mass = .1136 kg
g = 9.81 m/s^2
so
weight = .1136*9.81 = 1.114 Newtons
spring constant
k = 1.114 N/ .134 meters
= 8.32 N/m
T = 2 pi sqrt(m/k)
= 2 pi sqrt (.1136/8.32)
= 0.734 s = 7.34 * 10^-1 s
In a lab, a student measures the unstretched length of a spring as 12.9 cm. When a 113.6-g mass is hung from the spring, its length is 26.3 cm. The mass-spring system is set into oscillatory motion, and the student observes that the amplitude of the oscillation decreases by about a factor of 2 after five complete cycles.
a) Calculate the period of oscillation for this system, assuming no damping.
7.34×10-1 s
b) What would be the difference between the period with no damping and the period with damping?
1.79×10-4 s
Can anyone please explain how to solve these problems
2 answers
if damping proportional to v
F = -kx -bv
then x = A e^-(bt/2m) sin w' t
five cycles is about 3.67 seconds
so
1/2 = e^-(3.67 b /.226)
take ln both sides
-.693 = - -16.2 b
so
b = .0427
then w' = sqrt (k/m -b^2/4m^2)
= sqrt (73. - .035)
=8.542 = 2 pi/T
T = .7355 s
change in T = .0015 s =1.5*10^-4 s
You could do this more accurately either carrying more significant figures or using calculus to get dT/db
F = -kx -bv
then x = A e^-(bt/2m) sin w' t
five cycles is about 3.67 seconds
so
1/2 = e^-(3.67 b /.226)
take ln both sides
-.693 = - -16.2 b
so
b = .0427
then w' = sqrt (k/m -b^2/4m^2)
= sqrt (73. - .035)
=8.542 = 2 pi/T
T = .7355 s
change in T = .0015 s =1.5*10^-4 s
You could do this more accurately either carrying more significant figures or using calculus to get dT/db
1 answer