Please Answer it ASAP! Thanks By doubling the mass of the objet attached to the
ID: 586595 • Letter: P
Question
Please Answer it ASAP! Thanks
By doubling the mass of the objet attached to the spring the period of the oscillations will change by a factor of
0.5
1.4
4
2
1 points
QUESTION 2
According to Hook's Law, the force of a spring is proportional to
the change in the mass of the spring
the change in the length of the spring
the length of the spring
the mass of the spring
1 points
QUESTION 3
The units for the spring constant are
N/m
N
N/kg
m/s2
1 points
QUESTION 4
A 0.25kg object is placed at the end of a vertical spring. The unstretched (natural) length of the spring is 0.23m. After adding the mass the final length is 0.35m. What is the spring constant (in N/m)?
1 points
QUESTION 5
A 0.35kg object is placed at the end of a vertical spring with a spring constant k=21.85N/m . Assuming the spring is massless, what would be the period of oscillations (in s) ?
1 points
QUESTION 6
Assume that after completing the first activity in this lab you have this graph for the weight (force of the spring) in N as a function of change in the length in meters. What is the spring constant?
21.9 N/m
3.47 N/m
0.03 N/m
2.2 N/m
1 points
QUESTION 7
In the previous problem, what is the theoretical value for the period of oscillations, if the hanging mass is 0.2kg and the spring is massless?
2s
1s
0.6s
1.4s
1 points
QUESTION 8
In the previous problem what is the theoretical value for the period of the oscillations, if the mass is 0.2 kg and the mass of the spring is 0.15 kg?
2.1 s
1.2 s
0.67 s
1.5 s
1 points
QUESTION 9
The period of oscillations for a mass spring system would ___ if the experiment was done on the surface of the moon. NOTE: The acceleration due to gravity on the surface of the moon is 1/6th of the acceleration due to gravity on Earth.
increase by a factor of 6
double
quadruple
stay the same
1 points
QUESTION 10
The potential energy of a mass-spring system when the spring is fully compressed and the mass is at rest is 200 J. After releasing the mass, assuming there is no dissipative force, the system will oscillate. At a point during the oscillation the potential energy of the system is 50 J. What is the kinetic energy of the mass at that point, assuming the spring is massless?
50 J
250 J
150 J
200 J
0.5
1.4
4
2
Explanation / Answer
1) 1.4
we know, angular frequency, w = sqrt(k/m)
T = 2*pi/w
= 2*pi*sqrt(m/k)
when m becomes 2 times T becomes sqrt(2) times.
T' = 1.414*T
2) the change in the length of the spring
F = k*x
3) N/m
4) Apply, F_spring = k*x
Fg = m*g
in the equilibrium,
F_spring = Fg
k*x = m*g
k = m*g/x
= 0.25*9.8/(0.35 - 0.23)
= 20.42 N/ m