- #1
UrbanXrisis
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A 2 kg mass is attached to a spring aand placed on a horizontal, smooth surface. A force of 20N is required to hold the mass at reast when it is pulled .200m from the equilibrium position. The mass is released and undergoes simple harmoic motion.
a) what is the force constant?
b) frequency?
c) max speed and where does this occur?
d) mass acceleration and where does this occur?
e) total energy
f) speed when displacement equals one third of the max value
g) acceleration when displacement equals one third of the max value
a)
N/m=20/0.2m=100N/m
b)
T=2\pi\sqrt{\frac{m}{k}}
T=2\pi\sqrt{\frac{2kg}{100N/m}}
T=0.8886s
f=1/T=1.1125Hz
c)
v_{max}=\sqrt{\frac{m}{k}}*A
v_{max}=\sqrt{\frac{2kg}{100N/m}}*0.2m
v_{max}=0.0283m/s@x=0m
d)
a_{max}=\frac{mA}{k}}
a_{max}=\frac{2kg*0.2m}{100N/m}}
a_{max}=0.004m/s^2@x=+/-0.2m
e)
E=0.5kA^2
E=0.5(100N/m)(0.2m)^2
E=2J
f)
d_{max}/3=0.2/3=\frac{1}{15}m
v=\sqrt{\frac{k}{m} (A^2-x^2)}
v=\sqrt{\frac{100N/m}{2kg} ((0.2m)^2-(\frac{1}{15}m)^2)}
v=+/- \frac{4}{3}m/s
g)
v=-\omega A sin \omega t
\frac{4}{3}m/s=-\sqrt{\frac{100N/m}{2kg}} *0.2m sin \sqrt{\frac{100N/m}{2kg}} t
\frac{4}{3}m/s=-1.414213562sin7.071067812t
t=-0.1740839504s
a=-\omega^2 A cos \omega t
a=- \frac{100N/m*0.2m}{2kg} cos 7.071067812*-0.1740839504s
a=+/- \frac{10}{3} m/s^2
did I do this right? especially letter (g) thanks
a) what is the force constant?
b) frequency?
c) max speed and where does this occur?
d) mass acceleration and where does this occur?
e) total energy
f) speed when displacement equals one third of the max value
g) acceleration when displacement equals one third of the max value
a)
N/m=20/0.2m=100N/m
b)
T=2\pi\sqrt{\frac{m}{k}}
T=2\pi\sqrt{\frac{2kg}{100N/m}}
T=0.8886s
f=1/T=1.1125Hz
c)
v_{max}=\sqrt{\frac{m}{k}}*A
v_{max}=\sqrt{\frac{2kg}{100N/m}}*0.2m
v_{max}=0.0283m/s@x=0m
d)
a_{max}=\frac{mA}{k}}
a_{max}=\frac{2kg*0.2m}{100N/m}}
a_{max}=0.004m/s^2@x=+/-0.2m
e)
E=0.5kA^2
E=0.5(100N/m)(0.2m)^2
E=2J
f)
d_{max}/3=0.2/3=\frac{1}{15}m
v=\sqrt{\frac{k}{m} (A^2-x^2)}
v=\sqrt{\frac{100N/m}{2kg} ((0.2m)^2-(\frac{1}{15}m)^2)}
v=+/- \frac{4}{3}m/s
g)
v=-\omega A sin \omega t
\frac{4}{3}m/s=-\sqrt{\frac{100N/m}{2kg}} *0.2m sin \sqrt{\frac{100N/m}{2kg}} t
\frac{4}{3}m/s=-1.414213562sin7.071067812t
t=-0.1740839504s
a=-\omega^2 A cos \omega t
a=- \frac{100N/m*0.2m}{2kg} cos 7.071067812*-0.1740839504s
a=+/- \frac{10}{3} m/s^2
did I do this right? especially letter (g) thanks
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