An ideal spring has spring constant ks (to distinguish it from the electrostatic constant k) and equilibrium length l. Then, you
glue two identical negative point charges to the ends of the spring and observe that the equilibrium length doubles. Determine the amount of charge on each end of the spring
1 answer:
Answer:
q = square root (4KsL³/k)
The force of extension of the spring is equal to the force of repulsion between the two like charges. Two like charges(positive or negative) would always repel each other and two unlike charges would always attract each other. This electric force between the charges is what is responsible for the stretching of the spring. The electric force causes the spring to increase in length from L to 2L. Equating these forces, that is the electric force between the charges and the elastic force of the spring and rearranging the variables gives the expression to obtain q.
Explanation:
See the attachment below for full solution.
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Answer:
0.51 m
Explanation:
Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.
Kinetic energy, KE=½kx²
Where k is spring constant and x is the compression of spring
Potential energy, PE=mgh
Where g is acceleration due to gravity, h is height and m is mass
Equating KE=PE
mgh=½kx²
Making x the subject of formula
Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then
(1) The position around equilibrium of an object in simple harmonic motion is described by
where
A is the amplitude of the motion
is the angular frequency.
The velocity is the derivative of the position:
where
is the maximum velocity of the object.
The acceleration is the derivative of the velocity:
where
is the maximum acceleration of the object.
We know from the problem both maximum velocity and maximum acceleration:
From the first equation, we get
(1)
and if we substitute this into the second equation, we find the angular fequency
while the amplitude is (using (1)):
(b) We found in the previous step that the angular frequency of the motion is
But the angular frequency is related to the period by
and so, the period is
where
A is the amplitude of the motion
The velocity is the derivative of the position:
where
The acceleration is the derivative of the velocity:
where
We know from the problem both maximum velocity and maximum acceleration:
From the first equation, we get
and if we substitute this into the second equation, we find the angular fequency
while the amplitude is (using (1)):
(b) We found in the previous step that the angular frequency of the motion is
But the angular frequency is related to the period by
and so, the period is