A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? Wha
1 answer:
Answer:
1) Hence, the period is 0.33 s.
2) The amplitude is 10 cm.
Explanation:
1) The period is given by:
Where:
f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz
Hence, the period is 0.33 s.
2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:
Therefore, the amplitude is 10 cm.
I hope it helps you!
You might be interested in
Answer:
39.7 m
Explanation:
First, we conside only the last second of fall of the body. We can apply the following suvat equation:
where, taking downward as positive direction:
s = 23 m is the displacement of the body
t = 1 s is the time interval considered
is the acceleration
u is the velocity of the body at the beginning of that second
Solving for u, we find:
Now we can call this velocity that we found v,
v = 18 m/s
And we can now consider the first part of the fall, where we can apply the following suvat equation:
where
v = 18 m/s
u = 0 (the body falls from rest)
s' is the displacement of the body before the last second
Solving for s',
Therefore, the total heigth of the building is the sum of s and s':
h = s + s' = 23 m + 16.7 m = 39.7 m