m}{4 k g}} = 10 \sqrt {50} \, s^{-1} \approx 70.7 \, s^{-1}A=8mm=8103mA = 8 \, \text{mm} = 8 \cdot 10^{-3} \, \text{m}φ(phase angle)=π2\varphi (\text{phase angle}) = - \frac {\pi}{2}


Phase angle take in such way, that at the moment (t=0)(t=0) we have displacement equal to A-A.


X=8103msin(70.7s1tπ2)X = 8 \cdot 10^{-3} \, \text{m} \cdot \sin \left(70.7 \, s^{-1} \cdot t - \frac {\pi}{2}\right)dXdt=V=70.7s18103mcos(70.7s1tπ2)=70.7s18103mcos(70.7s1tπ2)=0.5656cos(70.7s1tπ2)m/s\begin{array}{l} \frac {d X}{d t} = V = 70.7 \, s^{-1} \cdot 8 \cdot 10^{-3} \, \text{m} \cdot \cos \left(70.7 \, s^{-1} \cdot t - \frac {\pi}{2}\right) \\ = 70.7 \, s^{-1} \cdot 8 \cdot 10^{-3} \, \text{m} \cdot \cos \left(70.7 \, s^{-1} \cdot t - \frac {\pi}{2}\right) \\ = 0.5656 \cdot \cos \left(70.7 \, s^{-1} \cdot t - \frac {\pi}{2}\right) \, \text{m/s} \end{array}dVdt=a=0.565670.7s1sin(70.7s1tπ2)m/s=40sin(70.7s1tπ2)ms2\frac {d V}{d t} = a = - 0.5656 \cdot 70.7 \, s^{-1} \cdot \sin \left(70.7 \, s^{-1} \cdot t - \frac {\pi}{2}\right) \, \text{m/s} = - 40 \cdot \sin \left(70.7 \, s^{-1} \cdot t - \frac {\pi}{2}\right) \frac {\text{m}}{s^{2}}


X, m



V, m|s



a, m|s^2

And


X(t=0.05s)=0.00738886607mX(t = 0.05\,s) = 0.00738886607\,m


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