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Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π × ................2
put here value
period of oscillations = 2π ×
period of oscillations = 0.6953
so period of oscillations is 0.695 second
1)
2) 13.9 s
Explanation:
1)
The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.
It can be calculated using the equation:
(1)
where
G is the gravitational constant
is the Earth's mass
r is the distance of the object from the Earth's center
The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is
where
is the Earth's radius
Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):
2)
The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula
where
L is the length of the pendulum
g is the acceleration due to gravity at the location of the pendulum
Note that the period of a pendulum does not depend on its mass.
For the pendulum in this problem, we have:
L = 3 m is its length
is the acceleration due to gravity (calculated in part 1)
Therefore, the period of the pendulum is: